# On the order of magnitude of Sudler products II

Sigrid Grepstad  Department of Mathematical Sciences, Norwegian University of
Science and Technology, NO-7491 Trondheim, Norway; E-mail address:
<EMAIL_ADDRESS>,  Mario Neumüller  Johannes Kepler Universität,
Linz, Austria; E-mail address<EMAIL_ADDRESS>and  Agamemnon
Zafeiropoulos  Department of Mathematical Sciences, Norwegian University of
Science and Technology, NO-7491 Trondheim, Norway; E-mail address:
<EMAIL_ADDRESS>
###### Abstract.

We study the asymptotic behavior of Sudler products
$P_{N}(\alpha)=\prod_{r=1}^{N}2|\sin\pi r\alpha|$ for quadratic irrationals
$\alpha\in\mathbb{R}$. In particular, we verify the convergence of certain
perturbed Sudler products along subsequences, and show that
$\liminf_{N}P_{N}(\alpha)=0$ and $\limsup_{N}P_{N}(\alpha)/N=\infty$ whenever
the maximal digit in the continued fraction expansion of $\alpha$ exceeds
$23$. This generalizes results obtained for the period one case
$\alpha=[0;\overline{a}]$ in [2].

SG and AZ are supported by Grant 275113 of the Research Council of Norway.

MN is funded by FWF projects F5509-N26, F5512-N26 and P29910-N35.

##  1\. Introduction and Main Results

###  1.1. Introduction

Let $\alpha\in\mathbb{R}$ and $N\geq 1$ be an integer. The Sudler product at
stage $N$ and with parameter $\alpha$ is defined as

$P_{N}(\alpha)=\prod_{r=1}^{N}|2\sin\pi r\alpha|.$

Sudler products have been studied extensively, as they bear connections with
several areas of research; we mention partition theory, Padé approximants and
dynamical systems, and refer to [12] and references therein for further
examples and details. In the present paper our main focus will be on the
asymptotic order of magnitude of $P_{N}(\alpha)$. This topic has received much
attention in recent years, and we begin by briefly reviewing key results
relevant to the main results of this paper. For a more detailed overview of
the asymptotic behavior of $P_{N}(\alpha)$, we refer the reader to the survey
paper [7].

Erdős and Szekeres showed in [5] that $\liminf_{N\to\infty}P_{N}(\alpha)=0$
for almost all $\alpha$, and conjectured that this result is true for _all_
values of $\alpha$. Lubinsky [10] later confirmed that
$\liminf_{N\to\infty}P_{N}(\alpha)=0$ whenever $\alpha$ has unbounded partial
quotients in its continued fraction expansion $\alpha=[0;a_{1},a_{2},\ldots]$.
In fact, he showed a slightly stronger statement, to which we will return
later on.

More recently, Mestel and Verschueren [12] studied the behavior of Sudler
products $P_{N}(\phi)$, where $\phi=[0;1,1,\ldots]$ is the fractional part of
the golden ratio. Their precise result was the following.

Theorem (Mestel, Verschueren): Let $\phi=(\sqrt{5}-1)/2$ be the fractional
part of the golden ratio and $(F_{n})_{n=0}^{\infty}$ be the sequence of
Fibonacci numbers. Then there exists a constant $C>0$ such that

$\lim_{n\to\infty}P_{F_{n}}(\phi)=C.$

Moreover, for the same constant $C$ we have
$\lim\limits_{n\to\infty}\dfrac{P_{F_{n}-1}(\phi)}{F_{n}}=\dfrac{C\sqrt{5}}{2\pi}\cdot$

Here the appearance of the Fibonacci sequence is not at all surprising, as it
is the sequence of denominators associated with the continued fraction
expansion of $\phi$. The proof of the result relies on the specific continued
fraction expansion $\phi=[0;1,1,\ldots]$ and the algebraic properties of the
sequence $(F_{n})_{n=0}^{\infty}$. Grepstad and Neumüller [8] later showed
that this convergence property of $P_{F_{n}}(\phi)$ is a special case of a
phenomenon exhibited by all quadratic irrationals $\alpha$.

Theorem (Grepstad, Neumüller): Let
$\alpha=[0;\overline{a_{1},a_{2},\ldots,a_{\ell}}]$ be a purely periodic
quadratic irrational, where $\ell\geq 1$ and
$a_{1},\ldots,a_{\ell}\in\mathbb{N}$, and let $(q_{n})_{n=1}^{\infty}$ be the
sequence of denominators of convergents of $\alpha$. Then there exist
constants $C_{1},C_{2},\ldots,C_{\ell}>0$ such that

(1.1) $\lim_{m\to\infty}P_{q_{m\ell+k}}(\alpha)=C_{k},\qquad
k=1,2,\ldots,\ell.$

Moreover, Grepstad, Kaltenböck and Neumüller [6] employed the factorisation
technique used in the proof of Mestel and Verschueren’s result to show that
$\liminf P_{N}(\phi)>0,$ finally disproving the conjecture in [5].

The proof of the lower bound on $P_{N}(\phi)$ given in [6] involved studying a
perturbed Sudler product $\prod_{r=1}^{N}2|\sin\pi(r\alpha+\varepsilon)|.$ The
authors of [2] then generalised the result of Mestel and Verschueren to
quadratic irrationals $\beta=[0;b,b,\ldots]$ and perturbed products defined by

(1.2)
$P_{q_{n}}(\beta,\varepsilon)=\prod_{r=1}^{q_{n}}2\Big{|}\sin\pi\Big{(}r\beta+(-1)^{n}\frac{\varepsilon}{q_{n}}\Big{)}\\!\Big{|},$

showing that for each digit $b\geq 1$, the sequence
$P_{q_{n}}(\beta,\varepsilon)$ of functions of the variable $\varepsilon$
defined in (1.2) converges locally uniformly to an explicitly defined function
$G_{b}(\varepsilon)$. In the same paper, the authors used this convergence to
show the following result on the asymptotic behavior of $P_{N}(\beta)$.

Theorem (Aistleitner, Technau, Zafeiropoulos): Let $\beta=[0;b,b,\ldots]$,
where $b\geq 1$. The following holds.

  * (i)

If $b\leq 5$, then $\liminf\limits_{N\to\infty}P_{N}(\beta)>0$ and
$\limsup\limits_{N\to\infty}\dfrac{P_{N}(\beta)}{N}<\infty.$

  * (ii)

If $b\geq 6$, then $\liminf\limits_{N\to\infty}P_{N}(\beta)=0$ and
$\limsup\limits_{N\to\infty}\dfrac{P_{N}(\beta)}{N}=\infty.$

Since the asymptotic order of magnitude of $P_{N}(\alpha)$ has been completely
investigated when $\alpha$ is a quadratic irrational of the specific form
$\alpha=[0;b,b,\ldots]$, it is reasonable to ask what happens for quadratic
irrationals with period length $\ell\geq 2$ in their continued fraction
expansion. The asymptotic behaviour of $P_{N}(\alpha)$ for such irrationals
$\alpha$ will be the main objective of this paper.

###  1.2. The irrational $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$

Let us briefly recall the basic properties of the convergents of the continued
fraction of an arbitrary quadratic irrational $\alpha$. The $n$-th convergent
of $\alpha$ is the number $p_{n}/q_{n}$, where

$\displaystyle p_{n+1}$ $\displaystyle=$ $\displaystyle
a_{n}p_{n}+p_{n-1},\qquad p_{0}=1,\quad p_{1}=0,$ $\displaystyle q_{n+1}$
$\displaystyle=$ $\displaystyle a_{n}q_{n}+q_{n-1},\qquad q_{0}=0,\quad
q_{1}=1.$

We mention that the dependence of $p_{n}$ and $q_{n}$ on $\alpha$ is not
explicitly stated, but if necessary we will write $p_{n}(\alpha)$ and
$q_{n}(\alpha)$ to make this dependence explicit. The sequence of convergents
satisfies

$\frac{p_{1}}{q_{1}}<\frac{p_{3}}{q_{3}}<\cdots<\alpha<\cdots<\frac{p_{4}}{q_{4}}<\frac{p_{2}}{q_{2}}\,$

and

$\frac{1}{2q_{n+1}q_{n}}\,<\,\left|\alpha-\frac{p_{n}}{q_{n}}\right|\,<\,\frac{1}{q_{n+1}q_{n}},\quad
n\geq 1.$

