RISC Report Series 22–13

# Error bounds for the asymptotic expansion of the partition function

Koustav Banerjee  Research Institute for Symbolic Computation (RISC), Johannes
Kepler University Linz, Altenberger Straße 69, 4040 Linz, Austria
<EMAIL_ADDRESS>,  Peter Paule  Research Institute for Symbolic
Computation (RISC), Johannes Kepler University Linz, Altenberger Straße 69,
4040 Linz, Austria<EMAIL_ADDRESS>,  Cristian-Silviu Radu  Research
Institute for Symbolic Computation (RISC), Johannes Kepler University Linz,
Altenberger Straße 69, 4040 Linz, Austria<EMAIL_ADDRESS>and
Carsten Schneider  Research Institute for Symbolic Computation (RISC),
Johannes Kepler University Linz, Altenberger Straße 69, 4040 Linz, Austria
<EMAIL_ADDRESS>
###### Abstract.

Asymptotic study on the partition function $p(n)$ began with the work of Hardy
and Ramanujan. Later Rademacher obtained a convergent series for $p(n)$ and an
error bound was given by Lehmer. Despite having this, a full asymptotic
expansion for $p(n)$ with an explicit error bound is not known. Recently
O’Sullivan studied the asymptotic expansion of $p^{k}(n)$-partitions into
$k$th powers, initiated by Wright, and consequently obtained an asymptotic
expansion for $p(n)$ along with a concise description of the coefficients
involved in the expansion but without any estimation of the error term. Here
we consider a detailed and comprehensive analysis on an estimation of the
error term obtained by truncating the asymptotic expansion for $p(n)$ at any
positive integer $n$. This gives rise to an infinite family of inequalities
for $p(n)$ which finally answers to a question proposed by Chen. Our error
term estimation predominantly relies on applications of algorithmic methods
from symbolic summation.

###### Key words and phrases:

The partition function, the Hardy-Ramanujan-Rademacher formula, asymptotics,
error bound

###### 1991 Mathematics Subject Classification:

05A16, 11P82, 68W30

###### Contents

  1. 1 Introduction
  2. 2 A roadmap for the reader
  3. 3 Estimation of the coefficients $g(t)$
  4. 4 Preliminary Lemmas
  5. 5 Estimation of $\bigl{(}S_{i}(t)\bigr{)}$
    1. 5.1 The Lemmas 5.1 to 5.4
    2. 5.2 The Proofs of Lemma 5.1 to 5.4
  6. 6 Error bounds
  7. 7 Main Theorem
  8. 8 Appendix
    1. 8.1 Proofs of the lemmas presented in Section 4.
    2. 8.2 The Sigma simplification of $S_{3}(t,u)$ in Lemma 5.3
  9. 9 Concluding remarks
  10. 10 Conflict of interest statement
  11. 11 Data availability statement

##  1\. Introduction

A partition of a positive integer $n$ is a non-increasing sequence of positive
integers which sum to $n$, and the partition function $p(n)$ counts the number
of partitions of $n$. In their epoch-making breakthrough work in the theory of
partitions, Hardy and Ramanujan [13] proved that

$p(n)\sim\dfrac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}\ \ \text{as}\
n\rightarrow\infty.$ (1.1)

They also proved that $p(n)$ is the integer nearest to

$\dfrac{1}{2\sqrt{2}}\sum_{q=1}^{\nu}\sqrt{q}A_{q}(n)\psi_{q}(n),$ (1.2)

where $A_{q}(n)$ is a certain exponential sum, $\nu=\nu(n)$ is of the order of
$\sqrt{n}$, and

$\psi_{q}(n)=\dfrac{d}{dn}\Biggl{(}\exp\Bigl{\\{}{\dfrac{C}{q}\lambda_{n}}\Bigr{\\}}\Biggr{)},\
\lambda_{n}=\sqrt{n-\frac{1}{24}},\ C=\pi\sqrt{\frac{2n}{3}}.$

Extending $\nu$ to infinity, Lehmer [15] proved that (1.2) is a divergent
series. Rademacher [24, 25, 26] considered a modification of (1.2) that
presents a convergent series for $p(n)$ which reads:

$p(n)=\dfrac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)\dfrac{d}{dn}\Biggl{(}\dfrac{\sinh\Bigl{(}C\lambda_{n}/k\Bigr{)}}{\lambda_{n}}\Biggr{)}.$
(1.3)

Lehmer [16, 17] obtained an error bound after subtraction of the $N$th partial
sum from the convergent series (1.3).

The study of a full asymptotic expansion for $p(n)$ can be traced in two
directions by considering two different classes that arise from imposing
restrictions on parts of partitions. The two restricted families are
$p^{s}(n)$, the number of partitions of $n$ into perfect $s$th powers, and
$p(n,k)$, the number of partitions of $n$ into at most $k$ parts. As an
application of the “circle method”, Hardy and Ramanujan [13, Section 7, 7.3]
obtained the main term in the asymptotic expansion of $p^{s}(n)$. This of
course retrieves (1.1) when we take $s=1$. Wright [32, 33] extended the work
of Hardy and Ramanujan and obtained a full asymptotic expansion for
$p^{s}(n)$. Recently O’Sullivan [21] proposed a simplified proof of Wright’s
results on the asymptotic expansion of $p^{s}(n)$, and consequently obtained
an asymptotic formula for $p(n)$.

######  Theorem 1.1.

[21, Proposition 4.4] Let $n$ and $R$ be positive integers. As
$n\rightarrow\infty$,

$p(n)=\dfrac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}1+\sum_{t=1}^{R-1}\dfrac{\omega_{t}}{\sqrt{n}^{t}}+O\Bigl{(}n^{-R/2}\Bigr{)}\Biggr{)},$
(1.4)

with an implied constant depending only on $R$, where

$\omega_{t}=\dfrac{1}{(-4\sqrt{6})^{t}}\sum_{k=0}^{\frac{t+1}{2}}\binom{t+1}{k}\dfrac{t+1-k}{(t+1-2k)!}\Bigl{(}\frac{\pi}{6}\Bigr{)}^{t-2k}.$
(1.5)

The binomial coefficient is defined as $\binom{x}{k}:=x(x-1)\dots(x-k+1)/k!$
if $k\in\mathbb{Z}_{\geq 0}$, $\binom{x}{0}:=1$, and $\binom{x}{k}:=0$ if
$k\in\mathbb{Z}_{<0}$. Szekeres [31] proposed an asymptotic expansion for
$p(n,k)$ for $n$ and $k$ sufficiently large and considering $k=n$, one obtains
the expansion for $p(n)$ as $p(n,k)=p(n)$. Canfield [5] proved Szekeres’
result by using a recursion satisfied by $p(n,k)$ without using theory of
complex functions and as a corollary, obtained the main term of the Hardy-
Ramanujan formulas for $p(n)$, see (1.1). For a probabilistic approach to the
asymptotic expansion of $p(n)$, we refer to [4].

The primary objective of this paper is to obtain an explicit and computable
error bound for the asymptotic expansion of $p(n)$. A main motivation to
consider such a problem is that from the literature, including the works [32,
31, 5, 4, 21], we could not retrieve any information on the error bound for
asymptotic expansion of $p(n)$. An advantage of getting a control over the
error bound is that one can prove the $\log$-concavity property of $p(n)$
directly from the asymptotic expansion as speculated by Chen [6, p. 121]. In
the language of Theorem 1.4, Chen’s question can be formulated as follows:

######  Question 1.2.

Do there exists $d$ and $n_{0}$ such that

$\dfrac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}1+\sum_{t=1}^{3}\dfrac{\omega_{t}}{\sqrt{n}^{t}}-\dfrac{d}{n^{2}}\Biggr{)}<p(n)<\dfrac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}1+\sum_{t=1}^{3}\dfrac{\omega_{t}}{\sqrt{n}^{t}}+\dfrac{d}{n^{2}}\Biggr{)}$
(1.6)

holds for all $n>n_{0}$?

Chen remarked that (1.6) implies that $p(n)$ is $\log$-concave for
sufficiently large $n$. Now in order to demystify the phrase “sufficiently
large”, explicit information about $n_{0}$ is required; a question being
intricately connected with the computation of the error bound $d$. A similar
phenomena can be found in O’Sullivan’s work:

######  Theorem 1.3.

[21, Theorem 1.3, (1.15)] For each positive integer $k$ there exists
$\mathcal{D}_{k}$ so that for all $n\geq\mathcal{D}_{k}$,

$p^{k}(n)^{2}\geq p^{k}(n+1)\cdot p^{k}(n-1)\cdot(1+n^{-2}).$ (1.7)

For $k=1$, Theorem 1.3 merely implies that $p(n)$ is $\log$-concave for
sufficiently large $n$ although we know that $(p(n))_{n\geq 26}$ is
$\log$-concave due to [20, 11]. Moreover, O’Sullivan [21, (5.17)] proved that
for large enough $n$,

$\dfrac{p(n+1)p(n-1)}{p(n)^{2}}\Biggl{(}1+\dfrac{\pi}{\sqrt{24}n^{3/2}}\Biggr{)}>1,$

settled by Chen, Wang, and Xie [8]. The first three authors and Zeng [3,
Theorem 7.6] proved a stronger version of (1.7) using an infinite familiy of
inequalities for $\log p(n)$.

We conclude this section by discussing the novelty of this paper in brevity.
In order to elucidate the term $O\Bigl{(}n^{-R/2}\Bigr{)}$ in (1.4),
determination of the asymptotic growth of the coefficients $\omega_{t}$ in
(1.5) is required; a task which looks deceptively simple. Our representation
of $\omega_{t}$ is of the following form:

$\omega_{t}=\sum_{u=0}^{t}\gamma(u)\sum_{s=0}^{u}\psi(s).$

In an effort to estimate the inner sum $\sum_{s=0}^{u}\psi(s)$, the use of the
symbolic summation tool Sigma [27] was essential. Schneider considered [27,
28, 29] a broader algorithmic framework that subsumes the theory of difference
field and ring extensions together with the method of creative telescoping.
This algorithmic tool began to be aimed at a wider class of multi-sums, most
frequently encountered in problems of enumerative combinatorics. For example,
in Andrews, Paule, and Schneider [2] we can see how Sigma assists to solve the
TSPP-problem in an LU-reformulation by Andrews. Beyond the world of
combinatorics, applications of Sigma transcends to solve a very general class
of Feynman integrals which are of relevance for manifold physical processes in
quantum field theory, see [1]. This paper adds a new facet to the regime of
applications of Sigma; in particular, its foray into asymptotic estimation for
partition-like functions seems to begin with this work.

##  2\. A roadmap for the reader

In this section we will provide a roadmap on the structure of this paper;
i.e., a navigation from the starting point to the final goal of this paper, to
facilitate for the reader to follow.

Using the Hardy-Ramanujan-Rademacher formula for $p(n)$ and Lehmer’s error
bound, Chen, Jia, and Wang [7, Lemma 2.2] proved that for all $n\geq 1207$,

$\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}-\dfrac{1}{\mu(n)^{10}}\Biggr{)}<p(n)<\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}+\dfrac{1}{\mu(n)^{10}}\Biggr{)},$
(2.1)

where for $n\geq 1$, $\mu(n):=\frac{\pi}{6}\sqrt{24n-1}$; a definition which
is kept throughout this paper. More generally, due to the first three authors
and Zeng, we have the following result.

######  Theorem 2.1.

[3, Theorem 4.4] For $k\in\mathbb{Z}_{\geq 2}$, define

$\widehat{g}(k):=\dfrac{1}{24}\Biggl{(}\dfrac{36}{\pi^{2}}\cdot\nu(k)^{2}+1\Biggr{)},$

where $\nu(k):=2\log 6+(2\log 2)k+2k\log k+2k\log\log k+\dfrac{5k\log\log
k}{\log k}$. Then for all $k\in\mathbb{Z}_{\geq 2}$ and $n>\widehat{g}(k)$
such that $(n,k)\neq(6,2)$, we have

$\frac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\frac{1}{\mu(n)}-\frac{1}{\mu(n)^{k}}\Biggr{)}<p(n)<\frac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\frac{1}{\mu(n)}+\frac{1}{\mu(n)^{k}}\Biggr{)}.$
(2.2)

The goal of this paper is to derive an inequality of the form

$\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{k-1}\frac{g(t)}{\sqrt{n}^{j}}+\frac{L(k)}{\sqrt{n}^{k}}\Biggr{)}<p(n)<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{k-1}\frac{g(t)}{\sqrt{n}^{t}}+\frac{U(k)}{\sqrt{n}^{k}}\Biggr{)},$
(2.3)

stated precisely in Theorem 7.5, starting from the inequality (2.2). As a
consequence we obtain Corollary 7.6 which will give an explicit answer to the
problem stated in Question 1.2 and which, as a further consequence reveals
that $p(n)$ is $\log$-concave for all $n\geq 26$, see Remark 7.7.

The first step is to find explicitly the coefficients $g(t)$ such that

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}=\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\sum_{t=0}^{\infty}\frac{g(t)}{\sqrt{n}^{t}}.$

This is done in Section 3 by computing separately $g(2t)$ and $g(2t+1)$. In
spite of having a double sum representation for $g(t)$, we will see that the
coefficients $g(t)$ are indeed equal to $\omega_{t}$ as in Theorem 1.1.

The next step is to estimate the number $g(t)$ in the following form:

$f(t)-l(t)\leq g(t)\leq f(t)+u(t).$ (2.4)

Here $f(t)$ has the property that $\lim_{t\to\infty}\frac{g(t)}{f(t)}=1$,
$\lim_{t\to\infty}\frac{l(t)}{f(t)}=0$, and
$\lim_{t\to\infty}\frac{u(t)}{f(t)}=0$. Precise descriptions for $f(t)$,
$u(t)$, and $l(t)$ are given in Section 5 along with the inequalities of the
form (2.4). In order to prove such inequalities, we will use the preliminary
lemmas from Section 4 and the summation package Sigma.

Finally in Section 6, applying the bounds for $g(t)$, given in Section 5, we
find $\widehat{L}_{1}(k),\widehat{U}_{1}(k)$ such that

$\frac{\widehat{L}_{1}(k)}{\sqrt{n}^{k}}<\sum_{t=k}^{\infty}\frac{g(t)}{\sqrt{n}^{t}}<\frac{\widehat{U}_{1}(k)}{\sqrt{n}^{k}}.$

Also we compute explicitly $\widehat{L}_{2}(k)$ and $\widehat{U}_{2}(k)$ such
that

$\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{\widehat{L}_{2}(k)}{\sqrt{n}^{k}}<\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\frac{1}{\mu(n)^{k}}<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{\widehat{U}_{2}(k)}{\sqrt{n}^{k}}.$

Combining the error bounds as $L(k)=\widehat{L}_{1}(k)+\widehat{L}_{2}(k)$ and
$U(k)=\widehat{U}_{1}(k)+\widehat{U}_{2}(k)$, we arrive at the desired
inequality (2.3) for $p(n)$.

##  3\. Estimation of the coefficients $g(t)$

From Theorem 2.1, we have for all $k\in\mathbb{Z}_{\geq 2}$ and
$n>\widehat{g}(k)$ such that $(n,k)\neq(6,2)$,

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}-\dfrac{1}{\mu(n)^{k}}\Bigr{)}<p(n)<\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}+\dfrac{1}{\mu(n)^{k}}\Bigr{)}.$
(3.1)

Rewrite the major term $\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}$ in the following way:

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}=\dfrac{1}{4n\sqrt{3}}e^{\pi\sqrt{2n/3}}\
\underset{:=A_{1}(n)}{\underbrace{e^{\pi\sqrt{2n/3}\
\bigl{(}\sqrt{1-\frac{1}{24n}}-1\bigr{)}}}}\underset{:=A_{2}(n)}{\underbrace{\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}}}.$
(3.2)

Next we compute the Taylor expansion of the residue parts of $A_{1}(n)$ and
$A_{2}(n)$, defined in (3.2).

######  Definition 3.1.

For $t\in\mathbb{Z}_{\geq 0}$, define

$e_{1}(t):=\begin{cases}1,&\quad\text{if}\ t=0\\\
\dfrac{(-1)^{t}}{(24)^{t}}\dfrac{(1/2-t)_{t+1}}{t}\displaystyle\sum_{u=1}^{t}\dfrac{(-1)^{u}(-t)_{u}}{(t+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u},&\quad\text{otherwise}\end{cases},$
(3.3)

and

$E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=\sum_{t=0}^{\infty}e_{1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t},\
n\geq 1.$ (3.4)

######  Definition 3.2.

For $t\in\mathbb{Z}_{\geq 0}$, define

$o_{1}(t):=-\dfrac{\pi}{12\sqrt{6}}\Biggl{(}\dfrac{(-1)^{t}(1/2-t)_{t+1}}{(24)^{t}}\sum_{u=0}^{t}\dfrac{(-1)^{u}(-t)_{u}}{(t+u+1)!(2u)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u}\Biggr{)}$
(3.5)

and

$O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=\sum_{t=0}^{\infty}o_{1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1},\
n\geq 1.$ (3.6)

######  Lemma 3.3.

For $j$, $k\in\mathbb{Z}_{\geq 0}$,

$\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}\binom{i/2}{j}=\left\\{\begin{array}[]{cc}1,&j=k=0\\\
(-1)^{j}2^{k-2j}\frac{k}{j}\binom{2j-k-1}{j-k},&\text{otherwise}\end{array}\right..$
(3.7)

  * Proof.