In what follows, we restrict our attention to purely periodic quadratic
irrationals, i.e. those of the form
$\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$. We use the notation of [8] and
set

$\displaystyle c(\alpha)=c=q_{\ell+1}+p_{\ell},$ (1.3) $\displaystyle
a(\alpha)=a=\frac{c(\alpha)+\sqrt{c(\alpha)^{2}+4(-1)^{\ell-1}}}{2},$
$\displaystyle
b(\alpha)=b=\frac{c(\alpha)-\sqrt{c(\alpha)^{2}+4(-1)^{\ell-1}}}{2}.$

The sequence $(q_{n})_{n=1}^{\infty}$ of denominators satisfies the additional
recursive relation

(1.4) $\displaystyle q_{n+\ell}=c(\alpha)q_{n}+(-1)^{\ell-1}q_{n-\ell},\qquad
n\geq 2\ell.$

For $k=0,1,\ldots,\ell-1$ we set

(1.5) $c_{k}=\frac{q_{\ell+k}-bq_{k}}{a-b}\qquad\text{ and }\qquad
e_{k}=(-1)^{k-1}\frac{|aq_{k}-q_{\ell+k}|}{q_{\ell}}\cdot$

For notational convenience we extend the definitions of $c_{k}$ and $e_{k}$ to
all integers $k\geq 0$ periodically modulo $\ell,$ so that in particular we
have $c_{\ell}=c_{0}$ and $e_{\ell}=e_{0}.$ We also make use of the following
relations (for more details see e.g. [8]):

(1.6) $\begin{gathered}c_{k}>0,\\\ \Lambda_{m\ell+k}:=q_{m\ell+k}\alpha-
p_{m\ell+k}=e_{k}b^{m}=(-1)^{m\ell+k+1}|e_{k}b^{m}|,\\\
\frac{1}{q_{m\ell+k}}=\mathcal{O}(|b|^{m}),\quad m\to\infty\\\
q_{m\ell+k}|b|^{m}=c_{k}+\mathcal{O}(b^{2m}),\quad m\to\infty.\end{gathered}$

When studying the irrational $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$, it
is useful to consider two families of permutations on $\ell-$tuples of
positive integers $\boldsymbol{a}=(a_{1},\ldots,a_{\ell})$. For
$k=0,1,\ldots,\ell-1$ we define the permutation operator
$\tau_{k}:\mathbb{N}^{\ell}\rightarrow\mathbb{N}^{\ell}$ by

$\tau_{k}(\boldsymbol{a})=(a_{k+1},\ldots,a_{\ell},a_{1},\ldots,a_{k}).$

Likewise, we define the permutations
$\sigma_{k}:\mathbb{N}^{\ell}\rightarrow\mathbb{N}^{\ell}$ for
$k=2,\ldots,\ell$ by

$\sigma_{k}(\boldsymbol{a})=(a_{k-1},\ldots,a_{1},a_{\ell},\ldots,a_{k})$

while for $k=1$ we set $\sigma_{1}(\boldsymbol{a})=(a_{\ell},\ldots,a_{1})$.
We can define $\tau_{k}$ and $\sigma_{k}$ for all $k\geq 1$ by extending the
previous definitions periodically modulo $\ell$. Given a purely periodic
irrational $\alpha$ with period $\boldsymbol{a}$, the corresponding purely
periodic irrationals with periods $\tau_{k}(\boldsymbol{a})$ and
$\sigma_{k}(\boldsymbol{a})$ will be denoted by

(1.7)
$\alpha_{\tau_{k}}=[0;\overline{a_{k+1},\ldots,a_{\ell},a_{1},\ldots,a_{k}}]\quad\text{
and }\quad\alpha_{\sigma_{k}}=[0;\overline{a_{k-1},\ldots
a_{1},a_{\ell},\ldots,a_{k}}].$

The significance of the permutations $\tau_{k}$ and $\sigma_{k}$ when studying
the approximation properties of $\alpha$ is indicated by the following
relations, which hold for any index $k=0,1,\ldots,\ell-1:$

(1.8)
$\begin{gathered}c(\alpha)\,=\,c(\alpha_{\tau_{k}})\,=\,c(\alpha_{\sigma_{k}}),\\\
q_{\ell}(\alpha_{\tau_{k}})\,=\,q_{\ell}(\alpha_{\sigma_{k}}),\\\
p_{\ell}(\alpha_{\tau_{k}})=q_{\ell-1}(\alpha_{\sigma_{k}})\,\text{ and
}\,p_{\ell}(\alpha_{\sigma_{k}})=q_{\ell-1}(\alpha_{\tau_{k}}),\\\
\frac{q_{\ell+1}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}=a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})},\\\
|c_{k}e_{k}|=\frac{q_{\ell}(\alpha_{\tau_{k}})}{c(\alpha_{\tau_{k}})-2b}\,\cdot\end{gathered}$

###  1.3. Main Results

As indicated above, our main goal is to generalize the convergence result of
[2] for perturbed products when $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$.
For any $\varepsilon\in\mathbb{R}$ we define

(1.9)
$P_{q_{n}}(\alpha,\varepsilon):=\prod_{r=1}^{q_{n}}2\big{|}\sin\pi\big{(}r\alpha+(-1)^{n+1}\frac{\varepsilon}{q_{n}}\big{)}\big{|}.$

In view of the aforementioned theorem in [8], one would expect the perturbed
products to converge along specific subsequences. We show that this is indeed
the case. For the sake of convenience, we introduce the notation

(1.10)
$u_{k}(t)=2\left(\frac{t}{|e_{k}c_{k}|}-\\{t\alpha_{\sigma_{k}}\\}+\frac{1}{2}\right),\quad
t=1,2,\ldots$

for each $k=1,\ldots,\ell$, where $\alpha_{\sigma_{k}}$ is as in (1.7).

######  Theorem 1.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ where
$\ell,a_{1},\ldots,a_{\ell}\in\mathbb{N}$ and $P_{q_{n}}(\alpha,\varepsilon)$
is the sequence of perturbed Sudler products defined in (1.9). Then for each
$k=1,\ldots,\ell$ the subsequence $P_{q_{m\ell+k}}(\alpha,\varepsilon)$
converges locally uniformly to a function $G_{k}(\alpha,\varepsilon)$. The
limit function satisfies

$\displaystyle G_{k}(\alpha,\varepsilon)=$
$\displaystyle\left|1+\frac{\varepsilon}{|c_{k}e_{k}|}\right|\left(1+\frac{1}{|b|^{2}}\right)^{\frac{1}{c-2}}\frac{1}{(c!)^{1/(c-2)}}\times$
(1.11)
$\displaystyle\times\prod_{t=1}^{\infty}\left|\left(1-\frac{\left(1+\frac{2\varepsilon}{|e_{k}c_{k}|}\right)^{2}}{u_{k}(t)^{2}}\right)\left(1-\frac{\left(1+\frac{2}{|b|^{2}}\right)^{2}}{u_{k}(t)^{2}}\right)^{\frac{1}{c-2}}\prod_{s=1}^{c-1}\left(1-\frac{(1+2s)^{2}}{u_{k}(t)^{2}}\right)^{-\frac{1}{c-2}}\right|$

when $\ell$ is even, and

(1.12)
$G_{k}(\alpha,\varepsilon)=\frac{\left|1+\dfrac{\varepsilon}{|c_{k}e_{k}|}\right|}{\prod\limits_{s=1}^{c}\left|s-a\right|^{\frac{1}{c}}}\times\prod_{t=1}^{\infty}\left|\left(1-\frac{\left(1+\frac{2\varepsilon}{|e_{k}c_{k}|}\right)^{2}}{u_{k}(t)^{2}}\right)\prod_{s=0}^{c-1}\left(1-\frac{\left(1+2(s-\frac{1}{|b|})\right)^{2}}{u_{k}(t)^{2}}\right)^{-\frac{1}{c}}\right|$

when $\ell$ is odd. The sequence $(u_{k}(t))_{t=1}^{\infty}$ is given in
(1.10), and the constants $a$, $b$, $c$, $c_{k}$ and $e_{k}$ are defined in
Section 1.2. In both cases, the functions $G_{k}(\alpha,\cdot)$,
$k=1,\ldots,\ell$, are continuous and $C^{\infty}$ on every interval where
they are non-zero.

######  Remark 1.

Note that we have altered the definition of $P_{q_{n}}(\alpha,\varepsilon)$
compared to [2], i.e. we use $(-1)^{n+1}$ instead of $(-1)^{n}$. This relates
to the fact that in [8] the denominator of the $n$–th convergent was defined
as $q_{n+1}$ while in [2] the denominator of the $n$–th convergent is $q_{n}$.

######  Remark 2.

An alternative proof of Theorem 1 has recently appeared in [1]. There the
limit function is given in a different form and additionally an explicit
approximation error is obtained.

Since the constants $C_{1},\ldots,C_{\ell}$ in (1.1) satisfy
$C_{k}=G_{k}(\alpha,0),(1\leq k\leq\ell)$ Theorem 1 allows us to explicitly
calculate their values.

######  Corollary 1.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ and
$C_{1},\ldots,C_{\ell}>0$ be the constants in (1.1). Then for
$k=1,2,\ldots,\ell$ we have

(1.13)
$C_{k}=\left(\frac{1+a^{2}}{c!}\right)^{\frac{1}{c-2}}\prod_{t=1}^{\infty}\left(1-\frac{1}{u_{k}(t)^{2}}\right)\left|1-\frac{(1+2a^{2})^{2}}{u_{k}(t)^{2}}\right|^{\frac{1}{c-2}}\prod_{s=1}^{c-1}\left|1-\frac{(1+2s)^{2}}{u_{k}(t)^{2}}\right|^{-\frac{1}{c-2}}$

when $\ell$ is even, and

(1.14)
$C_{k}=\frac{1}{\prod\limits_{s=1}^{c}\left|s-a\right|^{\frac{1}{c}}}\prod_{t=1}^{\infty}\left(1-\frac{1}{u_{k}(t)^{2}}\right)\prod_{s=0}^{c-1}\left|1-\frac{(1+2s-2a)^{2}}{u_{k}(t)^{2}}\right|^{-\frac{1}{c}}$

when $\ell$ is odd.

######  Remark 3.

Given $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$, consider the irrational
$\beta=[0;\overline{a_{2},\ldots,a_{\ell},a_{1}}]$, which is the image of
$\alpha$ via the Gauss map [4, Chapter 3]. Relations (1) and (1.12) imply that

$G_{1}(\beta,\varepsilon)=G_{2}(\alpha,\varepsilon),\quad
G_{2}(\beta,\varepsilon)=G_{3}(\alpha,\varepsilon),\quad\ldots\,,\,\quad
G_{0}(\beta,\varepsilon)=G_{1}(\alpha,\varepsilon).$

This means that the asymptotic behavior of $P_{q_{n}}(\alpha,\varepsilon)$ is
invariant under the action of the Gauss map on the parameter $\alpha$. The
same invariance is trivially observed under the action of the transformation
$\alpha\mapsto\alpha+1$, so the asymptotic behavior of
$P_{q_{n}}(\alpha,\varepsilon)$ is actually invariant under the action of any
modular transformation on $\alpha$. Note that for the specific value
$\varepsilon=0$, this observation follows from [8, Corollary 1.3].

The following Theorem relates the asymptotic size of $P_{N}(\alpha)$ with the
size of the constants $C_{1},\ldots,C_{\ell}$ in (1.1). This is the analogue
of Lemma $1$ in [2], and the proof is nearly identical. Nevertheless, we
include the proof later in the text for the sake of completeness.

######  Theorem 2.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ and $(C_{k})_{k=1}^{\ell}$
be the constants in (1.1). If $C_{k_{0}}<1$ for some index $1\leq
k_{0}\leq\ell$ then

(1.15) $\liminf_{N\to\infty}P_{N}(\alpha)=0\qquad\text{ and
}\qquad\limsup_{N\to\infty}\frac{P_{N}(\alpha)}{N}=\infty.$

Theorem 2 tells us that as long as _one_ of the constants $C_{k}\,\,(1\leq
k\leq\ell)$ defined in (1.1) is less than $1$, the Sudler product
corresponding to the irrational $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$
satisfies (1.15). This raises the question of which out of the $\ell$
constants $C_{k}$ associated with $\alpha$ is expected to be minimal.

The plots in Figures 1 and 2 show the graphs of the functions
$G_{k}(\alpha,\varepsilon)$ for specific choices of
$\alpha=[0;\overline{a_{1},a_{2}}]$ and $k\in\\{1,2\\}$. Since
$C_{k}=G_{k}(\alpha,0),$ the value of $C_{k}$ is the ordinate of the point of
intersection of the graph with the vertical axis. These graphs seem to suggest
that the bigger the digit $a_{k}$ is, the smaller the constant $C_{k}$
becomes.

(a) $\alpha=[0;\overline{1,2}]$ (b) $\alpha=[0;\overline{1,3}]$ (c)
$\alpha=[0;\overline{1,4}]$ (d) $\alpha=[0;\overline{1,5}]$

Figure 1. Plots of the limit functions $G_{2}(\alpha,\varepsilon)$ for the
stated values of $\alpha=[0;\overline{1,a_{2}}]$. It appears that
$C_{2}=G_{2}(\alpha,0)<1$ whenever $a_{2}\geq 4$.

(a) $\alpha=[0;\overline{2,3}]$ (b) $\alpha=[0;\overline{2,4}]$ (c)
$\alpha=[0;\overline{2,5}]$ (d) $\alpha=[0;\overline{2,6}]$

Figure 2. Plots of the limit functions $G_{2}(\alpha,\varepsilon)$ for the
stated values of $\alpha=[0;\overline{2,a_{2}}]$. It appears that
$C_{2}=G_{2}(\alpha,0)<1$ whenever $a_{2}\geq 5$.

In spite of the hints provided by the plots, it remains to verify rigorously
that $C_{k}=G_{k}(\alpha,0)$ decreases with increasing values for the digit
$a_{k}$. Moreover, it should be pointed out that the given plots provide no
information on the significance of the period length $\ell$ in the continued
fraction expansion of $\alpha$. The period length can be chosen arbitrarily
large, and it might be that the size of $\ell$ has an impact on the overall
sizes of the constants $C_{k}$. Moving forward, we will thus focus on three
main questions:

  1. $\bullet$

Is the phenomenon implied by the graphs in Figures 1 and 2 indeed true, i.e.
for any index $1\leq k\leq\ell$, is $C_{k}$ decreasing as a function of the
digit $a_{k}$?

  2. $\bullet$

Suppose we fix some period length $\ell\geq 2$. Does there exist an integer
$K=K_{\ell}\geq 1$ such that for any irrational
$\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ with $\max_{1\leq
i\leq\ell}a_{i}\geq K$ the Sudler product $P_{N}(\alpha)$ satisfies (1.15)?

  3. $\bullet$

If such an integer exists, can it be chosen independently of the period length
$\ell$?

A careful analysis of the product formulas established in Corollary 1 leads to
a positive answer to the first question for the index $k$ corresponding to the
maximal digit $a_{k}$ in $\alpha=[0;\overline{a_{1},\ldots a_{\ell}}]$. In
turn, this leads to the following result, which provides a positive answer to
questions two and three.

######  Theorem 3.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ with period length $\ell\geq
2$, and say $a_{k}=\max_{j}a_{j}$. Then

$\liminf_{N\to\infty}P_{N}(\alpha)=0\qquad\text{ and
}\qquad\limsup_{N\to\infty}\frac{P_{N}(\alpha)}{N}=\infty,$

whenever $a_{k}\geq 23$.

######  Remark 4.

Recall that it was shown by Lubinsky in [10] that $\liminf P_{N}(\alpha)=0$
whenever $\alpha$ has unbounded coefficients in its continued fraction
expansion. In fact, Lubinsky made the more striking observation that there
exists a cutoff value $a_{k}\geq K$ for which Theorem 3 is true, not only for
quadratic irrationals but for any irrational $\alpha$. As such, this is not a
new result. However, as illustrated in [7], Lubinsky’s approach merely tells
us that $K\approx e^{800}$ will suffice. Theorem 3 is thus a significant
improvement of the best known cutoff value $K$ for quadratic irrationals.