The case $j=k=0$ is trivial. By the inversion relation

$f(k)=\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}g(i)\Leftrightarrow
g(k)=\sum_{i=0}^{k}(-1)^{i}\binom{k}{i}f(i),$

(3.7) for $j\neq 0$ is equivalent to

$\sum_{i=0}^{k}(-1)^{i+j}2^{i-2j}\frac{i}{j}\binom{k}{i}\binom{2j-i-1}{j-i}=\binom{k/2}{j};$

which can be proved (and derived) by any standard summation method, resp.
algorithm. ∎

######  Lemma 3.4.

Let $A_{1}(n)$ be defined as in (3.2). Let $E_{1}(n)$ be as in Definition 3.1
and $O_{1}(n)$ as in Definition 3.2. Then

$A_{1}(n)=E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}+O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}.$
(3.8)

  * Proof.

From Equation (3.2), we get

$\displaystyle A_{1}(n)$ $\displaystyle=$ $\displaystyle e^{\pi\sqrt{2n/3}\
\bigl{(}\sqrt{1-\frac{1}{24n}}-1\bigr{)}}$ (3.9) $\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\dfrac{(\pi\sqrt{2n/3})^{k}}{k!}\Biggl{(}\sqrt{1-\frac{1}{24n}}-1\Biggr{)}^{k}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\dfrac{(\pi\sqrt{2/3})^{k}}{k!}(\sqrt{n})^{k}\sum_{i=0}^{k}\binom{k}{i}(-1)^{k-i}\Biggl{(}\sqrt{1-\frac{1}{24n}}\Biggr{)}^{i}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\dfrac{(\pi\sqrt{2/3})^{k}}{k!}(\sqrt{n})^{k}\sum_{i=0}^{k}\binom{k}{i}(-1)^{k-i}\sum_{j=0}^{\infty}\binom{i/2}{j}\dfrac{(-1)^{j}}{(24n)^{j}}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\sum_{i=0}^{k}\sum_{j=0}^{\infty}\dfrac{(\pi\sqrt{2/3})^{k}}{k!}\dfrac{(-1)^{k-i+j}}{(24)^{j}}\binom{k}{i}\binom{i/2}{j}(\sqrt{n})^{k-2j}.$

Define $S:=\bigl{\\{}(k,i,j)\in\mathbb{Z}^{3}_{\geq 0}:0\leq i\leq
k\bigr{\\}}$. In order to express $A_{1}(n)$ in the form
$\sum_{m=0}^{\infty}a_{m}(\frac{1}{\sqrt{n}})^{m}$, we split the set $S$ into
a disjoint union of subsets; i.e., $S:=\underset{t\in\mathbb{Z}_{\geq
0}}{\bigcup}V(t)$, where for each $t\in\mathbb{Z}_{\geq 0}$,
$V(t):=\bigl{\\{}(k,i,j)\in\mathbb{Z}^{3}_{\geq 0}:k-2j=-t\bigr{\\}}.$

Notice that for $k>j$, by Lemma 3.3,
$\sum_{i=0}^{k}\binom{k}{i}\binom{i/2}{j}=0$. Furthermore, for each element
$r=(k,i,j)\in S$, we define

$S(r):=\dfrac{(\pi\sqrt{2/3})^{k}}{k!}\dfrac{(-1)^{k-i+j}}{(24)^{j}}\binom{k}{i}\binom{i/2}{j}\
\ \text{and}\ \ f(r):=k-2j.$

Rewrite (3.9) as

$\displaystyle A_{1}(n)$ $\displaystyle=$ $\displaystyle\sum_{r\in
S}S(r)(\sqrt{n})^{f(r)}=\sum_{t=0}^{\infty}\sum_{r\in
V(t)}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}$ (3.10) $\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}\underset{r\in
V(2t)}{\sum}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}+\sum_{t=0}^{\infty}\underset{r\in
V(2t+1)}{\sum}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}.$

Now

$\displaystyle V(2t)$ $\displaystyle=$ $\displaystyle\bigl{\\{}(k,i,j)\in
S:k-2j=-2t\bigr{\\}}$ (3.11) $\displaystyle=$
$\displaystyle\bigl{\\{}(k,i,j)\in S:k\equiv 0\ (\text{mod}\ 2)\ \text{and}\
k-2j=-2t\bigr{\\}}$ $\displaystyle=$ $\displaystyle\bigl{\\{}(2u,i,j)\in
S:j=u+t\bigr{\\}}=\bigl{\\{}(2u,i,u+t)\in\mathbb{Z}^{3}_{\geq 0}:0\leq i\leq
2u\bigr{\\}}.$

From (3.11), it follows that

$\displaystyle\sum_{t=0}^{\infty}\underset{r\in
V(2t)}{\sum}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}$ $\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}\dfrac{(-1)^{t}}{(24)^{t}}\Biggl{(}\sum_{u=0}^{\infty}\dfrac{(2\pi^{2}/3)^{u}}{(2u)!}\dfrac{(-1)^{u}}{(24)^{u}}\sum_{i=0}^{2u}(-1)^{i}\binom{2u}{i}\binom{i/2}{u+t}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}\dfrac{(-1)^{t}}{(24)^{t}}\Biggl{(}\sum_{u=0}^{\infty}\dfrac{(-1)^{u}}{(2u)!}\Bigl{(}\dfrac{\pi}{6}\Bigr{)}^{2u}\underset{:=\mathcal{E}_{1}(u,t)}{\underbrace{\sum_{i=0}^{2u}(-1)^{i}\binom{2u}{i}\binom{i/2}{u+t}}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}.$

By Lemma 3.3,

$\mathcal{E}_{1}(u,t)=\left\\{\begin{array}[]{cc}1,&\text{if $u=t=0$}\\\
0,&\text{if $u>t$}\\\
\frac{2u(1/2-t)_{t+1}(-t)_{u}}{t(t+u)!},&\text{otherwise}\end{array}\right..$

Consequently, for all $t\geq 1$,

$\sum_{u=0}^{t}\dfrac{(-1)^{u}}{(2u)!}\Bigl{(}\dfrac{\pi}{6}\Bigr{)}^{2u}\mathcal{E}_{1}(u,t)=\dfrac{(1/2-t)_{t+1}}{t}\sum_{u=1}^{t}\dfrac{(-1)^{u}(-t)_{u}}{(t+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u}.$
(3.13)

It follows that

$\begin{split}\sum_{t=0}^{\infty}\underset{r\in
V(2t)}{\sum}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}&=1+\sum_{t=1}^{\infty}\Biggl{(}\dfrac{(-1)^{t}}{(24)^{t}}\dfrac{(1/2-t)_{t+1}}{t}\sum_{u=1}^{t}\dfrac{(-1)^{u}(-t)_{u}}{(t+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\\\
&=E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}.\end{split}$ (3.14)

Similar to (3.11), we have

$V(2t+1)=\bigl{\\{}(2u+1,i,u+t+1)\in\mathbb{Z}^{3}_{\geq 0}:0\leq i\leq
2u+1\bigr{\\}},$ (3.15)

and consequently, it follows that

$\displaystyle\sum_{t=0}^{\infty}\underset{r\in
V(2t+1)}{\sum}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$ $\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}\dfrac{(-1)^{t}}{(24)^{t}}\Biggl{(}\sum_{u=0}^{\infty}\dfrac{(\pi\sqrt{2/3})^{2u+1}}{(2u+1)!}\dfrac{(-1)^{u}}{(24)^{u+1}}\underset{:=\mathcal{O}_{1}(u,t)}{\underbrace{\sum_{i=0}^{2u+1}(-1)^{i}\binom{2u+1}{i}\binom{i/2}{u+t+1}}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}.$

By Lemma 3.3,

$\mathcal{O}_{1}(u,t)=\left\\{\begin{array}[]{cc}0,&\text{if $u>t$}\\\
-\frac{(2u+1)(1/2-t)_{t+1}(-t)_{u}}{(t+u+1)!},&\text{otherwise}\end{array}\right..$

It follows that

$\displaystyle\sum_{t=0}^{\infty}\underset{r\in
V(2t+1)}{\sum}S(r)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$
$\displaystyle=-\dfrac{\pi}{12\sqrt{6}}\sum_{t=0}^{\infty}\Biggl{(}\dfrac{(-1)^{t}(1/2-t)_{t+1}}{(24)^{t}}\sum_{u=0}^{t}\dfrac{(-1)^{u}(-t)_{u}}{(t+u+1)!(2u)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$
$\displaystyle=O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}.$ (3.17)

From (3.10), (3.14), and (3), we get (3.8). ∎

######  Definition 3.5.

For $t\in\mathbb{Z}_{\geq 0}$, define

$E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=\sum_{t=0}^{\infty}e_{2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\
\text{with}\ \ e_{2}(t):=\dfrac{1}{(24)^{t}}.$ (3.18)

######  Definition 3.6.

For $t\in\mathbb{Z}_{\geq 0}$, define

$O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=\sum_{t=0}^{\infty}o_{2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}\
\text{with}\ \
o_{2}(t):=-\dfrac{6}{\pi\sqrt{24}}\binom{-3/2}{t}\dfrac{(-1)^{t}}{(24)^{t}}.$
(3.19)

######  Lemma 3.7.

Let $A_{2}(n)$ be defined as in (3.2). Let $E_{2}(n)$ be as in Definition 3.5
and $O_{2}(n)$ as in Definition 3.6. Then

$A_{2}(n)=E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}+O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}.$
(3.20)

  * Proof.

Recall the definition of $A_{2}(n)$ from (3.2) and expand it in the following
way:

$\displaystyle A_{2}(n)$ $\displaystyle=$
$\displaystyle\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}=\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-1}-\dfrac{6}{\pi\sqrt{24}}\dfrac{1}{\sqrt{n}}\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-3/2}$
(3.21) $\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}\dfrac{1}{(24)^{t}}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}-\dfrac{6}{\pi\sqrt{24}}\sum_{t=0}^{\infty}\binom{-3/2}{t}\dfrac{(-1)^{t}}{(24)^{t}}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$
$\displaystyle=$ $\displaystyle
E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}+O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}.$

This completes the proof of (3.20). ∎

######  Definition 3.8.

In view of the Definitions 3.1-3.6, we define

$S_{e,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)},$
(3.22)
$S_{e,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)},$
(3.23)
$S_{o,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)},$
(3.24)

and

$S_{o,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}:=E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}.$
(3.25)

######  Lemma 3.9.

For each $i\in\\{1,2\\}$, let $S_{e,i}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ and
$S_{o,i}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ be as in Definition 3.8. Then

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}=\dfrac{1}{4n\sqrt{3}}e^{\pi\sqrt{2n/3}}\sum_{i=1}^{2}\Biggl{(}S_{e,i}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}+S_{o,i}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}\Biggr{)}.$
(3.26)

  * Proof.

The proof follows immediately by applying Lemmas 3.4 and 3.7 to (3.2). ∎

######  Definition 3.10.

For $t\in\mathbb{Z}_{\geq 0}$, define

$S_{1}(t):=\sum_{s=1}^{t}\dfrac{(-1)^{s}(1/2-s)_{s+1}}{s}\sum_{u=1}^{s}\dfrac{(-1)^{u}(-s)_{u}}{(s+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u},$
(3.27)

and

$g_{e,1}(t):=\dfrac{1}{(24)^{t}}\Bigl{(}1+S_{1}(t)\Bigr{)}.$ (3.28)

######  Lemma 3.11.

Let $S_{e,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ be as in (3.22). Let
$g_{e,1}(t)$ be as in Definition 3.10. Then

$S_{e,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}=\sum_{t=0}^{\infty}g_{e,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}.$
(3.29)

  * Proof.

From (3.4), (3.18), and (3.22), we have

$\displaystyle S_{e,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ $\displaystyle=$
$\displaystyle
E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}=\Biggl{(}1+\sum_{t=1}^{\infty}e_{1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\Biggr{)}\Biggl{(}1+\sum_{t=1}^{\infty}e_{2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\Biggr{)}$
$\displaystyle=$ $\displaystyle
1+\sum_{t=1}^{\infty}\Bigl{(}e_{1}(t)+e_{2}(t)\Bigr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}+\sum_{t=2}^{\infty}\Biggl{(}\sum_{s=1}^{t-1}e_{1}(s)e_{2}(t-s)\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}$
$\displaystyle=$ $\displaystyle
1+\sum_{t=1}^{\infty}\Biggl{(}e_{1}(t)+e_{2}(t)+\sum_{s=1}^{t-1}e_{1}(s)e_{2}(t-s)\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}.$

Combining (3.3) and (3.18), we obtain

$\displaystyle e_{1}(t)+e_{2}(t)+\sum_{s=1}^{t-1}e_{1}(s)e_{2}(t-s)$
$\displaystyle=$ $\displaystyle\dfrac{(-1)^{t}(1/2-t)_{t+1}}{(24)^{t}\
t}\sum_{u=1}^{t}\dfrac{(-1)^{u}(-t)_{u}}{(t+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u}+\dfrac{1}{(24)^{t}}$
(3.31) $\displaystyle+$
$\displaystyle\frac{1}{24^{t}}\sum_{s=1}^{t-1}\Biggl{(}\dfrac{(-1)^{s}(1/2-s)_{s+1}}{s}\sum_{u=1}^{s}\dfrac{(-1)^{u}(-s)_{u}}{(s+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u}\Biggr{)}$
$\displaystyle=$
$\displaystyle\dfrac{1}{(24)^{t}}\Bigl{(}1+S_{1}(t)\Bigr{)}=g_{e,1}(t),$

which concludes the proof of (3.29). ∎

######  Definition 3.12.

For $t\in\mathbb{Z}_{\geq 1}$, define

$S_{2}(t):=\sum_{s=0}^{t-1}(1/2-s)_{s+1}\binom{-3/2}{t-s-1}\sum_{u=0}^{s}\dfrac{(-1)^{u}(-s)_{u}}{(s+u+1)!(2u)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u},$
(3.32)

and

$g_{e,2}(t):=\dfrac{(-1)^{t-1}}{(24)^{t}}S_{2}(t).$ (3.33)

######  Lemma 3.13.

Let $S_{e,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ as in (3.23) and $g_{e,2}(t)$
as in Definition 3.12. Then

$S_{e,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}=\sum_{t=1}^{\infty}g_{e,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}.$
(3.34)

  * Proof.

From (3.6), (3.20) and (3.23), we have

$\displaystyle S_{e,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ $\displaystyle=$
$\displaystyle
O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$
(3.35) $\displaystyle=$
$\displaystyle\Biggl{(}\sum_{t=0}^{\infty}o_{1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}\Biggr{)}\Biggl{(}\sum_{t=0}^{\infty}o_{2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}\Biggr{)}$
$\displaystyle=$
$\displaystyle\sum_{t=1}^{\infty}\Biggl{(}\sum_{s=0}^{t-1}o_{1}(s)o_{2}(t-s-1)\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}$
$\displaystyle=$
$\displaystyle\sum_{t=1}^{\infty}g_{e,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\
(\text{by}\ \ \eqref{defeqn2}\ \text{and}\ \eqref{eqn19}).$

∎

######  Definition 3.14.

For $t\in\mathbb{Z}_{\geq 2}$, define

$S_{3}(t):=\sum_{s=1}^{t}\dfrac{(1/2-s)_{s+1}\binom{-3/2}{t-s}}{s}\sum_{u=1}^{s}\dfrac{(-1)^{u}(-s)_{u}}{(s+u)!(2u-1)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u},$
(3.36)

and

$g_{o,1}(t):=\begin{cases}-\dfrac{6}{\pi\sqrt{24}}\dfrac{(-1)^{t}}{(24)^{t}}\biggl{(}\binom{-3/2}{t}+S_{3}(t)\biggr{)},&\quad\text{if\
$t\geq 2$}\\\ -\dfrac{432+\pi^{2}}{2304\sqrt{6}\pi},&\quad\text{if\ $t=1$}\\\
-\dfrac{6}{\pi\sqrt{24}},&\quad\text{if\ $t=0$}\end{cases}.$ (3.37)

######  Lemma 3.15.

Let $S_{o,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ as in (3.24) and $g_{o,1}(t)$
be as in Definition 3.14. Then

$S_{o,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}=\sum_{t=0}^{\infty}g_{o,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}.$
(3.38)

  * Proof.

From (3.4), (3.19) and (3.24), it follows that

$\displaystyle S_{o,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ $\displaystyle=$
$\displaystyle
E_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}O_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$
$\displaystyle=$
$\displaystyle\dfrac{1}{\sqrt{n}}\Biggl{(}1+\sum_{t=1}^{\infty}e_{1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\Biggr{)}\Biggl{(}-\dfrac{6}{\pi\sqrt{24}}+\sum_{t=1}^{\infty}o_{2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}\Biggr{)}$
$\displaystyle=$
$\displaystyle-\dfrac{6}{\pi\sqrt{24}}\dfrac{1}{\sqrt{n}}-\dfrac{432+\pi^{2}}{2304\sqrt{6}\pi}\dfrac{1}{\sqrt{n}^{3}}+\sum_{t=2}^{\infty}\Biggl{(}o_{2}(t)+\sum_{s=1}^{t}e_{1}(s)o_{2}(t-s)\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$
$\displaystyle=$
$\displaystyle-\dfrac{6}{\pi\sqrt{24}}\dfrac{1}{\sqrt{n}}-\dfrac{432+\pi^{2}}{2304\sqrt{6}\pi}\dfrac{1}{\sqrt{n}^{3}}+\sum_{t=2}^{\infty}g_{o,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}(\text{by}\
\ \eqref{defeqn0}\ \text{and}\ \eqref{eqn19}).$

∎

######  Definition 3.16.

For $t\in\mathbb{Z}_{\geq 1}$, define

$S_{4}(t):=\sum_{s=0}^{t}(-1)^{s}(1/2-s)_{s+1}\sum_{u=0}^{s}\dfrac{(-1)^{u}(-s)_{u}}{(s+u+1)!(2u)!}\Bigl{(}\dfrac{\pi^{2}}{36}\Bigr{)}^{u},$
(3.40)

and

$g_{o,2}(t):=-\dfrac{\pi}{12\sqrt{6}}\dfrac{1}{(24)^{t}}S_{4}(t).$ (3.41)

######  Lemma 3.17.

Let $S_{o,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ be as in (3.25) and
$g_{o,2}(t)$ be as in Definition 3.16. Then

$S_{o,2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}=\sum_{t=0}^{\infty}g_{o,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}.$
(3.42)

  * Proof.

From (3.6), (3.18) and (3.25), it follows that

$\displaystyle S_{o,1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$ $\displaystyle=$
$\displaystyle
O_{1}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}E_{2}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}$
(3.43) $\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}\Biggl{(}\sum_{s=0}^{t}o_{1}(s)e_{2}(t-s)\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{\infty}g_{o,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}\
(\text{by}\ \ \eqref{defeqn3}\ \text{and}\ \eqref{eqn18}).$

∎

######  Definition 3.18.

For each $i\in\\{1,2\\}$, let $g_{e,i}(t)$ and $g_{o,i}(t)$ be as in
Definitions 3.10-3.16. We define a power series

$G(n):=\sum_{t=0}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}=\sum_{t=0}^{\infty}g(2t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}+\sum_{t=0}^{\infty}g(2t+1)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1},$

where

$g(2t):=g_{e,1}(t)+g_{e,2}(t)\ \ \text{and}\ \
g(2t+1):=g_{o,1}(t)+g_{o,2}(t).$ (3.44)

######  Lemma 3.19.

Let $G(n)$ be as in Definition 3.18. Then

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\Bigl{(}1-\dfrac{1}{\mu(n)}\Bigr{)}=\dfrac{1}{4n\sqrt{3}}e^{\pi\sqrt{2n/3}}\cdot
G(n).$ (3.45)

  * Proof.

Applying Lemmas 3.11-3.17 to Lemma 3.7, we immediately obtain (3.45). ∎

######  Remark 3.20.

Note that using Sigma and GeneratingFunctions due to Mallinger [18], we
observe that for all $t\geq 0$,

$g(2t)=g_{e,1}(t)+g_{e,2}(t)=\omega_{2t}\ \ \text{and}\ \
g(2t+1)=g_{o,1}(t)+g_{o,2}(t)=\omega_{2t+1},$ (3.46)

where $\omega_{t}$ is as in (1.5). Equivalently,

$g(t)=\omega_{t}=\dfrac{1}{(-4\sqrt{6})^{t}}\sum_{k=0}^{\frac{t+1}{2}}\binom{t+1}{k}\dfrac{t+1-k}{(t+1-2k)!}\Bigl{(}\frac{\pi}{6}\Bigr{)}^{t-2k}.$
(3.47)

However this was already clear from the uniqueness of the asymptotic expansion
for $p(n)$ and its proof can be considered as an additional verification of
our computations. The reader might wonder at this point why we did not use the
single sum expression found by O’Sullivan to bound the remainder of the
asymptotic expansion for $p(n)$. We tried this indeed, but could not obtain
from $\omega_{t}$ an effective upper and lower bound. The summation package
Sigma could not rewrite $\omega_{t}$ as a definite sum which is crucial for
our estimations. However going to the double sum expression $g(t)$, Sigma was
able to give a definite sum expression for the inner sum as we will see later,
and this enabled us to obtain effective upper and lower bounds in the sense
that we described earlier. Namely, $l(t)<g(t)<u(t)$ and
$\lim_{t\to\infty}\frac{l(t)}{g(t)}=\lim_{t\to\infty}\frac{u(t)}{g(t)}=1$.

##  4\. Preliminary Lemmas

This section presents all the preliminary facts needed for the proofs of the
lemmas stated in Section 5. The proofs of Lemmas 4.1 to 4.6, except 4.4, are
presented in Subsection 8.1.

######  Lemma 4.1.

Let $x_{1},x_{2},\dots,x_{n}\leq 1$ and $y_{1},\dots,y_{1}$ be non-negative
real numbers. Then

$\frac{(1-x_{1})(1-x_{2})\cdots(1-x_{n})}{(1+y_{1})(1+y_{2})\cdots(1+y_{n})}\geq
1-\sum_{j=1}^{n}x_{j}-\sum_{j=1}^{n}y_{j}.$

######  Lemma 4.2.

For $t\geq 1$ and non-negative integer $u\leq t$, we have

$\frac{1}{2t}\geq\frac{t(-t)_{u}(-1)^{u}}{(1+2t)(t+u)(t)_{u}}\geq\frac{1}{2t}\Biggl{(}1-\frac{u^{2}+\frac{1}{2}}{t}\Biggr{)}.$

######  Lemma 4.3.

For $t\geq 1$ and non-negative integer $u\leq t$, we have

$\frac{2u+1}{2t}\geq\frac{1}{1+2t}+\frac{2t}{1+2t}\sum_{i=1}^{u}\frac{(-t)_{i}(-1)^{i}}{(t+i)(t)_{i}}\geq\frac{2u+1}{2t}-\frac{4u^{3}+6u^{2}+8u+3}{12t^{2}}.$

Throughout the rest of this paper,

$\alpha:=\frac{\pi}{6}.$

######  Lemma 4.4.

We have

$\begin{split}&\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}=\cosh(\alpha),\
\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u)!}=\frac{1}{2}\alpha\sinh(\alpha),\
\sum_{u=0}^{\infty}\frac{u^{2}\alpha^{2u}}{(2u)!}=\frac{\alpha^{2}}{4}\cosh(\alpha)+\frac{\alpha}{4}\sinh(\alpha),\\\
\intertext{and}&\sum_{u=0}^{\infty}\frac{u^{3}\alpha^{2u}}{(2u)!}=\frac{3\alpha^{2}}{8}\cosh(\alpha)+\frac{\alpha(\alpha^{2}+1)}{8}\sinh(\alpha).\end{split}$

######  Lemma 4.5.

Let $u\in\mathbb{Z}_{\geq 0}$. Assume that $a_{n+1}-a_{n}\geq b_{n+1}-b_{n}$
for all $n\geq u$, and $\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}b_{n}=0$. Then

$b_{n}\geq a_{n}\text{ for all $n\geq u$}.$

######  Lemma 4.6.

For $t\geq 1$ and $k\in\\{0,1,2,3\\}$ we have

$\sum_{u=t+1}^{\infty}\frac{u^{k}\alpha^{2u}}{(2u)!}\leq\frac{C_{k}}{t^{2}}\ \
\text{with}\ \ C_{k}=\frac{\alpha^{4}2^{k}}{18}.$

######  Lemma 4.7.

[3, Equation 7.5, Lemma 7.3] For $n,k,s\in\mathbb{Z}_{\geq 1}$ and $n>2s$ let

$b_{k,n}(s):=\dfrac{4\sqrt{s}}{\sqrt{s+k-1}}\binom{s+k-1}{s-1}\dfrac{1}{n^{k}},$

then

$0<\sum_{t=k}^{\infty}\binom{-\frac{2s-1}{2}}{t}\dfrac{(-1)^{k}}{n^{k}}<b_{k,n}(s).$
(4.1)

######  Lemma 4.8.

[3, Equation 7.9, Lemma 7.5] For $m,n,s\in\mathbb{Z}_{\geq 1}$ and $n>2s$ let

$c_{m,n}(s):=\dfrac{2}{m}\dfrac{s^{m}}{n^{m}},$

then

$-\dfrac{c_{m,n}(s)}{\sqrt{m}}<\sum_{k=m}^{\infty}\binom{1/2}{k}\dfrac{(-1)^{k}s^{k}}{n^{k}}<0.$
(4.2)

######  Lemma 4.9.

[3, Equation 7.7, Lemma 7.4] For $n,s\in\mathbb{Z}_{\geq 1}$, $m\in\mathbb{N}$
and $n>2s$ let

$\beta_{m,n}(s):=\dfrac{2}{n^{m}}\binom{s+m-1}{s-1},$

then

$0<\sum_{k=m}^{\infty}\binom{-s}{k}\dfrac{(-1)^{k}}{n^{k}}<\beta_{m,n}(s).$
(4.3)

##  5\. Estimation of $\bigl{(}S_{i}(t)\bigr{)}$

For the sake of a compact representation the organization of this section is
as follows. We first present the statements of the lemmas needed; then, in a
separate subsection we present the proofs.

###  5.1. The Lemmas 5.1 to 5.4

######  Lemma 5.1.

Let $S_{1}(t)$ be as in Definition 3.10. Then for all $t\geq 1$,

$-\frac{1}{8t^{2}}<\frac{S_{1}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\bigl{(}\cosh(\alpha)-1\bigr{)}+\frac{1}{2t}\alpha\sinh(\alpha)<\frac{13}{25t^{2}}.$
(5.1)

######  Lemma 5.2.

Let $S_{2}(t)$ be as in Definition 3.12. Then for all $t\geq 1$,

$-\frac{11}{10t}<\frac{S_{2}(t)}{\binom{-\frac{3}{2}}{t}}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{\sinh(\alpha)}{\alpha}<\frac{1}{t}.$
(5.2)

######  Lemma 5.3.

Let $S_{3}(t)$ be as in Definition 3.14. Then for all $t\geq 2$,

$-\frac{71}{100t}<\frac{S_{3}(t)}{\binom{-\frac{3}{2}}{t}}+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\alpha\sinh(\alpha)+1-\cosh(\alpha)<\frac{12}{25t}.$
(5.3)

######  Lemma 5.4.

Let $S_{4}(t)$ be as in Definition 3.16. Then for $t\geq 1$,

$-\frac{1}{3t^{2}}<\frac{S_{4}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\frac{\sinh(\alpha)}{\alpha}+\frac{1}{2t}\cosh(\alpha)<\frac{13}{20t^{2}}.$
(5.4)

###  5.2. The Proofs of Lemma 5.1 to 5.4

_Proof of Lemma 5.1:_ We rewrite $S_{1}(t)$ as follows:

$\displaystyle S_{1}(t)$ $\displaystyle=$
$\displaystyle\sum_{u=1}^{t}\frac{(-1)^{u}\alpha^{2u}}{(2u-1)!}\sum_{s=u}^{t}\frac{(-1)^{s}}{s}\Bigl{(}\frac{1}{2}-s\Bigr{)}_{s+1}\frac{(-s)_{u}}{(s+u)!}$
$\displaystyle=$
$\displaystyle\sum_{u=1}^{t}\frac{(-1)^{u}\alpha^{2u}}{(2u-1)!}\underbrace{\sum_{s=0}^{t-u}\frac{(-1)^{s+u}}{s+u}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\frac{(-s-u)_{u}}{(s+2u)!}}_{=:S_{1}(t,u)}.$

We use the summation package Sigma (and its mechanization by
EvaluateMultiSums)222For further explanations of this rigorous computer
derviation we refer to Appendix 8.2 and Remark 8.1., to derive and prove that

$S_{1}(t,u)=(-1)^{t}\binom{-\frac{3}{2}}{t}\frac{(-1)^{u}}{2u}A_{1}(t,u),$
(5.6)

where

$A_{1}(t,u)=\frac{t(-t)_{u}(-1)^{u}}{(1+2t)(t+u)(t)_{u}}-\Biggl{(}\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}+\frac{1}{(1+2t)}+\frac{2t}{1+2t}\sum_{i=1}^{u}\frac{(-t)_{i}(-1)^{i}}{(t+i)(t)_{i}}\Biggr{)}.$

Now by Lemmas 4.2 and 4.3,

$\frac{1}{2t}+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}-\frac{u^{2}+\frac{1}{2}}{2t^{2}}\leq
A_{1}(t,u)\leq\frac{1}{2t}+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}+\frac{4u^{3}+6u^{2}+8u+3}{12t^{2}}.$

It is convenient to reorder the terms in this inequality with respect to the
powers of $u$:

$\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}-\frac{u}{t}-\frac{u^{2}}{2t^{2}}\leq
A_{1}(t,u)\leq\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}+\frac{1}{4t^{2}}+u\Bigl{(}\frac{2}{3t^{2}}-\frac{1}{t}\Bigr{)}+\frac{u^{2}}{2t^{2}}+\frac{u^{3}}{3t^{2}}.$
(5.7)

Combining (5.2) and (5.6), if follows that

$S_{1}(t)=(-1)^{t}\binom{-\frac{3}{2}}{t}\sum_{u=1}^{t}\frac{\alpha^{2u}A_{1}(t,u)}{(2u)!}.$
(5.8)

To derive a lower bound, combine (5.7) with (5.8) to get

$\displaystyle\dfrac{S_{1}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}$
$\displaystyle\geq$
$\displaystyle\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}\Biggr{)}\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{t}\sum_{u=1}^{t}\frac{u\alpha^{2u}}{(2u)!}-\frac{1}{2t^{2}}\sum_{u=1}^{t}\frac{u^{2}\alpha^{2u}}{(2u)!}$
$\displaystyle\geq$
$\displaystyle\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}\Biggr{)}\Biggl{(}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}-1-\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u)!}\Biggr{)}-\frac{1}{t}\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u)!}-\frac{1}{2t^{2}}\sum_{u=0}^{\infty}\frac{u^{2}\alpha^{2u}}{(2u)!}.$
$\displaystyle>$
$\displaystyle\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}\Biggr{)}\Biggl{(}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}-1-\frac{\alpha^{4}}{18t^{2}}\Biggr{)}-\frac{1}{t}\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u)!}-\frac{1}{2t^{2}}\sum_{u=0}^{\infty}\frac{u^{2}\alpha^{2u}}{(2u)!}$
$\displaystyle\hskip 128.0374pt\Biggl{(}\text{by Lemma\ \ref{helperC}\ \ and}\
\ \frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}>\frac{1}{4t^{2}}\ \ \text{for all}\
\ t\geq 1\Biggr{)}$ $\displaystyle=$
$\displaystyle\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}\Biggr{)}\Bigl{(}\cosh(\alpha)-1-\frac{\alpha^{4}}{18t^{2}}\Bigr{)}-\frac{1}{2t}\alpha\sinh(\alpha)-\frac{1}{2t^{2}}\Bigl{(}\frac{\alpha^{2}}{4}\cosh(\alpha)+\frac{\alpha}{4}\sinh(\alpha)\Bigr{)}$
$\displaystyle\hskip 284.52756pt\ \ \Bigl{(}\text{by Lemma\
\ref{helperC2}}\Bigr{)}$ $\displaystyle>$
$\displaystyle\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}\Biggr{)}\Bigl{(}\cosh(\alpha)-1\Bigr{)}-\frac{\alpha^{4}}{18t^{2}}-\frac{1}{2t}\alpha\sinh(\alpha)-\frac{1}{2t^{2}}\Bigl{(}\frac{\alpha^{2}}{4}\cosh(\alpha)+\frac{\alpha}{4}\sinh(\alpha)\Bigr{)}$
$\displaystyle\hskip 199.16928pt\Biggl{(}\text{as}\ \
\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{4t^{2}}<1\ \ \text{for all}\
\ t\geq 1\Biggr{)}$ $\displaystyle=$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)-\frac{1}{2t}\alpha\sinh(\alpha)-\frac{1}{2t^{2}}\Biggl{(}\frac{\cosh(\alpha)-1}{2}+\frac{\alpha^{4}}{9}+\frac{\alpha^{2}}{4}\cosh(\alpha)+\frac{\alpha}{4}\sinh(\alpha)\Biggr{)}.$
$\displaystyle>$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)-\frac{1}{2t}\alpha\sinh(\alpha)-\frac{1}{8t^{2}}$
$\displaystyle\hskip 128.0374pt\Biggl{(}\text{as}\
\frac{\cosh(\alpha)-1}{2}+\frac{\alpha^{4}}{9}+\frac{\alpha^{2}}{4}\cosh(\alpha)+\frac{\alpha}{4}\sinh(\alpha)<\frac{1}{4}\Biggr{)}.$

Similarly, for the upper bound, we have for all $t\geq 1$,

$\displaystyle\frac{S_{1}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}$ (5.10)
$\displaystyle\leq$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{t}\sum_{u=1}^{t}\frac{u\alpha^{2u}}{(2u)!}+\frac{1}{4t^{2}}\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u)!}+\frac{2}{3t^{2}}\sum_{u=1}^{t}\frac{u\alpha^{2u}}{(2u)!}+\frac{1}{2t^{2}}\sum_{u=1}^{t}\frac{u^{2}\alpha^{2u}}{(2u)!}+\frac{1}{3t^{2}}\sum_{u=1}^{t}\frac{u^{3}\alpha^{2u}}{(2u)!}$
$\displaystyle=$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{\infty}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{t}\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u)!}+\frac{1}{t}\sum_{u=t+1}^{\infty}\frac{u\alpha^{2u}}{(2u)!}+\frac{1}{4t^{2}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}$
$\displaystyle\hskip
142.26378pt+\frac{2}{3t^{2}}\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u)!}+\frac{1}{2t^{2}}\sum_{u=0}^{\infty}\frac{u^{2}\alpha^{2u}}{(2u)!}+\frac{1}{3t^{2}}\sum_{u=0}^{\infty}\frac{u^{3}\alpha^{2u}}{(2u)!}$
$\displaystyle\leq$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)-\frac{1}{2t}\alpha\sinh(\alpha)+\frac{\alpha^{4}}{9t^{3}}+\frac{1}{4t^{2}}\cosh(\alpha)+\frac{1}{3t^{2}}\alpha\sinh(\alpha)$
$\displaystyle\hskip
56.9055pt+\frac{1}{2t^{2}}\Biggl{(}\frac{\alpha^{2}}{4}\cosh(\alpha)+\frac{\alpha}{4}\sinh(\alpha)\Biggr{)}+\frac{1}{3t^{2}}\Biggl{(}\frac{3\alpha^{2}}{8}\cosh(\alpha)+\frac{\alpha(\alpha^{2}+1)}{8}\sinh(\alpha)\Biggr{)}$
$\displaystyle\hskip 270.30118pt\Bigl{(}\text{by Lemmas\ \ref{helperC2}\ and\
\ \ \ref{helperC}}\Bigr{)}$ $\displaystyle=$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)-\frac{1}{2t}\alpha\sinh(\alpha)+\frac{1}{t^{2}}\Biggl{(}\frac{\alpha^{4}}{9t}+\frac{\alpha^{2}+1}{4}\cosh(\alpha)+\frac{\alpha(\alpha^{2}+12)}{24}\sinh(\alpha)\Biggr{)}$
$\displaystyle\leq$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)-\frac{1}{2t}\alpha\sinh(\alpha)+\frac{1}{t^{2}}\Biggl{(}\frac{\alpha^{4}}{9}+\frac{\alpha^{2}+1}{4}\cosh(\alpha)+\frac{\alpha(\alpha^{2}+12)}{24}\sinh(\alpha)\Biggr{)}$
$\displaystyle<$
$\displaystyle\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)-\frac{1}{2t}\alpha\sinh(\alpha)+\frac{13}{25t^{2}}$
$\displaystyle\hskip 142.26378pt\Biggl{(}\text{as}\
\frac{\alpha^{4}}{9}+\frac{\alpha^{2}+1}{4}\cosh(\alpha)+\frac{\alpha(\alpha^{2}+12)}{24}\sinh(\alpha)<\frac{13}{25}\Biggr{)}.$