Theorem 3 can be seen as a partial analogue of the second part of the
aforementioned theorem by Aistleitner, Technau and Zafeiropoulos. We will not
attempt to imitate the first part of their result, stating that $\liminf
P_{N}(\alpha)>0$ for sufficiently small values of $\max_{i}a_{i}$. This is
because the case $\ell\geq 2$ appears to be far more intricate than the case
$\ell=1$. In fact, as will be clear from the proof of Theorem 3, we do not
fully understand the role played by the period length $\ell$ and its effect on
the asymptotic behaviour of the Sudler product $P_{N}(\alpha)$. We thus leave
the following open problems for further discussion.

######  Questions.

Let $\alpha=[0;\overline{a_{1},a_{2},\ldots,a_{\ell}}]$ be a quadratic
irrational with $a_{k}=\max_{i}a_{i}$.

  * •

According to Theorem 3, there exists an integer $K=K_{\ell}\geq 1$ such that
(1.15) holds whenever $a_{k}\geq K_{\ell}$, and $K_{\ell}\leq 23$ for all
period lengths $\ell$. However, if we fix $\ell$, what is then the optimal
value of $K_{\ell}$? We will see in the proof (see Section 6) that for odd
periods $\ell$, Theorem 3 holds for $K=22$. Moreover, for the special case
when $\alpha=[0;\overline{1,a_{2}}]$, Theorem 3 holds for $K=21$, and the
plots in Figure 1 and 2 suggest that we can actually do _much_ better. This
brings us to the following question.

  * •

Is it possibly true that if $a_{k}\geq 6$ then
$\liminf\limits_{N\to\infty}P_{N}(\alpha)=0$ and
$\limsup\limits_{N\to\infty}\dfrac{P_{N}(\alpha)}{N}=\infty$? In other words,
is $K_{\ell}\leq 6$ for all $\ell\geq 1$? Numerical evidence (beyond the plots
included in this paper) seems to suggest that the answer is positive, and that
the threshold value $K=6$ established for irrationals $\beta=[0;b,b,\ldots]$
in [2] might in fact be a universal bound for all quadratic irrationals
$\alpha=[0;\overline{a_{1},a_{2},\ldots,a_{\ell}}]$.

Finally, we point out that Aistleitner and Borda have shown in [1] the
following duality statement: for any badly approximable $\alpha,$ we have

$\liminf_{N\to\infty}P_{N}(\alpha)=0\quad\text{ if and only if
}\quad\limsup_{N\to\infty}\frac{P_{N}(\alpha)}{N}=\infty.$

Therefore for a fixed period length $\ell\geq 2,$ giving a complete
characterisation of the quadratic irrationals
$\alpha=[0;a_{1},a_{2},\ldots,a_{\ell}]$ for which (1.15) holds would also
determine those irrationals $\alpha$ for which
$\liminf\limits_{N\to\infty}P_{N}(\alpha)>0$ and
$\limsup\limits_{N\to\infty}P_{N}(\alpha)/N<\infty.$

###  1.4. Oragnization of the paper

The remainder of the paper is organized as follows. Theorems 1 and 2 are
proved in Sections 2 and 3, respectively. In Section 4, we analyse the product

$G(x)=\prod_{t=1}^{\infty}\left(1-\frac{x^{2}}{u_{k}(t)^{2}}\right),$

with $u_{k}(t)$ as defined in (1.10). This product plays a crucial role in the
expressions for $C_{k}$ in Corollary 1. Note that the sequence
$\mathcal{U}=(u_{k}(t))_{t\in\mathbb{N}}$ can be viewed as a perturbation of
the arithmetic progression $(2t/|e_{k}c_{k}|)_{t\in\mathbb{N}}$, and it is
thus natural to compare $G(x)$ to the product

$\prod_{t=1}^{\infty}\left(1-\frac{x^{2}}{\left(2t/|e_{k}c_{k}|\right)^{2}}\right).$

In doing so, we obtain Theorem 4, which tells us that

(1.16) $K_{1}\frac{\text{dist}(x,\mathcal{U})}{|x|}\leq|G(x)|\leq
K_{2}\frac{1}{|x|},$

for appropriate constants $K_{1}$ and $K_{2}$.

In Section 5 we use Corollary 1 and Theorem 4 to find an upper bound on
$C_{k}$ in (1.1) for both odd and even periods $\ell$. It turns out that
$C_{k}$ can be bounded by expressions which clearly decrease to zero as
$a_{k}\to\infty$ (see Theorems 5 and 6). The speed of decay depends on the
constants $K_{1}$ and $K_{2}$ in (1.16), and for this reason we treat
separately the case $\alpha=[0;\overline{a_{1},a_{2}}]$ where either $a_{1}=1$
or $a_{2}=1$ (as better constants $K_{1}$ and $K_{2}$ can then be found).

Finally, in Section 6 we use the bounds obtained for $C_{k}$ in Section 5 to
show that $C_{k}<1$ whenever $\alpha=[0;\overline{a_{1},a_{2},\ldots,a_{k}}]$
with $\max_{j}a_{j}\geq 23$. By Theorem 2, this proves Theorem 3.

##  2\. Proof of Theorem 1

Theorem 1 states that the perturbed Sudler products
$P_{q_{n}}(\alpha,\varepsilon)$ given in (1.9) converge along subsequences
$(q_{m\ell+k})_{m=1}^{\infty}$ to explicit limit functions

$G_{k}(\alpha,\varepsilon)=\lim_{m\to\infty}P_{q_{m\ell+k}}(\alpha,\varepsilon),$

for each $k=1,\ldots,\ell$. We begin by observing that
$P_{q_{n}}(\alpha,\varepsilon)$ can be decomposed into a product of three
factors.

######  Lemma 1.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ and let $q_{n}$ denote the
denominator of the $n$-th convergent of $\alpha$. Then for any
$\varepsilon\in\mathbb{R}$,

$P_{q_{n}}(\alpha,\varepsilon)=A_{n}(\varepsilon)\cdot B_{n}\cdot
C_{n}(\varepsilon),$

where

$\displaystyle A_{n}(\varepsilon)$
$\displaystyle=2q_{n}\Big{|}\sin\pi\Big{(}\Lambda_{n}+(-1)^{n+1}\frac{\varepsilon}{q_{n}}\Big{)}\Big{|}$
$\displaystyle B_{n}$
$\displaystyle=\left|\prod_{t=1}^{q_{n}-1}\frac{s_{n}(t)}{2\sin(\pi
t/q_{n})}\right|,$ $\displaystyle C_{n}(\varepsilon)$
$\displaystyle=\prod_{t=1}^{q_{n}-1}\left(1-\frac{s^{2}_{n}(0,\varepsilon)}{s^{2}_{n}(t)}\right)^{\frac{1}{2}},$

and

$\displaystyle
s_{n}(0,\varepsilon)=2\sin\pi\left(\frac{\Lambda_{n}}{2}+(-1)^{n+1}\frac{\varepsilon}{q_{n}}\right),\quad
s_{n}(t):=2\sin\pi\left(\frac{t}{q_{n}}-|\Lambda_{n}|\left(\left\\{\frac{tq_{n-1}}{q_{n}}\right\\}-\frac{1}{2}\right)\right).$

Lemma 1 is the natural analogue of Lemma $5.1$ of [12] for the product
$P_{q_{n}}(\phi)$ and Lemma $4.2$ in [8] for $P_{q_{n}}(\alpha)$. We omit the
proof since it is nearly identical, the only difference being that it involves
an additional term within the argument of the sine. We continue by analysing
the behaviour of each of the three factors $A_{n}(\varepsilon),B_{n}\text{ and
}C_{n}(\varepsilon)$.