By (5.2) and (5.10), for all $t\geq 1$, it follows that

$-\frac{1}{8t^{2}}<\frac{S_{1}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}(\cosh(\alpha)-1)+\frac{1}{2t}\alpha\sinh(\alpha)<\frac{13}{25t^{2}},$
(5.11)

which concludes the proof. ∎

_Proof of Lemma 5.2:_ Rewrite $S_{2}(t)$ as follows:

$\displaystyle S_{2}(t)$ $\displaystyle=$
$\displaystyle\sum_{u=0}^{t-1}\frac{(-1)^{u}\alpha^{2u}}{(2u)!}\sum_{s=u}^{t-1}\Bigl{(}\frac{1}{2}-s\Bigr{)}_{s+1}\binom{-\frac{3}{2}}{t-s-1}\frac{(-s)_{u}}{(s+u+1)!}$
$\displaystyle=$
$\displaystyle\sum_{u=0}^{t-1}\frac{(-1)^{u}\alpha^{2u}}{(2u)!}\underbrace{\sum_{s=0}^{t-u-1}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\binom{-\frac{3}{2}}{t-s-u-1}\frac{(-s-u)_{u}}{(s+2u+1)!}}_{=:S_{2}(t,u)}$

Using the summation package Sigma (and its mechanization by
EvaluateMultiSums)333We refer again to Appendix 8.2 and Remark 8.1 to see the
underlying machinery in action. we derive and prove that

$S_{2}(t,u)=\binom{-\frac{3}{2}}{t}(-1)^{u+1}\Bigl{(}A_{2,1}(t,u)+A_{2,2}(t,u)\Bigr{)},$
(5.13)

where

$A_{2,1}(t,u)=\frac{2t(t-u)(-t)_{u}(-1)^{u}}{(1+2t)(1+2u)(t+u)(t)_{u}}$

and

$A_{2,2}(t,u)=\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}+\frac{1}{1+2t}+\frac{2t}{1+2t}\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(t+i)(t)_{i}}.$

From (5.2) and (5.13) it follows that

$S_{2}(t)=-\binom{-\frac{3}{2}}{t}\Bigl{(}s_{2,1}(t)+s_{2,2}(t)\Bigr{)},$
(5.14)

where

$s_{2,1}(t)=\sum_{u=0}^{t-1}\frac{\alpha^{2u}}{(2u)!}A_{2,1}(t,u)\ \
\text{and}\ \
s_{2,2}(t)=\sum_{u=0}^{t-1}\frac{\alpha^{2u}}{(2u)!}A_{2,2}(t,u).$ (5.15)

By Lemma 4.2, we have

$\frac{1}{1+2u}-\frac{u^{2}+u+\frac{1}{2}}{t(1+2u)}\leq\frac{t-u}{t(1+2u)}\Biggl{(}1-\frac{u^{2}+\frac{1}{2}}{t}\Biggr{)}\leq
A_{2,1}(t,u)\leq\frac{t-u}{t(1+2u)}.$ (5.16)

Plugging (5.16) into (5.15) we obtain

$\sum_{u=0}^{t-1}\frac{\alpha^{2u}}{(2u+1)!}-\frac{1}{t}\sum_{u=0}^{t-1}\frac{u^{2}+u+\frac{1}{2}}{(2u+1)!}\alpha^{2u}\leq
s_{2,1}(t)\leq\sum_{u=0}^{t-1}\frac{\alpha^{2u}}{(2u+1)!}-\frac{1}{t}\sum_{u=0}^{t-1}\frac{u\alpha^{2u}}{(2u+1)!},$

and consequently,

$\begin{split}&\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}-\sum_{u=t}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}-\frac{1}{t}\sum_{u=0}^{\infty}\frac{u^{2}+u+\frac{1}{2}}{(2u+1)!}\alpha^{2u}\leq
s_{2,1}(t)\leq\\\ &\hskip
142.26378pt\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}-\frac{1}{t}\Biggl{(}\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u+1)!}-\sum_{u=t}^{\infty}\frac{u\alpha^{2u}}{(2u+1)!}\Biggr{)}.\end{split}$
(5.17)

By Lemma 4.6,

$\sum_{u=t}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}=\frac{1}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!}=\frac{2}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{u\alpha^{2u}}{(2u)!}\leq\frac{2C_{1}}{t^{2}}=\dfrac{2\alpha^{2}}{9t^{2}},$
(5.18)

and

$\sum_{u=t}^{\infty}\frac{u\alpha^{2u}}{(2u+1)!}=\frac{2}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{u(u-1)\alpha^{2u}}{(2u)!}\leq\frac{2}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{u^{2}\alpha^{2u}}{(2u)!}\leq\frac{2C_{2}}{\alpha^{2}t^{2}}=\frac{4\alpha^{2}}{9t^{2}}.$
(5.19)

Plugging (5.18) and (5.19) into (5.17) gives

$\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}-\frac{2\alpha^{2}}{9t^{2}}-\frac{1}{t}\sum_{u=0}^{\infty}\frac{u^{2}+u+\frac{1}{2}}{(2u+1)!}\alpha^{2u}\leq
s_{2,1}(t)\leq\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}-\frac{1}{t}\sum_{u=0}^{\infty}\frac{u\alpha^{2u}}{(2u+1)!}+\frac{4\alpha^{2}}{9t^{3}}.$
(5.20)

Using Lemma 4.4, (5.20) further reduces to

$\begin{split}&\frac{\sinh(\alpha)}{\alpha}-\frac{1}{t}\Biggl{(}\frac{\cosh(\alpha)}{4}+\frac{\sinh(\alpha)}{4\alpha}+\frac{\alpha\sinh(\alpha)}{4}+\frac{2\alpha^{2}}{9}\Biggr{)}\leq
s_{2,1}(t)\leq\\\ &\hskip
170.71652pt\frac{\sinh(\alpha)}{\alpha}-\frac{1}{t}\Biggl{(}\frac{\cosh(\alpha)}{2}-\frac{\sinh(\alpha)}{2\alpha}-\frac{4\alpha^{2}}{9}\Biggr{)}.\end{split}$
(5.21)

A numerical check shows that

$\frac{\cosh(\alpha)}{4}+\frac{\sinh(\alpha)}{4\alpha}+\frac{\alpha\sinh(\alpha)}{4}+\frac{2\alpha^{2}}{9}<\frac{7}{10}\
\ \text{and}\ \
\frac{\cosh(\alpha)}{2}-\frac{\sinh(\alpha)}{2\alpha}-\frac{4\alpha^{2}}{9}>-\frac{3}{40}.$

This, along with (5.21), gives

$\frac{\sinh(\alpha)}{\alpha}-\frac{7}{10t}<s_{2,1}(t)<\frac{\sinh(\alpha)}{\alpha}+\frac{3}{40t}.$
(5.22)

Next we employ Lemma 4.3 and get

$\frac{2u+1}{2t}-\frac{4u^{3}+6u^{2}+8u+3}{12t^{2}}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\leq
A_{2,2}(t,u)\leq\frac{2u+1}{2t}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}.$
(5.23)

Plugging (5.23) into (5.15), we obtain

$\sum_{u=0}^{t-1}\frac{\alpha^{2u}}{(2u)!}\Biggl{(}\frac{2u+1}{2t}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}-\frac{4u^{3}+6u^{2}+8u+3}{12t^{2}}\Biggr{)}\leq
s_{2,2}(t)\leq\sum_{u=0}^{t-1}\frac{\alpha^{2u}}{(2u)!}\Biggl{(}\frac{2u+1}{2t}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\Biggr{)},$

which, using $p_{3}(u):=4u^{3}+6u^{2}+8u+3$, can be rewritten as

$\begin{split}&\frac{1}{2t}\sum_{u=0}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}-\frac{1}{2t}\sum_{u=t}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{12t^{2}}\sum_{u=0}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u)!}\\\
&\hskip 113.81102pt\leq
s_{2,2}(t)\leq\frac{1}{2t}\sum_{u=0}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}-\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\sum_{u=t}^{\infty}\frac{\alpha^{2u}}{(2u)!}.\end{split}$
(5.24)

By Lemma 4.6 we obtain

$\sum_{u=t}^{\infty}\frac{\alpha^{2u}}{(2u)!}=\frac{1}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{(2u-1)2u\alpha^{2u}}{(2u)!}\leq\frac{4}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{u^{2}\alpha^{2u}}{(2u)!}\leq\frac{4C_{2}}{\alpha^{2}t^{2}}=\frac{8\alpha^{2}}{9t^{2}}$
(5.25)

and

$\sum_{u=t}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}=\frac{1}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{2u(2u-1)^{2}\alpha^{2u}}{(2u)!}\leq\frac{8}{\alpha^{2}}\sum_{u=t+1}^{\infty}\frac{u^{3}\alpha^{2u}}{(2u)!}\leq\frac{8C_{3}}{\alpha^{2}t^{2}}=\frac{32\alpha^{2}}{9t^{2}}.$
(5.26)

Combining (5.25) and (5.26) with (5.24) gives

$\begin{split}&\frac{1}{2t}\sum_{u=0}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}-\frac{1}{2t}\frac{32\alpha^{2}}{9t^{2}}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{12t^{2}}\sum_{u=0}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u)!}\\\
&\hskip 96.73918pt\leq
s_{2,2}(t)\leq\frac{1}{2t}\sum_{u=0}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\frac{8\alpha^{2}}{9t^{2}}.\end{split}$
(5.27)

Furthermore, for all $t\geq 1$ we have
$\binom{2t}{t}\geq\frac{4^{t}}{2\sqrt{t}}$ which implies

$\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}=\frac{2^{2t+1}}{t+1}\frac{1}{\binom{2t+2}{t+1}}<1,\
\ \text{$t\geq 1$.}$ (5.28)

Applying (5.28) and Lemma 4.6 to (5.27), we obtain

$\begin{split}&\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{2t}\Bigl{(}\underbrace{\cosh(\alpha)+\alpha\sinh(\alpha)}_{=:\mathrm{csh}(\alpha)}\Bigr{)}-\frac{C_{2,2}(\alpha)}{t^{2}}\leq
s_{2,2}(t)\leq\\\ &\hskip
113.81102pt\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{2t}\mathrm{csh}(\alpha)+\frac{8\alpha^{2}}{9},\end{split}$
(5.29)

where

$C_{2,2}(\alpha)=\frac{16\alpha^{2}}{9}+\frac{\alpha^{2}\cosh(\alpha)}{4}+\frac{\alpha^{3}\sinh(\alpha)}{24}+\frac{\cosh(\alpha)}{4}+\frac{\alpha\sinh(\alpha)}{2}<1\
\ \text{and}\ \ \frac{8\alpha^{2}}{9}<\frac{1}{4}.$

Therefore

$\begin{split}&\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{2t}\mathrm{csh}(\alpha)-\frac{1}{t^{2}}\leq
s_{2,2}(t)\leq\\\ &\hskip
85.35826pt\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{2t}\mathrm{csh}(\alpha)+\frac{1}{4t^{2}}.\end{split}$
(5.30)

Applying (5.22) and (5.30) to (5.14) we obtain

$\begin{split}&\frac{\sinh(\alpha)}{\alpha}-\frac{7}{10t}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{2t}\mathrm{csh}(\alpha)-\frac{1}{t^{2}}\leq-\frac{S_{2}(t)}{\binom{\frac{-3}{2}}{t}}\leq\\\
&\hskip
56.9055pt\frac{\sinh(\alpha)}{\alpha}+\frac{3}{40t}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{2t}\mathrm{csh}(\alpha)+\frac{1}{4t^{2}},\end{split}$

which implies that for $t\geq 1$,

$\begin{split}&\frac{\sinh(\alpha)}{\alpha}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{t}\Biggl{(}-\frac{7}{10}+\frac{\mathrm{csh}(\alpha)}{2}-1\Biggr{)}\leq-\frac{S_{2}(t)}{\binom{\frac{-3}{2}}{t}}\leq\\\
&\frac{\sinh(\alpha)}{\alpha}+\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)+\frac{1}{t}\Biggl{(}\frac{3}{40}+\frac{\mathrm{csh}(\alpha)}{2}+\frac{1}{4}\Biggr{)}.\end{split}$
(5.31)

Since

$-\frac{7}{10}+\frac{\mathrm{csh}(\alpha)}{2}-1>-1\ \ \text{and}\ \
\frac{3}{40}+\frac{\mathrm{csh}(\alpha)}{2}+\frac{1}{4}<\frac{11}{10},$

from (5.31), it follows that for all $t\geq 1$,

$-\frac{1}{t}<-\frac{\sinh(\alpha)}{\alpha}-\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}\cosh(\alpha)-\frac{S_{2}(t)}{\binom{-\frac{3}{2}}{t}}<\frac{11}{10t}.$
(5.32)

Multiplying by $-1$ on both sides of (5.32), we get (5.2). ∎

_Proof of Lemma 5.3:_ Rewrite $S_{3}(t)$ as follows:

$\displaystyle S_{3}(t)$ $\displaystyle=$
$\displaystyle\sum_{u=1}^{t}\frac{(-1)^{u}\alpha^{2u}}{(2u-1)!}\sum_{s=u}^{t}\frac{1}{s}\Bigl{(}\frac{1}{2}-s\Bigr{)}_{s+1}\binom{-\frac{3}{2}}{t-s}\frac{(-s)_{u}}{(s+u)!}$
$\displaystyle=$
$\displaystyle\sum_{u=1}^{t}\frac{(-1)^{u}\alpha^{2u}}{(2u-1)!}\underbrace{\sum_{s=0}^{t-u}\frac{1}{s+u}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\binom{-\frac{3}{2}}{t-s-u}\frac{(-s-u)_{u}}{(s+2u)!}}_{=:S_{3}(t,u)}$

Using the summation package Sigma (and its mechanization by
EvaluateMultiSums), the sum $S_{3}(t,u)$ can be rewritten444We refer again to
Appendix 8.2 and Remark 8.1 to see the underlying machinery in action. as an
indefinite sum

$S_{3}(t,u)=\binom{-\frac{3}{2}}{t}(-1)^{u}\Bigl{(}A_{3,1}(t,u)+A_{3,2}(t,u)\Bigr{)},$
(5.34)

where

$A_{3,1}(t,u)=\frac{t(1+2t-2u)(-t)_{u}(-1)^{u}}{2(1+2t)u(t+u)(t)_{u}}$

and

$\hskip
71.13188ptA_{3,2}(t,u)=\frac{(-1)^{t+1}}{\binom{-\frac{3}{2}}{t}}+\frac{1}{1+2t}+\frac{2t}{1+2t}\sum_{i=1}^{u}\frac{(-t)_{i}(-1)^{i}}{(t+i)(t)_{i}}.$

From (5.2) and (5.34), it follows that

$S_{3}(t)=\binom{-\frac{3}{2}}{t}\Bigl{(}s_{3,1}(t)+s_{3,2}(t)\Bigr{)},$
(5.35)

where

$s_{3,1}(t)=\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u-1)!}A_{3,1}(t,u)\ \
\text{and}\ \
s_{3,2}(t)=\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u-1)!}A_{3,2}(t,u).$ (5.36)

By Lemma 4.2, we have

$-\frac{1+2t-2u}{2u}\frac{1}{2t}\frac{u^{2}+\frac{1}{2}}{t}\leq
A_{3,1}(t,u)-\frac{1+2t-2u}{2u}\frac{1}{2t}=A_{3,1}(t,u)-\frac{1}{2u}+\frac{2u-1}{4ut}\leq
0.$ (5.37)

Equation (5.37) implies that

$-\frac{3u^{2}+2u+\frac{1}{2}}{4ut}=-\frac{u^{2}+\frac{1}{2}}{2ut}-\frac{\frac{u^{2}}{2}+\frac{1}{4}}{2ut}-\frac{2u-1}{4ut}\leq
A_{3,1}(t,u)-\frac{1}{2u}\leq-\frac{2u-1}{4ut}\leq 0.$ (5.38)

Plugging (5.38) into (5.36), we obtain

$-\frac{1}{2t}\sum_{u=1}^{\infty}\frac{(3u^{2}+2u+\frac{1}{2})\alpha^{2u}}{(2u)!}\leq-\frac{1}{2t}\sum_{u=1}^{t}\frac{(3u^{2}+2u+\frac{1}{2})\alpha^{2u}}{(2u)!}\leq
s_{3,1}(t)-\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u)!}\leq 0,$

and consequently,

$-\frac{1}{2t}\sum_{u=1}^{\infty}\frac{(3u^{2}+2u+\frac{1}{2})\alpha^{2u}}{(2u)!}-\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u)!}\leq
s_{3,1}(t)-\sum_{u=1}^{\infty}\frac{\alpha^{2u}}{(2u)!}\leq-\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u)!}\leq
0.$ (5.39)