The factor $B_{n}$ is independent of the perturbation argument $\varepsilon$,
and it is shown in [8] that for each $k=1,2,\ldots,\ell$ the limit

$B^{(k)}\,=\,\lim_{m\to\infty}B_{m\ell+k}$

exists. Regarding the factor $A_{m\ell+k}(\varepsilon)$, we have

$A_{m\ell+k}(\varepsilon)=2\pi\big{|}|c_{k}e_{k}|+\varepsilon\big{|}+\mathcal{O}(b^{2m}),\quad
m\to\infty,$

therefore $A_{m\ell+k}(\varepsilon)$ converges to
$2\pi\big{|}|c_{k}e_{k}|+\varepsilon\big{|}.$ Finally we need to establish
convergence for the factor $C_{n}(\varepsilon)$. Here we can argue as in [12,
Section $6$], taking into account that the factor $s_{n}(0,\varepsilon)$
depends on the parameter $\varepsilon$ and satisfies

$|s_{m\ell+k}(0,\varepsilon)|\sim\pi\left||e_{k}b^{m}|+2\varepsilon\right|+\mathcal{O}(b^{2m}),\quad
m\to\infty,$

and also

$\displaystyle
s_{m\ell+k}(t)=\pi|e_{k}b^{m}|u_{k}(t)+\mathcal{O}(|b|^{m/5}),\quad
m\to\infty.$

In view of Lemma 1 we deduce that for $k=1,2,\ldots,\ell$ the limiting
function $G_{k}$ satisfies

(2.1)
$G_{k}(\alpha,\varepsilon)=2\pi\big{|}|e_{k}c_{k}|+\varepsilon\big{|}B^{(k)}\prod_{t=1}^{\infty}\left|1-\frac{\left(1+\frac{2\varepsilon}{|e_{k}c_{k}|}\right)^{2}}{u_{k}(t)^{2}}\right|,$

where $u_{k}(t)$ is defined in (1.10). Arguing as in the proof of Theorem $1$
in [2] we can show that the convergence is locally uniform. Now we fix a value
of $k=1,\ldots,\ell$ and consider indices $n=m\ell+k,m=1,2,\ldots$ In order to
determine the formula of $G_{k}$ we will use relation (1.4). We distinguish
two cases depending on the parity of the period length $\ell$.

$\bullet$ If $\ell\equiv 0\pmod{2}$, then (1.4) gives
$c(\alpha)q_{n}=q_{n+\ell}+q_{n-\ell}$, and thus

(2.2) $P_{cq_{n}}(\alpha)=P_{q_{n-\ell}+q_{n+\ell}}(\alpha)\,.$

The left hand side in (2.2) is

$\displaystyle P_{cq_{n}}(\alpha)$ $\displaystyle=$
$\displaystyle\prod_{r=1}^{cq_{n}}2|\sin\pi
r\alpha|=\prod_{s=0}^{c-1}\prod_{r=1+sq_{n}}^{(s+1)q_{n}}2|\sin\pi
r\alpha|=\prod_{s=0}^{c-1}\prod_{r=1}^{q_{n}}2|\sin\pi(r\alpha+sq_{n}\alpha)|$
$\displaystyle\stackrel{{\scriptstyle\eqref{ntherror}}}{{=}}$
$\displaystyle\prod_{s=0}^{c-1}\prod_{r=1}^{q_{n}}2|\sin\pi(r\alpha+(-1)^{n+1}s|e_{k}b^{m}|)|=\prod_{s=0}^{c-1}P_{q_{n}}(\alpha,sq_{n}|e_{k}b^{m}|)\,,$

while the right hand side is

$\displaystyle P_{q_{n-\ell}+q_{n+\ell}}(\alpha)$ $\displaystyle=$
$\displaystyle\\!\\!\\!\prod_{r=1}^{q_{n+\ell}+q_{n-\ell}}2|\sin\pi
r\alpha|=\prod_{r=1}^{q_{n-\ell}}2|\sin\pi
r\alpha|\cdot\prod_{r=1+q_{n-\ell}}^{q_{n+\ell}+q_{n-\ell}}2|\sin\pi r\alpha|$
$\displaystyle=$ $\displaystyle
P_{q_{n-\ell}}(\alpha)\cdot\prod_{r=1}^{q_{n+\ell}}2|\sin\pi(r\alpha+q_{n-\ell}\alpha)|$
$\displaystyle\stackrel{{\scriptstyle\eqref{ntherror}}}{{=}}$ $\displaystyle
P_{q_{n-\ell}}(\alpha)\cdot\prod_{r=1}^{q_{n+\ell}}2|\sin\pi(r\alpha+(-1)^{n+\ell+1}|e_{k}b^{m-1}|)|$
$\displaystyle=$ $\displaystyle
P_{q_{n-\ell}}(\alpha)P_{q_{n+\ell}}(\alpha,q_{n+\ell}|e_{k}b^{m-1}|).$

Since the functions $P_{q_{n}}(\alpha,\varepsilon)$ converge locally
uniformly, letting $m\to\infty$ in (2.2) and taking (1.6) into account we
obtain

(2.3)
$\prod_{s=0}^{c-1}G_{k}(\alpha,s|c_{k}e_{k}|)=G_{k}(\alpha,0)\,G_{k}\\!\left(\alpha,\frac{|c_{k}e_{k}|}{|b|^{2}}\right).$

Substituting $G_{k}$ from (2.1) in (2.3) we obtain (1).

$\bullet$ If $\ell\equiv 1\pmod{2}$, then (1.4) becomes
$q_{n+\ell}=c(\alpha)q_{n}+q_{n-\ell}$, so

$\displaystyle\frac{P_{q_{n+\ell}}(\alpha)}{P_{q_{n-\ell}}(\alpha)}$
$\displaystyle=$
$\displaystyle\prod_{r=1}^{q_{n+\ell}-q_{n-\ell}}2|\sin\pi(r\alpha+q_{n-\ell}\alpha)|=\prod_{r=1}^{c(\alpha)q_{n}}2|\sin\pi(r\alpha+q_{n-\ell}\alpha)|$
$\displaystyle=$
$\displaystyle\prod_{s=0}^{c(\alpha)-1}\prod_{r=1}^{q_{n}}2|\sin\pi(r\alpha+sq_{n}\alpha+q_{n-\ell}\alpha)|$
$\displaystyle\stackrel{{\scriptstyle\eqref{ntherror}}}{{=}}$
$\displaystyle\prod_{s=0}^{c(\alpha)-1}\prod_{r=1}^{q_{n}}2|\sin\pi(r\alpha+(-1)^{n+1}s|e_{k}b^{m}|-(-1)^{n+1}|e_{k}b^{m-1}|)|$
$\displaystyle=$
$\displaystyle\prod_{s=0}^{c(\alpha)-1}P_{q_{n}}\left(\alpha,q_{n}|e_{k}b^{m}|\left(s-\frac{1}{|b|}\right)\right).$

By the result of [8] we know that the sequence $P_{q_{n}}(\alpha)$ converges
to a limit $C_{k}>0$, hence letting $m\to\infty$ in the above equality we get

$\prod_{s=0}^{c(\alpha)-1}G_{k}\left(\alpha,|c_{k}e_{k}|(s-\tfrac{1}{|b|})\right)=1.$

Substituting this into (2.1) we obtain (1.12).

##  3\. Proof of Theorem 2

Let us now prove Theorem 2, namely show that if $G_{k}(\alpha,0)=C_{k}<1$ for
some $1\leq k\leq\ell$, then

$\liminf_{N\to\infty}P_{N}(\alpha)=0\quad\text{ and
}\quad\limsup_{N\to\infty}\frac{P_{N}(\alpha)}{N}=\infty.$

Let $k_{0}$ denote the index for which $C_{k_{0}}<1$, and fix some $\lambda$
such that $C_{k_{0}}<\lambda<1$. Since $G_{k_{0}}$ is continuous at $0$ and
$G_{k_{0}}(\alpha,0)=C_{k_{0}}<\lambda$, there exists $\eta>0$ such that
$G_{k_{0}}(\alpha,\varepsilon)<\lambda$ for all $|\varepsilon|<\eta.$ Consider
a subsequence $(m_{i})_{i=1}^{\infty}$ of $(q_{m\ell+k_{0}})_{m=1}^{\infty}$
such that

  1. (i)

$m_{i+1}\geq 2m_{i},\quad i=1,2,\ldots\quad$ and

  2. (ii)

$\|m_{i+1}\alpha\|<\dfrac{\eta}{4m_{i}},\quad i=1,2,\ldots$

where $\|x\|=\min\\{|x-k|:k\in\mathbb{Z}\\}$ denotes the distance of
$x\in\mathbb{R}$ to the nearest integer. We set

$N_{i}=m_{i}+\ldots+m_{1}\quad\text{and}\quad M_{j}=N_{i}-N_{j},\quad i\geq
1,\quad j=1,\ldots,i.$

Then

$\displaystyle P_{N_{i}}(\alpha)\,$ $\displaystyle=\,$
$\displaystyle\prod_{r=1}^{N_{i}}2|\sin\pi
r\alpha|\,\,=\,\,\prod_{j=1}^{i}\prod_{r=1}^{m_{j}}2|\sin\pi(r+M_{j})\alpha|$
$\displaystyle=\,$
$\displaystyle\prod_{j=1}^{i}\prod_{r=1}^{m_{j}}2\left|\sin\pi\left(r\alpha+\frac{\varepsilon_{j}}{m_{j}}\right)\right|\,\,=\,\,\prod_{j=1}^{i}P_{m_{j}}(\alpha,(-1)^{\delta_{j}}\varepsilon_{j}),$

where $\varepsilon_{j}=\pm m_{j}\|M_{j}\alpha\|$ and $\delta_{j}=0\text{ or
}1.$ We see that

$\displaystyle|\varepsilon_{j}|$ $\displaystyle\leq$ $\displaystyle
m_{j}\left(\|m_{j+1}\alpha\|+\ldots+\|m_{k}\alpha\|\right)<\frac{\eta}{2}\cdot$

By the choice of $\eta$, this implies that
$P_{m_{j}}(\alpha,(-1)^{\delta_{j}}\varepsilon_{j})<\lambda$ for all $j$ large
enough, hence $P_{N_{i}}(\alpha)\ll\lambda^{i},\,i\to\infty$. This shows that
$\liminf\limits_{N\to\infty}P_{N}(\alpha)=0$.

Now set also $T_{i}=m_{i+1}-(N_{i}+1),\,i\geq 1$, so that

(3.1)
$P_{T_{i}}(\alpha)=\frac{P_{m_{i+1}-1}(\alpha)}{\prod\limits_{r=N_{i}+1}^{m_{i+1}-1}\\!\\!2|\sin\pi
r\alpha|}=\frac{P_{m_{i+1}-1}(\alpha)}{\prod\limits_{j=1}^{i}\prod\limits_{r=1}^{m_{j}}2|\sin\pi(r+M_{j}-m_{i+1})\alpha|}\cdot$

At this point we need to point out a simple fact which follows from the proof
of Theorem $1.1$ in [8] but is not explicitly stated in the text. If
$(C_{k})_{k=1}^{\ell}$ are the constants in (1.1), then for each
$k=1,\ldots,\ell$ we have

$\lim_{m\to\infty}\frac{P_{q_{m\ell+k}-1}(\alpha)}{q_{m\ell+k}}=\frac{C_{k}}{2\pi|c_{k}e_{k}|}\,,$

where $(c_{k})_{k=1}^{\ell}$ and $(e_{k})_{k=1}^{\ell}$ are defined in (1.5).
Armed with this observation we deduce that the numerator in (3.1) is
$P_{m_{i+1}-1}(\alpha)\asymp m_{i+1}\asymp T_{i},\,i\to\infty$, while for the
denominator in (3.1) we can show arguing as in the previous step that

$\liminf_{i\to\infty}\prod\limits_{j=1}^{i}\prod\limits_{r=1}^{m_{j}}2|\sin\pi(r+M_{j}-m_{i+1})\alpha|=0.$

Therefore

$\limsup_{N\to\infty}\frac{P_{N}(\alpha)}{N}=\limsup_{i\to\infty}\frac{P_{T_{i}}(\alpha)}{T_{i}}=\infty.$

##  4\. A perturbed sinc product

In Sections 5 and 6, our aim will be to determine when the limit $C_{k}$ in
(1.1) is guaranteed to be less than one. Corollary 1 suggests that we will
need to differentiate between two cases, depending on the parity of the period
length $\ell$. Common to both cases is the need for appropriate upper and
lower bounds on the function

(4.1)
$G(x)=\prod_{t=1}^{\infty}\left(1-\frac{x^{2}}{u_{k}(t)^{2}}\right),\qquad
x\in\mathbb{R},$

where we recall from (1.10) that

$u_{k}(t)=2\left(\frac{t}{|e_{k}c_{k}|}-\\{t\alpha_{\sigma_{k}}\\}+\frac{1}{2}\right).$

Let us now set

(4.2) $A=2|e_{k}c_{k}|^{-1}\quad\text{ and
}\quad\delta_{t}=1-2\\{t\alpha_{\sigma_{k}}\\},$

so that $u_{k}(t)=At+\delta_{t}$. The function $G(x)$ in (4.1) can then be
seen as a perturbed version of the well-known product

$\frac{\sin(\pi A^{-1}x)}{\pi
A^{-1}x}=\prod_{t=1}^{\infty}\left(1-\frac{x^{2}}{(At)^{2}}\right).$

It is not difficult to show that if the perturbations $\delta_{t}$ satisfy
$|\delta_{t}|\leq\delta<A/4$ $(t\in\mathbb{N})$ for some $\delta>0$, then the
function $G$ obeys the bounds

$C_{1}\frac{\mathrm{dist}\left(x,\mathcal{U}\right)}{|x|^{1+4\delta/A}}\leq|G(x)|\leq
C_{2}\frac{1}{|x|^{1-4\delta/A}},$

for constants $C_{1}$ and $C_{2}$, where
$\mathrm{dist}\left(x,\mathcal{U}\right)=\min\\{|x-At-\delta_{t}|\,:\,t\in\mathbb{N}\\}$.
This is related to Kadec’s $1/4$-rule [9], and can e.g. be seen as a
consequence of [3, Lemma 4]. Due to the low-discrepancy property of Kronecker
sequences, the sequence $(\delta_{t})_{t}$ satisfies a much stronger
condition, which in turn enables us to establish stronger bounds on $G$.

######  Theorem 4.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$, and let $k\leq\ell$ be an
index satisfying $a_{k}=\max_{j}a_{j}$. Then for $|x|\geq
A/2=|e_{k}c_{k}|^{-1}$, the function $G$ in (4.1) satisfies

(4.3)
$\frac{2}{\pi}e^{-f(a_{k})}\left(1-\frac{2}{3Am}\right)\left(1-\frac{1}{Am}\right)^{2}\,\frac{\mathrm{dist}(x,\mathcal{U})}{|x|}\,\,\leq\,|G(x)|\,\,\leq\,\,\frac{14A}{9}e^{f(a_{k})}\frac{1}{|x|},$

where the positive integer $m=m(x)\geq 1$ is such that
$|Am-x|=\min\\{|An-x|:n\in\mathbb{N}\\}$ and

(4.4) $f(a_{k})=\frac{13.7}{a_{k}}+\frac{1}{20\log
a_{k}}+\frac{1}{100}+\frac{2}{a_{k}^{2}}.$

For $|x|<A/2$, we have the bound

(4.5) $\frac{2}{\pi}e^{-f(a_{k})}\leq|G(x)|\leq 1.$

######  Remark 5.

Notice that since $f(a_{k})\to 0.01$ as $a_{k}\to\infty$, equation (4.3) reads

$K_{1}\frac{\mathrm{dist}(x,\mathcal{U})}{|x|}\leq|G(x)|\leq
K_{2}\frac{1}{|x|},$

with $K_{1}\approx 2/\pi$ and $K_{2}\approx 14A/9$ whenever $a_{k}$ and $x$
are large. Similar bounds can be established when $k$ is _not_ the index of a
maximal continued fraction coefficient of $\alpha$, but the size of the
constants $K_{1}$ and $K_{2}$ will then depend both on $a_{k}$ and on the size
of $\max_{j}a_{j}$.

We will see that the following is an immediate consequence of the proof of
Theorem 4.

######  Corollary 2.

Suppose that we have $m(x)=1$ in Theorem 4, that is $A/2\leq x\leq 3A/2$. Then
$G(x)$ in (4.1) satisfies

$\frac{2}{\pi}e^{-g(a_{k})}\left(1-\frac{2}{3A}\right)\left(1-\frac{1}{A}\right)^{2}\,\frac{\mathrm{dist}(x,\mathcal{U})}{|x|}\,\,\leq\,|G(x)|\,\,\leq\,\,\frac{14A}{9}e^{g(a_{k})}\frac{1}{|x|},$

where

(4.6) $g(a_{k})=\frac{3.3}{a_{k}}+\frac{1}{80\log
a_{k}}+\frac{1}{400}+\frac{2}{a_{k}^{2}}\cdot$

Before we embark on the proof of Theorem 4, we establish two preliminary
results. The first concerns the size of $A=2|e_{k}c_{k}|^{-1}$.