Applying Lemmas 4.6 and 4.4 to (5.39) gives

$-\frac{1}{2t}<-\frac{1}{t}\Biggl{(}\frac{3\alpha^{2}\cosh(\alpha)+7\alpha\sinh(\alpha)+2\cosh(\alpha)-2}{8}+\frac{\alpha^{4}}{9}\Biggr{)}\leq
s_{3,1}(t)+1-\cosh(\alpha)\leq 0.$ (5.40)

Next, by Lemma 4.3, we obtain

$-\frac{4u^{3}+6u^{2}+8u+3}{12t^{2}}\leq
A_{3,2}(t,u)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}\leq 0.$
(5.41)

Applying (5.41) to (5.36), it follows that

$s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\sum_{u=1}^{t}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}\leq
0$ (5.42)

and

$\begin{split}s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{t}\frac{\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\sum_{u=1}^{t}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}&\geq-\frac{1}{12t^{2}}\sum_{u=1}^{t}\frac{p_{3}(u)\alpha^{2u}}{(2u-1)!}\\\
&\geq-\frac{1}{12t^{2}}\sum_{u=1}^{\infty}\frac{(p_{3}(u)\alpha^{2u}}{(2u-1)!},\end{split}$
(5.43)

where $p_{3}(u)=4u^{3}+6u^{2}+8u+3$ is as in (5.24). Equations (5.42) and
(5.43) imply that

$s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\sum_{u=1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}\leq\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!},$
(5.44)

and

$\begin{split}s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!}&-\frac{1}{2t}\sum_{u=1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}\geq\\\
&-\frac{1}{12t^{2}}\sum_{u=1}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\sum_{u=t+1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}.\end{split}$
(5.45)

By Lemma 4.6 we obtain

$\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!}=2\sum_{u=t+1}^{\infty}\frac{u\alpha^{2u}}{(2u)!}\leq\frac{4\alpha^{4}}{3\cdot
3!t^{2}}=\frac{2\alpha^{4}}{9t^{2}}$ (5.46)

and

$\sum_{u=t+1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}=2u\sum_{u=t+1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}\leq\frac{20\alpha^{4}}{3\cdot
3!t^{2}}=\frac{10\alpha^{4}}{9t^{2}}.$ (5.47)

Substituting (5.46)-(5.47) into (5.44) and (5.45), it follows that

$s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\sum_{u=1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}\leq\frac{3}{2}\cdot\frac{2\alpha^{4}}{9t^{2}}=\frac{\alpha^{4}}{3t^{2}}$
(5.48)

and

$\begin{split}-\frac{1}{12t^{2}}\sum_{u=1}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u-1)!}-\frac{\alpha^{4}}{3t^{2}}&\leq-\frac{1}{12t^{2}}\sum_{u=1}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\frac{10\alpha^{4}}{9t^{2}}\leq\\\
&s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=1}^{\infty}\frac{\alpha^{2u}}{(2u-1)!}-\frac{1}{2t}\sum_{u=1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u-1)!}.\end{split}$
(5.49)

Using Lemma 4.4 into (5.48) and (5.49), we obtain

$\begin{split}-\frac{61}{100t^{2}}&<-\frac{1}{t^{2}}\Biggl{(}\frac{3\alpha^{3}\sinh(\alpha)}{8}+\frac{(\alpha^{4}+24\alpha^{2})\cosh(\alpha)}{24}+\frac{3\alpha\sinh(\alpha)}{4}+\frac{5\alpha^{4}}{9}\Biggr{)}\leq\\\
&s_{3,2}(t)+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\alpha\sinh(\alpha)-\frac{1}{2t}\mathrm{sch}(\alpha)\leq\frac{\alpha^{4}}{3t^{2}}<\frac{3}{100t^{2}},\end{split}$
(5.50)

where $\mathrm{sch}(\alpha):=\alpha^{2}\cosh(\alpha)+2\alpha\sinh(\alpha)$.
Combining (5.40) and (5.50), and then plugging into (5.35) it follows that

$-\frac{1}{2t}-\frac{61}{100t^{2}}<\frac{S_{3}(t)}{\binom{-\frac{3}{2}}{t}}+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\alpha\sinh(\alpha)-\frac{1}{2t}\mathrm{sch}(\alpha)+1-\cosh(\alpha)<\frac{3}{100t^{2}}.$

Since for $t\geq 2$,

$-\frac{1}{2t}-\frac{61}{100t^{2}}+\frac{1}{2t}\mathrm{sch}(\alpha)>-\frac{71}{100t},$

and

$\frac{3}{100t^{2}}+\frac{1}{2t}\mathrm{sch}(\alpha)<\frac{12}{25t},$

we finally get

$\frac{12}{25t}>\frac{S_{3}(t)}{\binom{-\frac{3}{2}}{t}}+\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\alpha\sinh(\alpha)+1-\cosh(\alpha)>-\frac{71}{100t}.$
(5.51)

∎

_Proof of Lemma 5.4:_ Rewrite $S_{4}(t)$ as follows:

$\displaystyle S_{4}(t)$ $\displaystyle=$
$\displaystyle\sum_{u=0}^{t}\frac{(-1)^{u}\alpha^{2u}}{(2u)!}\sum_{s=u}^{t}(-1)^{s}\Bigl{(}\frac{1}{2}-s\Bigr{)}_{s+1}\frac{(-s)_{u}}{(s+u+1)!}$
$\displaystyle=$
$\displaystyle\sum_{u=0}^{t}\frac{(-1)^{u}\alpha^{2u}}{(2u)!}\underbrace{\sum_{s=0}^{t-u}(-1)^{s+u}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\frac{(-s-u)_{u}}{(s+2u+1)!}}_{=:S_{4}(t,u)}$

Using again the summation package Sigma (and its mechanization by
EvaluateMultiSums)555We refer again to Appendix 8.2 and Remark 8.1 to see the
underlying machinery in action., we rewrite $S_{4}(t,u)$ as an indefinite sum

$S_{4}(t,u)=\binom{-\frac{3}{2}}{t}(-1)^{u+t}\Bigl{(}A_{4,1}(t,u)+A_{4,2}(t,u)\Bigr{)},$
(5.53)

where

$A_{4,1}(t,u)=\frac{t(-t)_{u}(-1)^{u}}{2(1+2t)(t+u)(t+u+1)(t)_{u}}$

and

$\hskip
85.35826ptA_{4,2}(t,u)=\frac{1}{1+2u}\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{1}{1+2t}-\frac{2t}{1+2t}\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(t+i)(t)_{i}}\Biggr{)}.$

From (5.2) and (5.53) it follows that

$S_{4}(t)=(-1)^{t}\binom{-\frac{3}{2}}{t}\Bigl{(}s_{4,1}(t)+s_{4,2}(t)\Bigr{)},$
(5.54)

where

$s_{4,1}(t)=\sum_{u=0}^{t}\frac{\alpha^{2u}}{(2u)!}A_{4,1}(t,u)\ \ \text{and}\
\ s_{4,2}(t):=\sum_{u=0}^{t}\frac{\alpha^{2u}}{(2u)!}A_{4,2}(t).$ (5.55)

From Lemmas 4.1-4.2 we have

$\frac{1}{4t^{2}}\Biggl{(}1-\frac{u^{2}+u+\frac{3}{2}}{t}\Biggr{)}\leq\frac{1}{2(t+u+1)}\frac{1}{2t}\Biggl{(}1-\frac{u^{2}+\frac{1}{2}}{t}\Biggr{)}\leq
A_{4,1}(t,u)\leq\frac{1}{2(t+u+1)}\frac{1}{2t}\leq\frac{1}{4t^{2}}.$ (5.56)

Combining (5.56) with (5.55), we obtain

$\frac{1}{4t^{2}}\sum_{u=0}^{t}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{4t^{3}}\sum_{u=0}^{t}\frac{(u^{2}+u+\frac{3}{2})\alpha^{2u}}{(2u)!}\leq
s_{4,1}(t)\leq\frac{1}{4t^{2}}\sum_{u=0}^{t}\frac{\alpha^{2u}}{(2u)!},$

and consequently, we get

$\frac{1}{4t^{2}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{4t^{2}}\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u)!}-\frac{1}{4t^{3}}\sum_{u=0}^{\infty}\frac{(u^{2}+u+\frac{3}{2})\alpha^{2u}}{(2u)!}\leq
s_{4,1}(t)\leq\frac{1}{4t^{2}}\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u)!}.$
(5.57)

Equation (5.57) together with Lemmas 4.6-4.4 imply

$\begin{split}\frac{1}{4t^{2}}\cosh(\alpha)-\frac{3}{5t^{3}}\leq\frac{1}{4t^{2}}\cosh(\alpha)-\frac{1}{t^{3}}\Biggl{(}\frac{\alpha^{4}}{72}+\frac{(\alpha^{2}+6)\cosh(\alpha)}{16}+\frac{3\alpha\sinh(\alpha)}{16}\Biggr{)}&\leq
s_{4,1}(t)\leq\\\ &\frac{1}{4t^{2}}\cosh(\alpha).\end{split}$ (5.58)

Next, by Lemma 4.3, we obtain

$0\leq
A_{4,2}(t,u)-\frac{1}{1+2u}\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}\Biggr{)}\leq\frac{1}{1+2u}\frac{p_{3}(u)}{12t^{2}},$
(5.59)

where $p_{3}(u)$ is as in (5.24). Plugging (5.59) into (5.55), it follows that

$\begin{split}0\leq
s_{4,2}(t)-\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}\Biggr{)}+&\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}\Biggr{)}\\\
&\frac{1}{12t^{2}}\sum_{u=0}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u+1)!},\end{split}$

which implies that

$\begin{split}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}\leq&s_{4,2}(t)-\sum_{u=0}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}\Biggl{(}\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}-\frac{2u+1}{2t}\Biggr{)}\leq\\\
&\frac{1}{12t^{2}}\sum_{u=0}^{\infty}\frac{p_{3}(u)\alpha^{2u}}{(2u+1)!}+\frac{1}{2t}\sum_{u=t+1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u+1)!}.\end{split}$
(5.60)

By Lemma 4.6,

$\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u+1)!}\leq\sum_{u=t+1}^{\infty}\frac{\alpha^{2u}}{(2u)!}\leq\frac{\alpha^{4}}{3\cdot
3!t^{2}}=\frac{\alpha^{4}}{18t^{2}},$ (5.61)

and

$\sum_{u=t+1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u+1)!}\leq\sum_{u=t+1}^{\infty}\frac{(2u+1)\alpha^{2u}}{(2u)!}\leq\frac{C_{0}+2C_{1}}{t^{2}}=\frac{\alpha^{4}(1+4)}{3\cdot
3!t^{2}}=\frac{5\alpha^{4}}{18t^{2}}.$ (5.62)

Applying (5.61) and (5.62) to (5.60) and using Lemma 4.4, we finally obtain

$\begin{split}-\frac{3}{1000t^{2}}<-\frac{2}{3}\cdot\frac{\alpha^{4}}{18}\leq-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\frac{\alpha^{4}}{18t^{2}}&\leq
s_{4,2}(t)-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\frac{\sinh(\alpha)}{\alpha}+\frac{1}{2t}\cosh(\alpha)\leq\\\
&\frac{1}{t^{2}}\Biggl{(}\frac{(\alpha^{2}+6)\cosh(\alpha)}{24}+\frac{\alpha\sinh(\alpha)}{8}+\frac{5\alpha^{4}}{36}\Biggr{)}<\frac{7}{20t^{2}}.\end{split}$
(5.63)

From (5.58), (5.63), and (5.54), it follows that

$\begin{split}\frac{1}{t^{2}}\Biggl{(}-\frac{3}{1000}+\frac{1}{4}\cosh(\alpha)-\frac{3}{5}\Biggr{)}&\leq-\frac{3}{1000t^{2}}-\frac{3}{5t^{3}}+\frac{1}{4t^{2}}\cosh(\alpha)\leq\\\
&\frac{S_{4}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\frac{\sinh(\alpha)}{\alpha}+\frac{1}{2t}\cosh(\alpha)<\frac{1}{t^{2}}\Bigl{(}\frac{7}{20}+\frac{1}{4}\cosh(\alpha)\Bigr{)}.\end{split}$

This implies for $t\geq 1$,

$\frac{13}{20t^{2}}>\frac{S_{4}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}-\frac{(-1)^{t}}{\binom{-\frac{3}{2}}{t}}\frac{\sinh(\alpha)}{\alpha}+\frac{1}{2t}\cosh(\alpha)>-\frac{1}{3t^{2}}.$
(5.64)

∎

##  6\. Error bounds

######  Lemma 6.1.

For all $n,k\in\mathbb{Z}_{\geq 1}$,

$\dfrac{1}{(24n)^{k}}<\sum_{t=k}^{\infty}\dfrac{1}{(24n)^{t}}\leq\dfrac{24}{23}\dfrac{1}{(24n)^{k}}.$
(6.1)

  * Proof.

The statement follows from

$\sum_{t=k}^{\infty}\dfrac{1}{(24n)^{t}}=\dfrac{1}{(24n)^{k}}\dfrac{24n}{24n-1}\
\ \text{and}\ \ 1<\dfrac{24n}{24n-1}\leq\dfrac{24}{23}\ \ \text{for all}\
n\geq 1.$

∎

######  Lemma 6.2.

For all $n,k,s\in\mathbb{Z}_{\geq 1}$,

$\dfrac{1}{(k+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{k}}<\sum_{t=k}^{\infty}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{s}}\dfrac{1}{(24n)^{t}}<\dfrac{12}{5(k+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{k}}.$
(6.2)

  * Proof.

Rewrite the infinite sum as

$\sum_{t=k}^{\infty}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{s}}\dfrac{1}{(24n)^{t}}=\sum_{t=k}^{\infty}\dfrac{\binom{2t+2}{t+1}}{4^{t}}\dfrac{t+1}{2t^{s}}\dfrac{1}{(24n)^{t}}.$
(6.3)

For all $t\geq 1$,

$\dfrac{4^{t}}{2\sqrt{t}}\leq\binom{2t}{t}\leq\dfrac{4^{t}}{\sqrt{\pi t}}.$

From (6.3) we get

$\sum_{t=k}^{\infty}\dfrac{\sqrt{t+1}}{t^{s}}\dfrac{1}{(24n)^{t}}\leq\sum_{t=k}^{\infty}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{s}}\dfrac{1}{(24n)^{t}}\leq\dfrac{4}{\sqrt{\pi}}\sum_{t=k}^{\infty}\dfrac{\sqrt{t+1}}{2t^{s}}\dfrac{1}{(24n)^{t}}.$
(6.4)

For all $k\geq 1$,

$\sum_{t=k}^{\infty}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{s}}\dfrac{1}{(24n)^{t}}\geq\sum_{t=k}^{\infty}\dfrac{\sqrt{t+1}}{t^{s}}\dfrac{1}{(24n)^{t}}>\sum_{t=k}^{\infty}\dfrac{1}{(t+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{t}}>\dfrac{1}{(k+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{k}}$
(6.5)

and

$\displaystyle\sum_{t=k}^{\infty}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{s}}\dfrac{1}{(24n)^{t}}\leq\dfrac{4}{\sqrt{\pi}}\sum_{t=k}^{\infty}\dfrac{\sqrt{t+1}}{2t^{s}}\dfrac{1}{(24n)^{t}}$
$\displaystyle<$
$\displaystyle\dfrac{4}{\sqrt{\pi}}\sum_{t=k}^{\infty}\dfrac{1}{(t+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{t}}$
(6.6) $\displaystyle\leq$
$\displaystyle\dfrac{4}{\sqrt{\pi}(k+1)^{s-\frac{1}{2}}}\sum_{t=k}^{\infty}\dfrac{1}{(24n)^{t}}$
$\displaystyle<$ $\displaystyle\dfrac{4\cdot
24}{23\cdot\sqrt{\pi}}\dfrac{1}{(k+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{k}}\
(\text{by}\ \eqref{errorlem1eqn1}).$ $\displaystyle<$
$\displaystyle\frac{12}{5}\dfrac{1}{(k+1)^{s-\frac{1}{2}}}\dfrac{1}{(24n)^{k}}.$

Equations (6.5) and (6.6) imply (6.2). ∎

######  Lemma 6.3.

For $n\in\mathbb{Z}_{\geq 1}$and $k\in\mathbb{Z}_{\geq 0}$,

$0<\sum_{t=k}^{\infty}\binom{-\frac{3}{2}}{t}\dfrac{(-1)^{t}}{(24n)^{t}}<4\sqrt{2}\dfrac{\sqrt{k+1}}{(24n)^{k}}.$
(6.7)

  * Proof.

Setting $(n,s)\mapsto(24n,2)$ in (4.1), it follows that for all $n\geq 1$,

$0<\sum_{t=k}^{\infty}\binom{-\frac{3}{2}}{t}\dfrac{(-1)^{t}}{(24n)^{t}}<4\sqrt{2}\dfrac{\sqrt{k+1}}{(24n)^{k}}.$

∎

######  Definition 6.4.

For all $k\geq 1$ define

$\begin{split}L_{1}(k)&:=\Biggl{(}\cosh(\alpha)-\dfrac{6\alpha\sinh(\alpha)}{5\sqrt{k+1}}-\dfrac{3}{10(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k}\\\
\hskip 56.9055pt&\\\
U_{1}(k)&:=\Biggl{(}\dfrac{24\cosh(\alpha)}{23}-\dfrac{\alpha\sinh(\alpha)}{2\sqrt{k+1}}+\dfrac{5}{4(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k}.\end{split}$

######  Lemma 6.5.

Let $L_{1}(k)$ and $U_{1}(k)$ be as in Definition 6.4. Let $g_{e,1}(t)$ be as
in Definition 3.10. Then for all $n,k\in\mathbb{Z}_{\geq 1}$,

$L_{1}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k}<\sum_{t=k}^{\infty}g_{e,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}<U_{1}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k}.$
(6.8)

  * Proof.

From (3.28) and (5.1), it follows that for $t\geq 1$,

$\begin{split}\cosh(\alpha)-\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{2t}\alpha\sinh(\alpha)-\dfrac{1}{8}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{2}}&<(24)^{t}g_{e,1}(t)=1+S_{1}(t)\\\
<\cosh(\alpha)&-\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{2t}\alpha\sinh(\alpha)+\dfrac{13}{25}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{2}}.\end{split}$
(6.9)

Now, applying (6.1) and (6.2) with $s=1\ \text{and}\ 2$, respectively, to
(6.9), it follows that for all $k\geq 1$,

$\sum_{t=k}^{\infty}g_{e,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}>\Biggl{(}\cosh(\alpha)-\dfrac{6\alpha\sinh(\alpha)}{5\sqrt{k+1}}-\dfrac{3}{10(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k}$

and

$\begin{split}\sum_{t=k}^{\infty}g_{e,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}&<\Biggl{(}\dfrac{24\cosh(\alpha)}{23}-\dfrac{\alpha\sinh(\alpha)}{2\sqrt{k+1}}+\dfrac{13\cdot
12}{25\cdot
5}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k}\\\
&<\Biggl{(}\dfrac{24\cosh(\alpha)}{23}-\dfrac{\alpha\sinh(\alpha)}{2\sqrt{k+1}}+\frac{5}{4}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k}.\end{split}$

∎

######  Definition 6.6.

For all $k\geq 1$, define

$\begin{split}L_{2}(k)&:=\Biggl{(}-\dfrac{24\cosh(\alpha)}{23}-\dfrac{12}{5\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k}\\\
\hskip 56.9055pt&\\\
U_{2}(k)&:=\Biggl{(}-\cosh(\alpha)+\dfrac{4\sqrt{2}\sinh(\alpha)}{\alpha}\sqrt{k+1}+\dfrac{66}{25\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k}.\end{split}$

######  Lemma 6.7.

Let $L_{2}(k)$ and $U_{2}(k)$ be as in Definition 6.6. Let $g_{e,2}(t)$ be as
in Definition 3.12. Then for all $n,k\in\mathbb{Z}_{\geq 1}$,

$L_{2}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k}<\sum_{t=k}^{\infty}g_{e,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}<U_{2}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k}.$
(6.10)

  * Proof.

From (3.33) and (5.2), it follows that for $t\geq 1$,

$\begin{split}-\cosh(\alpha)+(-1)^{t}\binom{-\frac{3}{2}}{t}\dfrac{\sinh(\alpha)}{\alpha}-\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t}&<(24)^{t}g_{e,2}(t)=(-1)^{t-1}S_{2}(t)\\\
<-\cosh(\alpha)&+(-1)^{t}\binom{-\frac{3}{2}}{t}\dfrac{\sinh(\alpha)}{\alpha}+\dfrac{11}{10}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t}.\end{split}$
(6.11)

Now, applying (6.1), (6.2) with $s=1$ and (6.7) to (6.11), it follows that for
all $k\geq 1$,

$\sum_{t=k}^{\infty}g_{e,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}>\Biggl{(}-\dfrac{24\cosh(\alpha)}{23}-\dfrac{12}{5\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k}$

and

$\sum_{t=k}^{\infty}g_{e,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}<\Biggl{(}-\cosh(\alpha)+\dfrac{4\sqrt{2}\sinh(\alpha)}{\alpha}\sqrt{k+1}+\dfrac{66}{25}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k}.$

∎

######  Definition 6.8.

For all $k\geq 1$, define

$\begin{split}L_{3}(k)&:=\Biggl{(}\dfrac{19}{10}\alpha\sinh(\alpha)-\dfrac{109}{10}\cosh(\alpha)\sqrt{k+1}-\dfrac{23}{10}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k+1}\\\
\hskip 56.9055pt&\\\
U_{3}(k)&:=\Biggl{(}2\alpha\sinh(\alpha)+\dfrac{33}{10}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k+1}.\end{split}$

######  Lemma 6.9.

Let $L_{3}(k)$ and $U_{3}(k)$ be as in Definition 6.8. Let $g_{o,1}(t)$ be as
in Definition 3.14. Then for all $n,k\in\mathbb{Z}_{\geq 1}$,

$L_{3}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k+1}<\sum_{t=k}^{\infty}g_{o,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}<U_{3}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k+1}.$
(6.12)

  * Proof.

Define $c_{1}(t):=-\dfrac{6}{\pi}(-1)^{t}\binom{-\frac{3}{2}}{t}$. From (3.37)
and (5.3), it follows that for $t\geq 2$,

$\begin{split}\dfrac{6}{\pi}\alpha\sinh(\alpha)-\dfrac{6}{\pi}\cosh(\alpha)(-1)^{t}\binom{-\frac{3}{2}}{t}-\dfrac{12\cdot
6}{25\cdot\pi}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t}&<(\sqrt{24})^{2t+1}g_{o,1}(t)=c_{1}(t)\Biggl{(}1+\dfrac{S_{3}(t)}{\binom{-\frac{3}{2}}{t}}\Biggr{)}\\\
<\dfrac{6}{\pi}\alpha\sinh(\alpha)-\dfrac{6}{\pi}\cosh(\alpha)(-1)^{t}\binom{-\frac{3}{2}}{t}&+\dfrac{71\cdot
6}{100\cdot\pi}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t}.\end{split}$ (6.13)

A numerical check confirms that (6.13) also holds for $t=1$; see (3.37). Now,
applying (6.1), (6.2) with $s=1$, and (6.7) to (6.13), it follows that for all
$k\geq 1$,

$\begin{split}\sum_{t=k}^{\infty}g_{o,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}&>\Biggl{(}\dfrac{6}{\pi}\alpha\sinh(\alpha)-\dfrac{6\cdot
4\sqrt{2}}{\pi}\cosh(\alpha)\sqrt{k+1}-\dfrac{12\cdot 6\cdot 12}{25\cdot
5\cdot\pi}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}\\\
&>\Biggl{(}\dfrac{19}{10}\alpha\sinh(\alpha)-\dfrac{109}{10}\cosh(\alpha)\sqrt{k+1}-\dfrac{23}{10}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}\end{split}$

and

$\begin{split}\sum_{t=k}^{\infty}g_{o,1}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}&<\Biggl{(}\dfrac{6\cdot
24}{23\cdot\pi}\alpha\sinh(\alpha)+\dfrac{71\cdot 6\cdot 12}{100\cdot
5\cdot\pi}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}\\\
&<\Biggl{(}2\alpha\sinh(\alpha)+\dfrac{33}{10}\dfrac{1}{\sqrt{k+1}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}.\end{split}$

∎

######  Definition 6.10.

For all $k\geq 1$, define

$\begin{split}L_{4}(k)&:=\Biggl{(}\dfrac{1}{4}\dfrac{\cosh(\alpha)}{\sqrt{k+1}}-\dfrac{11}{20}\alpha\sinh(\alpha)-\dfrac{41}{50}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k+1}\\\
\hskip 56.9055pt&\\\
U_{4}(k)&:=\Biggl{(}\dfrac{63}{100}\dfrac{\cosh(\alpha)}{\sqrt{k+1}}-\dfrac{13}{25}\alpha\sinh(\alpha)+\dfrac{21}{50}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24}}\Bigr{)}^{2k+1}.\end{split}$

######  Lemma 6.11.

Let $L_{4}(k)$ and $U_{4}(k)$ be as in Definition 6.10. Let $g_{o,2}$ be as in
Definition 3.16. Then for all $n,k\in\mathbb{Z}_{\geq 1}$,

$L_{4}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k+1}<\sum_{t=k}^{\infty}g_{o,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}<U_{4}(k)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2k+1}.$
(6.14)

  * Proof.

Define $c_{2}(t):=-\dfrac{\pi}{6}(-1)^{t}\binom{-\frac{3}{2}}{t}$. From (3.41)
and (5.4), it follows that for $t\geq 1$,

$\begin{split}\dfrac{\pi}{6\cdot
2}\cosh(\alpha)\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t}-\dfrac{\pi}{6}\alpha\sinh(\alpha)-\dfrac{13\cdot\pi}{20\cdot
6}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{2}}&<(\sqrt{24})^{2t+1}g_{o,2}(t)=c_{2}(t)\dfrac{S_{4}(t)}{(-1)^{t}\binom{-\frac{3}{2}}{t}}\\\
<\dfrac{\pi}{6\cdot
2}\cosh(\alpha)\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t}-\dfrac{\pi}{6}\alpha\sinh(\alpha)&+\dfrac{\pi}{6\cdot
3}\dfrac{(-1)^{t}\binom{-\frac{3}{2}}{t}}{t^{2}}.\end{split}$ (6.15)

Now, applying (6.1) and (6.2) with $s=1\ \text{and}\ 2$, respectively, to
(6.13), it follows that for all $k\geq 1$,

$\begin{split}\sum_{t=k}^{\infty}g_{o,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}&>\Biggl{(}\dfrac{\pi}{12}\dfrac{\cosh(\alpha)}{\sqrt{k+1}}-\dfrac{24\cdot\pi}{23\cdot
6}\alpha\sinh(\alpha)-\dfrac{13\cdot 12\cdot\pi}{20\cdot 6\cdot
5}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}\\\
&>\Biggl{(}\dfrac{1}{4}\dfrac{\cosh(\alpha)}{\sqrt{k+1}}-\dfrac{11}{20}\alpha\sinh(\alpha)-\dfrac{41}{50}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}\end{split}$

and

$\begin{split}\sum_{t=k}^{\infty}g_{o,2}(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}&<\Biggl{(}\dfrac{12\cdot\pi}{12\cdot
5}\dfrac{\cosh(\alpha)}{\sqrt{k+1}}-\dfrac{\pi}{6}\alpha\sinh(\alpha)+\dfrac{12\cdot\pi}{6\cdot
3\cdot
5}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}\\\
&<\Biggl{(}\dfrac{63}{100}\dfrac{\cosh(\alpha)}{\sqrt{k+1}}-\dfrac{13}{25}\alpha\sinh(\alpha)+\dfrac{21}{50}\dfrac{1}{(k+1)^{3/2}}\Biggr{)}\Bigl{(}\dfrac{1}{\sqrt{24n}}\Bigr{)}^{2k+1}.\end{split}$

∎

######  Definition 6.12.

For $k\geq 1$, define

$\widehat{L}_{2}(k):=\dfrac{1}{\alpha^{k}}\dfrac{1}{\sqrt{24}^{k}}\Bigl{(}1-\dfrac{1}{4\sqrt{n}}\Bigr{)}\
\ \ \
\widehat{U}_{2}(k):=\dfrac{1}{\alpha^{k}}\dfrac{1}{\sqrt{24}^{k}}\Bigl{(}1+\dfrac{k}{3n}\Bigr{)}.$

######  Definition 6.13.

For $k\geq 1$, define

$n_{0}(k):=\dfrac{k+2}{24}.$

######  Lemma 6.14.

Let $\widehat{L}_{2}(k)$, and $\widehat{U}_{2}(k)$ be as in Definition 6.12.
Let $n_{0}(k)$ be as in Definition 6.13. Then for all $k\in\mathbb{Z}_{\geq
1}$ and $n>n_{0}(k)$,

$\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{\widehat{L}_{2}(k)}{\sqrt{n}^{k}}<\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\frac{1}{\mu(n)^{k}}<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{\widehat{U}_{2}(k)}{\sqrt{n}^{k}}.$
(6.16)

  * Proof.

Define

$\mathcal{E}(n,k):=\dfrac{\sqrt{12}\ e^{\mu(n)}}{24n-1}\frac{1}{\mu(n)^{k}},\
\mathcal{U}(n,k)=\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{1}{\sqrt{n}^{k}}$

and

$\mathcal{Q}(n,k):=\dfrac{\mathcal{E}(n,k)}{\mathcal{U}(n,k)}=\mathcal{Q}(n,k)=\dfrac{e^{\pi\sqrt{\frac{2n}{3}}\Bigl{(}\sqrt{1-\frac{1}{24n}}-1\Bigr{)}}}{\alpha^{k}}\dfrac{1}{\sqrt{24}^{k}}\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}.$

Using (4.2) with $(m,n,s)\mapsto(1,24n,1)$, we obtain for all $n\geq 1$,

$-\dfrac{1}{12n}<\sqrt{1-\dfrac{1}{24n}}-1=\sum_{m=1}^{\infty}\binom{1/2}{m}\dfrac{(-1)^{m}}{(24n)^{m}}<0,$

which implies that for $n\geq 1$,

$\Bigl{(}1-\dfrac{1}{4\sqrt{n}}\Bigr{)}<e^{-\frac{\pi}{12}\sqrt{\frac{2}{3n}}}<e^{\pi\sqrt{\frac{2n}{3}}\Bigl{(}\sqrt{1-\frac{1}{24n}}-1\Bigr{)}}<1.$
(6.17)

Hence

$\dfrac{1}{(\alpha\cdot\sqrt{24})^{k}}\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}\Bigl{(}1-\dfrac{1}{4\sqrt{n}}\Bigr{)}<\mathcal{Q}(n,k)<\dfrac{1}{(\alpha\cdot\sqrt{24})^{k}}\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}.$
(6.18)

In order to estimate $\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}$, we
need to split into two cases depending on $k$ is even or odd.

For $k=2\ell$ with $\ell\in\mathbb{Z}_{\geq 0}$:

$\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}=\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-(\ell+1)}=1+\sum_{j=1}^{\infty}\binom{-(\ell+1)}{j}\dfrac{(-1)^{j}}{(24n)^{j}}.$

From (4.3) with $(m,s,n)\mapsto(1,\ell+1,24n)$, for all $n>\frac{\ell+1}{12}$,
we get

$0<\sum_{j=1}^{\infty}\binom{-(\ell+1)}{j}\dfrac{(-1)^{j}}{(24n)^{j}}<\beta_{1,24n}(\ell+1)=\dfrac{\ell+1}{12n},$

which is equivalent to

$1<\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}<1+\dfrac{k+2}{24n}\ \
\text{for all}\ \ n>\frac{k+2}{24}.$ (6.19)

For $k=2\ell+1$ with $\ell\in\mathbb{Z}_{\geq 0}$:

$\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}=\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{2\ell+3}{2}}=1+\sum_{j=1}^{\infty}\binom{-\frac{2\ell+3}{2}}{j}\dfrac{(-1)^{j}}{(24n)^{j}}.$

Using (4.1) with $(m,s,n)\mapsto(1,\ell+2,24n)$, for all
$n>\frac{\ell+2}{24}$, we get

$0<\sum_{j=1}^{\infty}\binom{-\frac{2\ell+3}{2}}{j}\dfrac{(-1)^{j}}{(24n)^{j}}<b_{1,24n}(\ell+2)=\dfrac{\ell+2}{6n}$

which is equivalent to

$1<\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}<1+\dfrac{k+3}{12n}\leq
1+\dfrac{k}{3n}\ \ \text{for all}\ \ n>\frac{k+3}{48}.$ (6.20)

From (6.19) and (6.20), for all $n>\frac{k+2}{24}$ it follows that

$1<\Bigl{(}1-\dfrac{1}{24n}\Bigr{)}^{-\frac{k+2}{2}}<1+\dfrac{k+3}{12n}\leq
1+\dfrac{k}{3n}.$ (6.21)

Combining (6.18) and (6.21) concludes the proof.

∎

##  7\. Main Theorem

######  Definition 7.1.

For $w\in\mathbb{Z}_{\geq 1}$, define

$\bigl{(}\gamma_{0}(w),\gamma_{1}(w)\bigr{)}:=\begin{cases}(23,24),\quad\text{if
$w$ is even}\\\ (15,17),\quad\text{if $w$ is odd}\end{cases}.$

######  Definition 7.2.

Let $\gamma_{0}(w)$ and $\gamma_{1}(w)\bigr{)}$ be as in Definition 7.1. Then
for all $w\in\mathbb{Z}_{\geq 1}$, define

$L(w):=-\gamma_{0}(w)\dfrac{\sqrt{\lceil w/2\rceil+1}}{\sqrt{24}^{w}}\ \
\text{and}\ \ U(w):=\gamma_{1}(w)\dfrac{\sqrt{\lceil
w/2\rceil+1}}{\sqrt{24}^{w}}.$

######  Lemma 7.3.

Let $\widehat{g}(k)$ be as in Theorem 2.1 and $n_{0}(k)$ as in Definition
6.13. Let $g(t)$ be as in (3.47). Let $L(w)$ and $U(w)$ be as in Definition
7.2. If $m\in\mathbb{Z}_{\geq 1}$ and
$n>\max\bigl{\\{}1,n_{0}(2m),\widehat{g}(2m)\bigr{\\}}$, then

$\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{L(2m)}{\sqrt{n}^{2m}}\Biggr{)}<p(n)<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{U(2m)}{\sqrt{n}^{2m}}\Biggr{)}.$

  * Proof.