######  Lemma 2.

Let $1\leq k\leq\ell.$ We have

(4.7)
$\frac{1}{|c_{k}e_{k}|}=a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\,\cdot$

###### Proof.

By (1.8) we obtain

$\displaystyle\frac{1}{|c_{k}e_{k}|}=\frac{q_{\ell+1}(\alpha_{\tau_{k}})+p_{\ell}(\alpha_{\tau_{k}})-2b}{q_{\ell}(\alpha_{\tau_{k}})}=a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\cdot$

∎

######  Corollary 3.

For any $k=1,2,\ldots,\ell$ we have

$a_{k}<\frac{1}{|c_{k}e_{k}|}<a_{k}+2\,.$

By Corollary 3, it immediately follows that

(4.8) $2a_{k}<A<2(a_{k}+2).$

The second result concerns the perturbations $\delta_{t}$. We state it without
proof, as it is an easy consequence of [11, Corollary 3].

######  Lemma 3.

Let $\alpha=[0;a_{1},a_{2},\ldots]$ be an irrational with bounded continued
fraction coefficients. Then for any fixed $n\in\mathbb{N}$, we have

$\Big{|}\sum_{t=n+1}^{n+N}\\!\Big{(}\frac{1}{2}-\left\\{t\alpha\right\\}\Big{)}\Big{|}\leq\log
N\left(\frac{a}{8\log a}+6\right)+\frac{a}{8}+\frac{23}{4},$

for all $N\geq 1$, where $a=\max_{j}a_{j}$.

Recall from (4.2) that $\delta_{t}=1-2\\{t\alpha_{\sigma_{k}}\\}$, and thus by
Lemma 3 it follows that

(4.9) $\left|\sum_{t=n+1}^{n+N}\delta_{t}\right|\leq\min\left\\{N,\log
N\left(\frac{a_{k}}{4\log
a_{k}}+12\right)+\frac{a_{k}}{4}+\frac{23}{2}\right\\},$

where $a_{k}=\max_{j}a_{j}$ for the quadratic irrational $\alpha$. Note that

$\log N\left(\frac{a_{k}}{4\log
a_{k}}+12\right)+\frac{a_{k}}{4}+\frac{23}{2}\geq 12\log
N+\frac{23}{2}>N\quad\text{ for all }\quad N\leq 60,$

regardless of the value of $a_{k}$. Accordingly, we will use the bound
$\sum_{n+1}^{n+N}\delta_{t}\leq N$ whenever $N\leq 60$.

We are now equipped to prove Theorem 4.

###### Proof of Theorem 4.

Since $G$ in (4.1) is an even function, it suffices to consider $x\geq 0$. Let
$m=m(x)\geq 0$ be the non-negative integer satisfying

$|x-Am|\,=\,\min\left\\{|x-An|:n=0,1,2,\ldots\right\\},$

and let us first assume that $m\geq 1$, meaning that $x\geq A/2$. Excluding
the case of the golden ratio, we may safely assume that
$a_{k}=\max_{j}a_{j}\geq 2$, and thus by (4.8) we have $A\geq 4$. It follows
that

(4.10)
$|x-Am|\leq\frac{A}{2},\qquad|x-Am-\delta_{m}|\leq\frac{A}{2}+1\leq\frac{3A}{4},$

and

$|x-At|\geq\frac{A}{2}\quad\text{ and
}\quad|x-At-\delta_{t}|\geq\frac{A}{2}-1\geq\frac{A}{4},$

for any $t\neq m$.

The function $G$ may be split into three products
$G(x)=\Pi_{1}(x)\Pi_{2}(x)\Pi_{3}(x),$ where

$\displaystyle\Pi_{1}(x)$ $\displaystyle=$ $\displaystyle
1-\frac{x^{2}}{(Am+\delta_{m})^{2}},$ $\displaystyle\Pi_{2}(x)$
$\displaystyle=$ $\displaystyle\mathop{\prod_{t\geq 1}}_{t\neq
m}\left(1-\frac{x^{2}}{(At+\delta_{t})^{2}}\right)\left(1-\frac{x^{2}}{(At)^{2}}\right)^{-1},$
$\displaystyle\Pi_{3}(x)$ $\displaystyle=$ $\displaystyle\mathop{\prod_{t\geq
1}}_{t\neq m}\left(1-\frac{x^{2}}{(At)^{2}}\right).$

For the first product, we observe that

$\Pi_{1}(x)=1-\frac{x^{2}}{(Am+\delta_{m})^{2}}=\frac{(Am+\delta_{m}-x)(Am+\delta_{m}+x)}{(Am+\delta_{m})^{2}},$

and thus by (4.10) we get

(4.11)
$\text{dist}(x,\mathcal{U})\frac{Am+\delta_{m}+x}{(Am+\delta_{m})^{2}}\leq|\Pi_{1}(x)|\leq\frac{3A}{4}\frac{Am+\delta_{m}+x}{(Am+\delta_{m})^{2}}\,\cdot$

We then consider the second factor $\Pi_{2}(x)=\prod\limits_{t\neq
m}Q_{t}(x)$, where

$Q_{t}(x)=\left(1-\frac{x^{2}}{(At+\delta_{t})^{2}}\right)\left(1-\frac{x^{2}}{(At)^{2}}\right)^{-1},\quad
t\geq 1.$

We have

$\displaystyle Q_{t}(x)$ $\displaystyle=$
$\displaystyle\frac{1-\dfrac{x}{At+\delta_{t}}}{1-\dfrac{x}{At}}\cdot\frac{1+\dfrac{x}{At+\delta_{t}}}{1+\dfrac{x}{At}}\,\,=\,\,\frac{1+\dfrac{\delta_{t}}{At-x}}{1+\dfrac{\delta_{t}}{At}}\cdot\frac{1+\dfrac{\delta_{t}}{At+x}}{1+\dfrac{\delta_{t}}{At}}$
$\displaystyle=$
$\displaystyle\exp\left\\{\log\left(1+\dfrac{\delta_{t}}{At-x}\right)+\log\left(1+\dfrac{\delta_{t}}{At+x}\right)-2\log\left(1+\frac{\delta_{t}}{At}\right)\right\\}.$

Thus if we employ the inequality

$x-x^{2}\,<\,\log(1+x)\,<\,x\qquad\text{ for all }x>-\frac{1}{2}$

we obtain

(4.12)
$\displaystyle\quad\exp\left\\{\delta_{t}s_{t}-\delta_{t}^{2}\left(\frac{1}{(At-x)^{2}}+\frac{1}{(At+x)^{2}}\right)\right\\}<Q_{t}(x)<\exp\left\\{\delta_{t}s_{t}+\frac{2\delta_{t}^{2}}{(At)^{2}}\right\\},$

where we define

(4.13) $s_{t}=\frac{1}{At-x}+\frac{1}{At+x}-\frac{2}{At}\cdot$

The factors in (4.12) contributing significantly to $\Pi_{2}(x)=\prod_{t\neq
m}Q_{t}$ are the first-order terms $\delta_{t}s_{t}$. For the second-order
terms, we observe that the contribution on the left hand side is larger (in
absolute value) than that on the right hand side, and a straightforward
calculation verifies that

$\sum_{t\neq
m}\left(\frac{1}{(At+x)^{2}}+\frac{1}{(At-x)^{2}}\right)<\frac{8}{A^{2}}.$

We thus conclude that

(4.14) $\Pi_{2}(x)=\prod_{t\neq m}Q_{t}(x)=\exp\left(\sum_{t\neq
m}\delta_{t}s_{t}+E\right),$

where $s_{t}$ is given in (4.13) and $|E|<\dfrac{8}{A^{2}}$.