Recalling Definition 3.18, from Lemma 3.19, we have

$\displaystyle\sum_{t=0}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\sum_{t=2m}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\sum_{t=m}^{\infty}g(2t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}+\sum_{t=m}^{\infty}g(2t+1)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\sum_{t=m}^{\infty}(g_{e,1}(t)+g_{e,2}(t))\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}+\sum_{t=m}^{\infty}(g_{o,1}(t)+g_{o,2}(t))\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}.$

Using Lemmas 6.5-6.11 by assigning $k\mapsto m$, it follows that

$\displaystyle\dfrac{L_{1}(m)+L_{2}(m)}{\sqrt{n}^{2m}}+\dfrac{L_{3}(m)+L_{4}(m)}{\sqrt{n}^{2m+1}}$
$\displaystyle<$
$\displaystyle\sum_{t=2m}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}<\dfrac{U_{1}(m)+U_{2}(m)}{\sqrt{n}^{2m}}+\dfrac{U_{3}(m)+U_{4}(m)}{\sqrt{n}^{2m+1}}.$

Moreover, by Lemma 6.14 with $k=2m$, it follows that

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\frac{1}{\mu(n)^{2m}}<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{\widehat{U}_{2}(2m)}{\sqrt{n}^{2m}}.$
(7.3)

Finally, from (Proof.) and (7.3) along with the fact that
$U_{3}(m)+U_{4}(m)>0$, we obtain

$\begin{split}\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}+\dfrac{1}{\mu(n)^{2m}}\Biggr{)}&<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{\sum_{i=1}^{4}U_{i}(m)+\widehat{U}_{2}(2m)}{\sqrt{n}^{2m}}\Biggr{)}.\end{split}$
(7.4)

Since for all $m\geq 1$, $L_{3}(m)+L_{4}(m)<0$, it follows that

$\begin{split}\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}-\dfrac{1}{\mu(n)^{2m}}\Biggr{)}&>\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{\sum_{i=1}^{4}L_{i}(m)-\widehat{U}_{2}(2m)}{\sqrt{n}^{2m}}\Biggr{)}.\end{split}$
(7.5)

From Lemmas 6.5-6.11 and 6.14, for all $n\geq\max\\{1,n_{0}(2m)\\}$,

$\displaystyle\sum_{i=1}^{4}U_{i}(m)+\widehat{U}_{2}(2m)$ $\displaystyle<$
$\displaystyle\Biggl{(}4+\dfrac{4}{\sqrt{m+1}}+\dfrac{2}{(m+1)^{3/2}}+6\sqrt{m+1}+\dfrac{2m}{3\alpha^{2}n}\Biggr{)}\dfrac{1}{\sqrt{24}^{2m}}.$

For all $1\leq m\leq 10$ observe that $n_{0}(2m)<1$ and therefore,
$\dfrac{2m}{3\alpha^{2}n}<\dfrac{20}{3\alpha^{2}}<25$; whereas for $m\geq 11$,
$n_{0}(2m)>1$. Consequently,
$\dfrac{2m}{3\alpha^{2}n}<\dfrac{8m}{\alpha^{2}(m+1)}<\dfrac{8}{\alpha^{2}}<10$;
i.e., $\dfrac{2m}{3\alpha^{2}n}<25$.
Continuing our estimation

$\displaystyle\sum_{i=1}^{4}U_{i}(m)+\widehat{U}_{2}(2m)$ $\displaystyle<$
$\displaystyle\Biggl{(}29+\dfrac{4}{\sqrt{m+1}}+\dfrac{2}{(m+1)^{3/2}}+6\sqrt{m+1}\Biggr{)}\dfrac{1}{\sqrt{24}^{2m}}$
(7.6) $\displaystyle\leq$
$\displaystyle\dfrac{24\sqrt{m+1}}{\sqrt{24}^{2m}}=U(2m).$

Similarly, for all $n\geq\max\\{1,n_{0}(2m)\\}$,

$\displaystyle\sum_{i=1}^{4}L_{i}(m)-\widehat{U}_{2}(2m)$ $\displaystyle>$
$\displaystyle\Biggl{(}-29-\dfrac{4}{\sqrt{m+1}}-\dfrac{1}{2(m+1)^{3/2}}-3\sqrt{m+1}\Biggr{)}\dfrac{1}{\sqrt{24}^{2m}}$
(7.7) $\displaystyle\geq$
$\displaystyle-\dfrac{23\sqrt{m+1}}{\sqrt{24}^{2m}}=L(2m).$

Plugging (7.6) and (7.7) into (7.4) and (7.5), respectively, and applying
Theorem 2.1, we get

$p(n)<\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}+\dfrac{1}{\mu(n)^{2m}}\Biggr{)}<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{U(2m)}{\sqrt{n}^{2m}}\Biggr{)}$
(7.8)

and

$p(n)>\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}-\dfrac{1}{\mu(n)^{2m}}\Biggr{)}>\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{L(2m)}{\sqrt{n}^{2m}}\Biggr{)}.$
(7.9)

∎

######  Lemma 7.4.

Let $\widehat{g}(k)$ be as in Theorem 2.1 and $n_{0}(k)$ as in Definition
6.13. Let $g(t)$ be as in Equation (3.47). Let $L(w)$ and $U(w)$ be as in
Definition 7.2. If $m\in\mathbb{Z}_{\geq 0}$ and
$n>\max\bigl{\\{}1,n_{0}(2m+1),\widehat{g}(2m+1)\bigr{\\}}$, then

$\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{L(2m+1)}{\sqrt{n}^{2m+1}}\Biggr{)}<p(n)<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{U(2m+1)}{\sqrt{n}^{2m+1}}\Biggr{)}.$

  * Proof.

Recalling Definition 3.18, by Lemma 3.19 we have

$\displaystyle\sum_{t=0}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\sum_{t=2m+1}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\sum_{t=m}^{\infty}g(2t+1)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}+\sum_{t=m+1}^{\infty}g(2t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}$
$\displaystyle=$
$\displaystyle\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\sum_{t=m}^{\infty}(g_{o,1}(t)+g_{o,2}(t))\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t+1}+\sum_{t=m+1}^{\infty}(g_{e,1}(t)+g_{e,2}(t))\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{2t}.$

Using Lemmas 6.5-6.7 by substituting $k\mapsto m+1$ and Lemmas 6.9-6.11 by
substituting $k\mapsto m$, it follows that

$\displaystyle\dfrac{L_{1}(m+1)+L_{2}(m+1)}{\sqrt{n}^{2m+2}}+\dfrac{L_{3}(m)+L_{4}(m)}{\sqrt{n}^{2m+1}}$
$\displaystyle<$
$\displaystyle\sum_{t=2m+1}^{\infty}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}$
$\displaystyle<$
$\displaystyle\dfrac{U_{1}(m+1)+U_{2}(m+1)}{\sqrt{n}^{2m+2}}+\dfrac{U_{3}(m)+U_{4}(m)}{\sqrt{n}^{2m+1}}.$

By Lemma 6.14 with $k=2m+1$,

$\dfrac{\sqrt{12}\
e^{\mu(n)}}{24n-1}\frac{1}{\mu(n)^{2m+1}}<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\frac{\widehat{U}_{2}(2m+1)}{\sqrt{n}^{2m+1}}.$
(7.12)

From (Proof.) and (7.12) along with the fact that $U_{1}(m)+U_{2}(m)>0$, we
obtain

$\begin{split}\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}+\dfrac{1}{\mu(n)^{2m+1}}\Biggr{)}&<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{\widehat{U}(2m+1)}{\sqrt{n}^{2m+1}}\Biggr{)}\end{split},$
(7.13)

where

$\widehat{U}(2m+1)=U_{1}(m+1)+U_{2}(m+1)+U_{3}(m)+U_{4}(m)+\widehat{U}_{2}(2m+1).$

Since for all $m\geq 0$, $L_{1}(m)+L_{2}(m)<0$, it follows that

$\begin{split}\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}-\dfrac{1}{\mu(n)^{2m+1}}\Biggr{)}&>\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{\widehat{L}(2m+1)}{\sqrt{n}^{2m+1}}\Biggr{)}\end{split}$
(7.14)

with

$\widehat{L}(2m+1)=L_{1}(m+1)+L_{2}(m+1)+L_{3}(m)+L_{4}(m)-\widehat{U}_{2}(2m+1).$

Next, we estimate $\widehat{U}_{2}(2m+1)$. Recall from Lemma 6.14 that for all
$n>n_{0}(2m+1)$,

$\widehat{U}_{2}(2m+1)<\dfrac{1}{\alpha^{2m+1}}\Bigl{(}1+\dfrac{2m+1}{3n}\Bigr{)}\dfrac{1}{\sqrt{24}^{2m+1}}<\dfrac{1}{\alpha}\Bigl{(}1+\dfrac{2m+1}{3n}\Bigr{)}\dfrac{1}{\sqrt{24}^{2m+1}}.$

We note that for $0\leq m\leq 10$; $n\geq 1>n_{0}(2m+1)$, and therefore,
$\dfrac{1}{\alpha}\Bigl{(}1+\dfrac{2m+1}{3n}\Bigr{)}<\dfrac{8}{\alpha}$;
whereas for $m\geq 11$, $n>\dfrac{2m+3}{24}$. This implies that
$\dfrac{1}{\alpha}\Bigl{(}1+\dfrac{2m+1}{3n}\Bigr{)}<\dfrac{9}{\alpha}$.
Hence, for all $n\geq\max\\{1,n_{0}(2m+1)\\}$,

$\widehat{U}_{2}(2m+1)<\dfrac{9}{\alpha}\dfrac{1}{\sqrt{24}^{2m+1}}.$ (7.15)

From Lemmas 6.5-6.11 and 6.14, for all $n\geq\max\\{1,n_{0}(2m+1)\\}$, we get

$\displaystyle\widehat{U}(2m+1)$ $\displaystyle<$
$\displaystyle\Biggl{(}18+\dfrac{5}{\sqrt{m+1}}+\dfrac{1}{(m+1)^{3/2}}+2\sqrt{m+2}\Biggr{)}\dfrac{1}{\sqrt{24}^{2m+1}}$
(7.16) $\displaystyle\leq$
$\displaystyle\dfrac{17\sqrt{m+2}}{\sqrt{24}^{2m+1}}=U(2m+1).$

Similarly for all $n\geq\max\\{1,n_{0}(2m+1)\\}$, it follows that

$\displaystyle\widehat{L}(2m+1)$ $\displaystyle>$
$\displaystyle\Biggl{(}-17-\dfrac{3}{\sqrt{m+1}}-\dfrac{1}{(m+1)^{3/2}}-13\sqrt{m+2}\Biggr{)}\dfrac{1}{\sqrt{24}^{2m+1}}$
(7.17) $\displaystyle\geq$
$\displaystyle-\dfrac{15\sqrt{m+2}}{\sqrt{24}^{2m+1}}=L(2m+1).$

Plugging (7.16) and (7.17) into (7.13) and (7.14), respectively, and applying
Theorem 2.1, we get

$p(n)<\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}+\dfrac{1}{\mu(n)^{2m+1}}\Biggr{)}<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{U(2m+1)}{\sqrt{n}^{2m+1}}\Biggr{)}$
(7.18)

and

$p(n)>\dfrac{\sqrt{12}e^{\mu(n)}}{24n-1}\Biggl{(}1-\dfrac{1}{\mu(n)}-\dfrac{1}{\mu(n)^{2m+1}}\Biggr{)}>\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{2m}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{L(2m+1)}{\sqrt{n}^{2m+1}}\Biggr{)}.$
(7.19)

∎

######  Theorem 7.5.

Let $\widehat{g}(k)$ be as in Theorem 2.1 and $g(t)$ as in (3.47). Let $L(w)$
and $U(w)$ be as in Definition 7.2. If $w\in\mathbb{Z}_{\geq 1}$ with $\lceil
w/2\rceil\geq 1$ and $n>\widehat{g}(w)$, then

$\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{w-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{L(w)}{\sqrt{n}^{w}}\Biggr{)}<p(n)<\frac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{w-1}g(t)\Bigl{(}\dfrac{1}{\sqrt{n}}\Bigr{)}^{t}+\dfrac{U(w)}{\sqrt{n}^{w}}\Biggr{)}.$
(7.20)

  * Proof.

Combining Lemmas 7.3 and 7.4 together with the fact that
$\widehat{g}(k)>\max\bigl{\\{}n_{0}(k),1\bigr{\\}}$, we arrive at (7.20). ∎

######  Corollary 7.6.

For all $n\geq 116$, we have

$\dfrac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{3}\dfrac{g(t)}{\sqrt{n}^{t}}-\dfrac{1}{14n^{2}}\Biggr{)}<p(n)<\dfrac{e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}}\Biggl{(}\sum_{t=0}^{3}\dfrac{g(t)}{\sqrt{n}^{t}}+\dfrac{1}{13n^{2}}\Biggr{)},$
(7.21)

where

$g(0)=1,\ g(1)=-\dfrac{\pi^{2}+72}{24\sqrt{6}\pi},\
g(2)=\dfrac{\pi^{2}+432}{6912},\
g(3)=-\dfrac{\pi^{4}+1296\pi^{2}+93312}{497664\sqrt{6}\pi}.$

  * Proof.

Plugging $w=4$ into (7.20), we obtain the inequality (7.21). ∎

######  Remark 7.7.

Corollary 7.6 provides an answer to the Question 1.2, asked by Chen. As a
consequence from (7.21), one can derive that $p(n)$ is $\log$-concave for all
$n\geq 26$.

##  8\. Appendix

###  8.1. Proofs of the lemmas presented in Section 4.

_Proof of Lemma 4.1:_ For $n=1$ we have to prove

$\frac{1-x_{1}}{1+y_{1}}\geq 1-x_{1}-y_{1},$

which is equivalent to

$1-x_{1}\geq(1+y_{1})(1-x_{1}-y_{1})\geq
1-x_{1}-y_{1}^{2}-x_{1}y_{1}\Leftrightarrow 0\geq-y_{1}^{2}-x_{1}y_{1}.$

This is always true because $x_{1},y_{1}$ are non-negative real numbers. Now
assume by induction that the statement is true for $n=N$. Next we prove the
statement for $n=N+1$. For $n=N$, we have $P\geq(1-S)$ with

$P:=\frac{(1-x_{1})(1-x_{2})\cdots(1-x_{N})}{(1+y_{1})(1+y_{2})\cdots(1+y_{N})}\
\ \text{and}\ \ S:=\sum_{j=1}^{N}x_{j}+\sum_{j=1}^{N}y_{j}.$

This implies that

$P\frac{1-x_{N+1}}{1+y_{N+1}}\geq(1-S)\frac{1-x_{N+1}}{1+y_{N+1}}.$

Therefore it suffices to prove that

$(1-S)\frac{1-x_{N+1}}{1+y_{N+1}}\geq 1-S-y_{N+1}-x_{N+1},$

which is equivalent to

$(1-S)(1-x_{N+1})\geq(1-S-y_{N+1}-x_{N+1})(1+y_{N+1}).$

Equivalently,

$1-x_{N+1}-S+S\cdot x_{N+1}\geq 1-S-x_{N+1}-y^{2}_{N+1}-x_{N+1}y_{N+1}-S\cdot
y_{N+1},$

which amounts to say that

$S\cdot x_{N+1}\geq-y^{2}_{N+1}-x_{N+1}y_{N+1}-S\cdot y_{N+1},$

and this inequality holds because $x_{N+1},y_{N+1},S\geq 0$. ∎

_Proof of Lemma 4.2:_ Expanding the quotient
$\frac{(-1)^{i}(-t)_{i}}{(t)_{i}}$ as

$(-1)^{i}\frac{(-t)_{i}}{(t)_{i}}=(-1)^{i}\prod_{j=1}^{i}\frac{-t+j-1}{t+j-1}=\prod_{j=1}^{i}\frac{t-j+1}{t+j-1},$

we obtain

$\frac{t(-t)_{u}(-1)^{u}}{(1+2t)(t+u)(t)_{u}}=\frac{t}{2(t+\frac{1}{2})(t+u)}\prod_{j=1}^{u}\frac{t-(j-1)}{t+j-1}=\frac{1}{2t(1+\frac{1}{2t})(1+\frac{u}{t})}\prod_{j=1}^{u}\frac{1-\frac{j-1}{t}}{1+\frac{j-1}{t}}.$

Since $t\geq 1$ and $u<t$, it is clear that

$\frac{1}{2t(1+\frac{1}{2t})(1+\frac{u}{t})}\prod_{j=1}^{u}\frac{1-\frac{j-1}{t}}{1+\frac{j-1}{t}}\leq\dfrac{1}{2t}.$
(8.1)

By Lemma 4.1, it follows that

$\frac{1}{2t(1+\frac{1}{2t})(1+\frac{u}{t})}\prod_{j=1}^{u}\frac{1-\frac{j-1}{t}}{1+\frac{j-1}{t}}\geq\frac{1}{2t}\Bigl{(}1-\frac{\frac{1}{2}+u+2\sum_{j=1}^{u}(j-1)}{t}\Bigr{)}=\frac{1}{2t}\Bigl{(}1-\frac{u^{2}+\frac{1}{2}}{t}\Bigr{)}.$
(8.2)

Combining (8.1) and (8.2) concludes the proof. ∎

_Proof of Lemma 4.3:_ By Lemma 4.1,

$\frac{1}{2t}\geq\frac{1}{1+2t}=\frac{1}{2t}\frac{1}{(1+\frac{1}{2t})}\geq\frac{1}{2t}\Bigl{(}1-\frac{1}{2t}\Bigr{)}\geq\frac{1}{2t}-\frac{1}{4t^{2}}.$
(8.3)

Now

$\frac{2t\sum_{i=1}^{u}\frac{(-t)_{i}(-1)^{i}}{(t+i)(t)_{i}}}{1+2t}=\frac{1}{1+\frac{1}{2t}}\sum_{i=1}^{u}\frac{1}{t+i}\prod_{j=1}^{i}\frac{t-j+1}{t+j-1}=\frac{1}{t}\frac{1}{1+\frac{1}{2t}}\sum_{i=1}^{u}\frac{1}{1+\frac{i}{t}}\prod_{j=1}^{i}\frac{1-\frac{j-1}{t}}{1+\frac{j-1}{t}}.$

As $t\geq 1$ and $u<t$, it directly follows that

$\frac{1}{t}\frac{1}{1+\frac{1}{2t}}\sum_{i=1}^{u}\frac{1}{1+\frac{i}{t}}\prod_{j=1}^{i}\frac{1-\frac{j-1}{t}}{1+\frac{j-1}{t}}\leq\dfrac{u}{t}.$
(8.4)

Applying Lemma 4.1, we obtain

$\frac{1}{t}\frac{1}{1+\frac{1}{2t}}\sum_{i=1}^{u}\frac{1}{1+\frac{i}{t}}\prod_{j=1}^{i}\frac{1-\frac{j-1}{t}}{1+\frac{j-1}{t}}\geq\frac{1}{t}\sum_{i=1}^{u}1-\frac{\frac{1}{2}+i+2\sum_{j=1}^{i}(j-1)}{t}=\frac{u}{t}-\frac{u(2u^{2}+3u+4)}{6t^{2}}.$
(8.5)

Finally, (8.3), (8.4), and (8.5) imply the desired inequality. ∎

_Proof of Lemma 4.5:_ Let $n\geq u$ be fixed. We have to show that $b_{n}\geq
a_{n}$. First we note that

$a_{k+1}-a_{n}=\sum_{j=n}^{k}(a_{j+1}-a_{j})\geq\sum_{j=n}^{k}(b_{j+1}-b_{j})=b_{k+1}-b_{n}.$

Consequently, for all $k\geq n$ we have

$a_{k+1}-a_{n}\geq b_{k+1}-b_{n}\Leftrightarrow b_{n}-b_{k+1}\geq
a_{n}-a_{k+1}.$

This implies that

$b_{n}=\lim_{k\to\infty}(b_{n}-b_{k+1})\geq\lim_{k\to\infty}(a_{n}-a_{k+1})=a_{n}.$

∎

_Proof of Lemma 4.6:_ We apply Lemma 4.5 with
$a_{n}=\sum_{u=n+1}^{\infty}\frac{u^{k}\alpha^{2u}}{(2u)!}$ and
$b_{n}=\frac{C_{k}}{n^{2}}$:

$a_{n+1}-a_{n}=-\frac{(n+1)^{k}\alpha^{2n+2}}{(2n+2)!}\ \ \text{and}\ \
b_{n+1}-b_{n}=-\frac{C_{k}(2n+1)}{n^{2}(n+1)^{2}}.$

Therefore $b_{n+1}-b_{n}\leq a_{n+1}-a_{n}$ is equivalent to

$\frac{(n+1)^{k}\alpha^{2n+2}}{(2n+2)!}\leq\frac{C_{k}(2n+1)}{n^{2}(n+1)^{2}}\Leftrightarrow
f(n):=\frac{n^{2}(n+1)^{k+2}\alpha^{2n+2}}{(2n+1)(2n+2)!}\leq C_{k}.$

In order to prove $f(n)\leq C_{k}$, it suffices to prove $f(m)\leq C_{k}$,
where $m$ is such that $f(m)$ is maximal. Hence in order to find such a $m$,
we find the first $m$ such that $f(m+1)\leq f(m)$. This is equivalent to
finding $\frac{f(m+1)}{f(m)}\leq 1$, also as we will see there is only one
such maximum. Then $\underset{n\in\mathbb{N}}{\max}\ f(n)=f(m)$. Now

$\frac{f(n+1)}{f(n)}=\frac{(n+1)^{2}(n+2)^{k+2}\alpha^{2n+4}}{(2n+3)(2n+4)!}\frac{(2n+1)(2n+2)!}{n^{2}(n+1)^{k+2}\alpha^{2n+2}}=\frac{\alpha^{2}(n+2)^{k+2}(2n+1)}{(2n+4)(2n+3)^{2}(n+1)^{k}n^{2}}.$

Using Mathematica’s implementation of Cylindrical Algebraic Decomposition
[10], we obtain that

$\frac{\alpha^{2}(n+2)^{k+2}(2n+1)}{(2n+4)(2n+3)^{2}(n+1)^{k}n^{2}}\leq
1,\text{ for all $\alpha^{2}\leq\frac{800}{729}$.}$

As $\alpha^{2}=\frac{\pi^{2}}{36}<\frac{800}{729}$,
$\underset{n\in\mathbb{N}}{\max}\ f(n)=f(1)$; i.e., $f(n)\leq f(1)=C_{k}$.

∎

###  8.2. The Sigma simplification of $S_{3}(t,u)$ in Lemma 5.3

Using the symbolic summation package Sigma [27] and its underlying machinery
in the setting of difference rings [29] the inner sum $S_{3}(t,u)$ can be
simplified as follows. Recall from (5.2) that

$S_{3}(t,u)=\sum_{s=0}^{t-u}\frac{1}{s+u}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\binom{-\frac{3}{2}}{t-s-u}\frac{(-s-u)_{u}}{(s+2u)!}.$

After loading Sigma into the computer algebra system Mathematica

In[1]:=$\tt\bf\displaystyle<<Sigma.m$

Sigma - A summation package by Carsten Schneider © RISC-JKU

we input the sum under consideration

In[2]:=$\tt\bf\displaystyle
mySum3=\sum_{s=0}^{t-u}\frac{1}{s+u}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\binom{-\frac{3}{2}}{t-s-u}\frac{(-s-u)_{u}}{(s+2u)!};$

and compute a recurrence of it by executing

In[3]:=$\tt\bf\displaystyle rec3=GenerateRecurrence[mySum3]$

Out[3]=
$\displaystyle\tt(t-u)uSUM[u]+2(2+t)(1+u)SUM[u+1]+(2+u)(2+t+u)SUM[u+2]==0$

As a result we get a homogeneous linear recurrence of order $2$ for
$S_{3}(t,u)=\texttt{SUM[u](=mySum3)}$. Internally, Zeilberger’s creative
telescoping paradigm [23] is applied which not only provides a recurrence but
delivers simultaneously a proof certificate that guarantees the correctness of
the result.

Verification of the recurrence. Denote the summand of $S_{3}(t,u)$ by
$f(t,u,s)$; i.e. set

$f(t,u,s)=\frac{1}{s+u}\Bigl{(}\frac{1}{2}-s-u\Bigr{)}_{s+u+1}\binom{-\frac{3}{2}}{t-s-u}\frac{(-s-u)_{u}}{(s+2u)!}.$

Then one can verify that the polynomials $a_{0}(t,u)=u(t-u)$,
$a_{1}(t,u)=2(2+t)(1+u)$ and $a_{2}(t,u)=(2+u)(2+t+u)$ (free of the summation
variable $s$) and the expression

$g(t,u,s)=-\frac{\gamma(t,u,s)s\binom{-\frac{3}{2}}{-s+t-u}(-s-u)_{u}\big{(}\frac{1}{2}-s-u\big{)}_{1+s+u}}{(s+2u)!(s+u)(1+s+2u)(2+s+2u)(3+s+2u)(-1+2s-2t+2u)}$

with

$\displaystyle\gamma(t,u,s)=$
$\displaystyle-6s-6s^{2}+16s^{3}+8s^{4}+6t-6st-46s^{2}t-20s^{3}t+12t^{2}+30st^{2}+12s^{2}t^{2}$
$\displaystyle-12u-22su+66s^{2}u+64s^{3}u+8s^{4}u-7tu-138stu-126s^{2}tu-16s^{3}tu$
$\displaystyle+52t^{2}u+57st^{2}u+8s^{2}t^{2}u-27u^{2}+88su^{2}+172s^{2}u^{2}+44s^{3}u^{2}-108tu^{2}$
$\displaystyle-235stu^{2}-68s^{2}tu^{2}+57t^{2}u^{2}+24st^{2}u^{2}+32u^{3}+192su^{3}+92s^{2}u^{3}$
$\displaystyle-140tu^{3}-98stu^{3}+18t^{2}u^{3}+75u^{4}+86su^{4}-48tu^{4}+30u^{5}$

satisfy the summand recurrence

$g(t,u,s+1)-g(t,u,s)=a_{0}(t,u)f(t,u,s)+a_{1}(t,u)f(t,u+1,s)+a_{2}(t,u)f(t,u+2,s)$
(8.6)

for all $0\leq s\leq t-u$ with $t\geq u$. The components of the summand
recurrence can be obtained with the function call CreativeTelescoping[mySum3].
Summing the verified equation (8.6) over $s$ from $0$ to $t-u$ yields the
output recurrence Out[8.2], which at the same time yields a proof for the
correctness of Out[8.2].∎

We remark that Sigma’s creative telescoping approach works not only for
hypergeometric sums (here one could use, for instance, also the Paule-Schorn
implementation [22] of Zeilberger’s algorithm [23]), but can be applied in the
general setting of difference rings which allows to treat summands built by
indefinite nested sums and products. More involved examples in the context of
plane partitions can be found, e.g., in [2].

We are now in the position to solve the output reccurrence Out[8.2] with the
function call

In[4]:=$\tt\bf\displaystyle recSol=SolveRecurrence[rec3,SUM[u]]$

Out[4]=
$\displaystyle\tt\bigg{\\{}\big{\\{}0,(-1)^{u}\big{\\}},\big{\\{}0,\frac{(2+t-u)}{u(2+t+u)}\frac{(-t)_{u}}{(2+t)_{u}}+2(-1)^{u}\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(2+i+t)(2+t)_{i}}\big{\\}},\big{\\{}1,0\big{\\}}\bigg{\\}}$

This means that we found two linearly independent solutions (the list entries
whose first entry is a zero) that span the full solution space, i.e., the
general solution to Out[8.2] is

$G(t,u)=c_{1}(t)\,(-1)^{u}+c_{2}\Big{(}\frac{(2+t-u)}{u(2+t+u)}\frac{(-t)_{u}}{(2+t)_{u}}+2(-1)^{u}\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(2+i+t)(2+t)_{i}}\Big{)}$
(8.7)

where the $c_{1},c_{2}$ are constants being free of $u$. For further details
on the underlying machinery (inspired by [23]) we refer to [29].

Verification of the general solution. The correctness of the solutions can be
verified by plugging them into the recurrence Out[8.2] and applying
(iteratively) the shift relations

$\displaystyle(-1)^{u+1}$ $\displaystyle=-(-1)^{u},$ $\displaystyle(-t)_{1+u}$
$\displaystyle=(-t+u)(-t)_{u},$ $\displaystyle(2+t)_{1+u}$
$\displaystyle=(2+t+u)(2+t)_{u},$
$\displaystyle\sum_{i=1}^{1+u}\frac{(-1)^{i}(-t)_{i}}{(2+i+t)(2+t)_{i}}$
$\displaystyle=\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(2+i+t)(2+t)_{i}}+\frac{(-1)^{u}(t-u)(-t)_{u}}{(2+t+u)(3+t+u)(2+t)_{u}}.$

Then simple rational function arithmetic shows that the obtained expression
collapses to zero.∎

Finally, we compute the first two initial values (by another round of symbolic
summation) and find that

$\begin{split}S_{3}(t,1)&=(-1)^{t}-\frac{(t+2)}{2(1+t)}\binom{-\frac{3}{2}}{t},\\\
S_{3}(t,2)&=-(-1)^{t}+\frac{\big{(}8+7t+t^{2}\big{)}}{4(1+t)(2+t)}\binom{-\frac{3}{2}}{t}.\end{split}$
(8.8)

With this information we can set
$c_{1}=-(-1)^{t}+\frac{(3t+4)}{2(t+1)(t+2)}\binom{-\frac{3}{2}}{t}$ and
$c_{2}=\frac{1}{2}\binom{-\frac{3}{2}}{t}$ so that the general solution (8.7)
agrees with $S_{3}(t,u)$ for $u=1,2$. Since $S_{3}(t,u)$ and the specialized
general solution are both solutions of the recurrence Out[8.2] and the first
two initial values agree, they are identical for all $u\geq 0$ with $u\leq t$.
This last step of combining the solutions accordingly can be accomplished by
inserting the list of two initial values

In[5]:=$\tt\bf\displaystyle
initialL=\bigg{\\{}(-1)^{t}-\frac{(t+2)}{2(1+t)}\binom{-\frac{3}{2}}{t},-(-1)^{t}+\frac{\big{(}8+7t+t^{2}\big{)}}{4(1+t)(2+t)}\binom{-\frac{3}{2}}{t}\bigg{\\}};$

and then executing the command

In[6]:=$\tt\bf\displaystyle
FindLinearCombination[recSol3,\\{1,initialL\\},u,2]$

Out[6]=
$\displaystyle\tt-(-1)^{t}(-1)^{u}+\frac{1}{2}\binom{-\frac{3}{2}}{t}\Big{(}\frac{(2+t-u)}{u(2+t+u)}\frac{(-t)_{u}}{(2+t)_{u}}+(-1)^{u}\Big{(}\frac{1}{1+t}+\frac{2}{2+t}+2\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(2+i+t)(2+t)_{i}}\Big{)}\Big{)}$

Carrying out all the steps above (including also the calculation of the
initial values) can be rather cumbersome. In order to support the user with
the simplification of such problems, the package

In[7]:=$\tt\bf\displaystyle<<EvaluateMultiSums.m$

EvaluateMultiSum by Carsten Schneider © RISC-JKU

has been developed. More precisely, by applying the command EvaluateMultiSums
to the input sum $\texttt{mySum3}(=S_{3}(t,u))$, all the above steps are
carried out automatically and one obtains in one stroke the desired result:

In[8]:=$\tt\bf\displaystyle
sol3=EvaluateMultiSum[mySum3,\\{\\},\\{u,t\\},\\{0,1\\},\\{t,\infty\\}]$

Out[8]=
$\displaystyle\tt-(-1)^{t}(-1)^{u}+\frac{1}{2}\binom{-\frac{3}{2}}{t}\Big{(}\frac{(2+t-u)}{u(2+t+u)}\frac{(-t)_{u}}{(2+t)_{u}}+(-1)^{u}\Big{(}\frac{1}{1+t}+\frac{2}{2+t}+2\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(2+i+t)(2+t)_{i}}\Big{)}\Big{)}$

Since we prefer to rewrite the found expression in terms of the Pochhammer
symbol $(t)_{u}$, we execute the final simplification step with the function
call

In[9]:=$\tt\bf\displaystyle SigmaReduce[sol3,u,Tower\to\\{(t)_{u}\\}]$

Out[9]=
$\displaystyle\tt-(-1)^{t}(-1)^{u}+\binom{-\frac{3}{2}}{t}\Big{(}\frac{t(1+2t-2u)}{2(1+2t)u(t+u)}\frac{(-t)_{u}}{(t)_{u}}+(-1)^{u}\Big{(}\frac{1}{1+2t}+\frac{2t}{1+2t}\sum_{i=1}^{u}\frac{(-1)^{i}(-t)_{i}}{(i+t)(t)_{i}}\Big{)}\Big{)}$

######  Remark 8.1.

We should mention that there is no particular reason for explaining the
details of Sigma application only for $S_{3}(t,u)$. The simplification of the
sums $S_{1}(t,u)$, $S_{2}(t,u)$, and $S_{4}(t,u)$, as in (5.2), (5.2), and
(5.2), respectively, works completely analogously.

##  9\. Concluding remarks

We conclude this paper with a list of possible future work based on the method
devised in this paper and its further applications.

  1. (1)

A prudent application of our method might lead to obtaining full asymptotic
expansion and respective error bounds for a broad class of functions; for
example: $q(n)$-partitions into distinct parts, $p^{s}(n)$-partitions into
perfect $s$th powers, $k$-colored partitions, $k$-regular partitions, Andrews’
spt-function, $\alpha(n)$-$n$th coefficient of Ramanujan’s third order mock
theta function $f(q)$, the coefficient sequence of Klein’s $j$-function, etc.

  2. (2)

More generally consider the class of Dedekind $\eta$-quotients which fit
perfectly into [9, Thm. 1.1] or [30, Thm. 1.1]. Therefore one can also obtain
a full asymptotic expansion and infinite families of inequalities for the
coefficient sequence arising from the Fourier expansion of the considered
Dedekind $\eta$-function.

  3. (3)

Theorem 7.5 can be utilized as a black box in order to prove inequalities
pertaining to the partition function by constructing an unified framework. A
major class of inequalities for $p(n)$ can be separated into the following two
categories among many others:

    1. (a)

Turán inequalities and its higher order analogues related to the real
rootedness of Jensen polynomials associated to $p(n)$, studied in [11], [7],
and [12].

    2. (b)

Linear homogeneous inequalities for $p(n)$; i.e.,

$\sum_{i=1}^{r}p(n+x_{i})\leq\sum_{i=1}^{s}p(n+y_{i}).$

For more details we refer to [14, 19].

  4. (4)

More generally, it would be interesting to design a constructive method to
decide whether for some positive integer $N$ a relation of the form

$\sum_{j=1}^{M_{1}}\prod_{i=1}^{T_{1}}p(n+s^{(j)}_{i})\leq\sum_{j=1}^{M_{2}}\prod_{i=1}^{T_{2}}p(n+r^{(j)}_{i})$

holds for all $n\geq N$ or not.

Acknowledgements

Banerjee was funded by the Austrian Science Fund (FWF): W1214-N15, project
DK6. Paule and Radu were supported by grant SFB F50-06 of the FWF. Schneider
has received funding from the Austrian Science Fund (FWF) grants SFB F50
(F5009-N15) and P33530.

##  10\. Conflict of interest statement

On behalf of all authors, the corresponding author states that there is no
conflict of interest.

##  11\. Data availability statement

No datasets were generated or analysed during the current study.

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