It remains to find an appropriate bound for

$\left|\sum_{t\neq
m}\delta_{t}s_{t}\right|\leq\left|\sum_{t<m}\delta_{t}s_{t}\right|+\left|\sum_{t>m}\delta_{t}s_{t}\right|.$

We first consider the final term on the right hand side above. Summation by
parts yields

$\sum_{t=m+1}^{m+M}\delta_{t}s_{t}=s_{m+M}\sum_{t=m+1}^{m+M}\delta_{t}+\sum_{t=m+1}^{m+M-1}(s_{t}-s_{t+1})\sum_{k=m+1}^{t}\delta_{k},$

for any $M\geq 1$. We observe that

$|s_{t+1}-s_{t}|\leq\frac{1}{At-x}-\frac{1}{A(t+1)-x}\leq\frac{1}{A(t-m-\frac{1}{2})(t-m+\frac{1}{2})},$

and

$|s_{m+M}|\leq\frac{1}{A(m+M)-x}\leq\frac{1}{A(M-\frac{1}{2})}\cdot$

Combining this with (4.9), we find that

$\left|\sum_{t=m+1}^{m+M}\delta_{t}s_{t}\right|\leq\varepsilon(M)+\sum_{t=m+1}^{m+M-1}\frac{\min\\{t-m,\,K\log(t-m)+C\\}}{A\left((t-m)^{2}-\frac{1}{4}\right)},$

where $K=12+a_{k}/(4\log a_{k})$, $C=23/2+a_{k}/4$, and where
$\varepsilon(M)\to 0$ as $M\to\infty$. Letting $M\to\infty$, we thus find

$\displaystyle\left|\sum_{t>m}\delta_{t}s_{t}\right|$ $\displaystyle\leq$
$\displaystyle\frac{1}{A}\left(\sum_{t=1}^{60}\frac{t}{(t^{2}-\frac{1}{4})}+K\sum_{t=61}^{\infty}\frac{\log
t}{(t^{2}-\frac{1}{4})}+C\sum_{t=61}^{\infty}\frac{1}{(t^{2}-\frac{1}{4})}\right)$
$\displaystyle\leq$ $\displaystyle\frac{1}{A}\left(5.1+0.1K+0.02C\right),$

and inserting values of $C$ and $K$, and recalling that $A>2a_{k}$, we get

$\left|\sum_{t>m}\delta_{t}s_{t}\right|\leq\frac{3.3}{a_{k}}+\frac{1}{80\log
a_{k}}+\frac{1}{400}\cdot$

By an analogous argument, one can show that

$\displaystyle\left|\sum_{t<m}\delta_{t}s_{t}\right|$ $\displaystyle\leq$
$\displaystyle\frac{10.4}{a_{k}}+\frac{3}{80\log a_{k}}+\frac{3}{400},$

and thus combined we have

$\left|\sum_{t\neq
m}\delta_{t}s_{t}\right|\leq\frac{13.7}{a_{k}}+\frac{1}{20\log
a_{k}}+\frac{1}{100}\cdot$

Inserting this in (4.14), we arrive at

(4.15) $e^{-f(a_{k})}\leq\left|\Pi_{2}(x)\right|\leq e^{f(a_{k})},$

with $f$ defined as in (4.4).

Finally, we observe that

$\Pi_{3}(x)=\frac{A^{2}m^{2}}{(Am-x)(Am+x)}\cdot\frac{\sin(\pi xA^{-1})}{\pi
xA^{-1}}$

and since $|\pi xA^{-1}-\pi m|\leq\dfrac{\pi}{2}$ we get

$\frac{2}{A}\,\leq\,\left|\frac{\sin(\pi
xA^{-1})}{Am-x}\right|\,\leq\,\frac{\pi}{A}\,\cdot$

This implies that

(4.16)
$\frac{2A^{2}m^{2}}{\pi(Am+x)x}\,\leq\,|\Pi_{3}(x)|\,\leq\,\frac{A^{2}m^{2}}{(Am+x)x}\,.$

Combining the bounds (4.11), (4.15) and (4.16), we find that

$|G(x)|\geq\frac{2}{\pi}e^{-f(a_{k})}\cdot\mathrm{dist}(x,\mathcal{U})\cdot\frac{1}{|x|}\cdot\frac{(Am)^{2}}{(Am+\delta_{m})^{2}}\cdot\frac{Am+\delta_{m}+x}{Am+x},$

and

$|G(x)|\leq\frac{3A}{4}e^{f(a_{k})}\cdot\frac{1}{|x|}\cdot\frac{(Am)^{2}}{(Am+\delta_{m})^{2}}\cdot\frac{Am+\delta_{m}+x}{Am+x}\,\cdot$

The common factor

$\frac{(Am)^{2}}{(Am+\delta_{m})^{2}}\cdot\frac{Am+\delta_{m}+x}{Am+x}=\left(1-\frac{\delta_{m}}{Am+\delta_{m}}\right)^{2}\left(1+\frac{\delta_{m}}{Am+x}\right)$

will necessarily tend to $1$ as $m(x)\to\infty$. Only the rate of convergence
from below will be important to us. We therefore apply the rough upper bound

$\frac{(Am)^{2}}{(Am+\delta_{m})^{2}}\cdot\frac{Am+\delta_{m}+x}{Am+x}\leq\left(\frac{4}{3}\right)^{2}\left(\frac{7}{6}\right),$

and the more precise lower bound

$\frac{(Am)^{2}}{(Am+\delta_{m})^{2}}\cdot\frac{Am+\delta_{m}+x}{Am+x}\geq\left(1-\frac{1}{Am}\right)^{2}\left(1-\frac{2}{3Am}\right).$

Inserting these bounds in the inequalities for $|G(x)|$ completes the proof of
Theorem 4 in the case $|x|\geq A/2$.

Finally, we consider the case $0\leq x<A/2$. As an upper bound, we use

$|G(x)|\leq 1.$

For the lower bound, we again split $G$ into the subproducts $\Pi_{1}$,
$\Pi_{2}$ and $\Pi_{3}$. Note that in this case, the product $\Pi_{1}$ is
empty, and the product $\Pi_{3}$ is simply a sinc function bounded by

$\frac{2}{\pi}\leq\Pi_{3}(x)\leq 1.$

For the product $\Pi_{2}$, we may use the bound (4.15) established for the
case $x\geq A/2$ (in fact we can do better, as will be argued below). Combined
we get

$\frac{2}{\pi}e^{-f(a_{k})}\leq|G(x)|\leq 1,$

and this completes the proof of Theorem 4. ∎

###### Proof of Corollary 2.

Retracing the proof of Theorem 4, we arrive at (4.14), and note that for
$m(x)=1$ we have

$\Pi_{2}(x)=\prod_{t\neq
m}Q_{t}(x)=\exp\left(\sum_{t>m}\delta_{t}s_{t}+\frac{8}{A^{2}}\right)\leq\exp\left(\frac{3.3}{a_{k}}+\frac{1}{80\log
a_{k}}+\frac{1}{400}+\frac{2}{a_{k}^{2}}\right),$

since the product $\sum_{t<m}\delta_{t}s_{t}$ is empty. Apart from this, the
proof remains unchanged. ∎

######  Remark 6.

Note that this bound on $\Pi_{2}(x)$ is clearly also valid for $x\leq A/2$.
Thus, the lower bound on $G(x)$ in (4.5) may be improved to

$\frac{2}{\pi}e^{-g(a_{k})}\leq|G(x)|\leq 1,\quad x\leq A/2,$

with $g$ given in (4.6).

######  Remark 7.

The lower bound for $G(x)$ in (4.3) is used in the following sections to
determine an upper bound for the constants $C_{k}.$ In turn, this upper bound
gives a threshold value $K>1$ such that $a_{k}=\max_{1\leq i\leq\ell}a_{i}\geq
K$ implies that $C_{k}<1.$

We believe that an improvement for the threshold value $K=23$ in Theorem 3
might be obtained as follows: in the estimates for $\Pi_{2}(x),$ we have
bounded $\Big{|}\sum\limits_{t\neq m(x)}\delta_{t}s_{t}(x)\Big{|}$ from above
uniformly for all $x>0$. In the proof of Theorem 3 we are actually interested
in the quantity $G(1)G(3)\cdots G(2c-1),$ which means that we need a bound for
the product $\prod_{x=1}^{c}\Pi_{2}(2x-1)$. Rather than using a uniform bound
for all terms in this product, we could seek an upper bound for the double sum

$\Big{|}\sum_{1\leq x\leq c}\sum_{\begin{subarray}{c}t\geq 1\\\ t\neq
m(2x-1)\end{subarray}}\delta_{t}s_{t}(2x-1)\Big{|}.$

This would take the specific range of $x$-values into account, and possibly
provide a substantial improvement in the lower bound on
$\prod_{x=1}^{c}\Pi_{2}(2x+1)$. We leave this task to the interested reader.

##  5\. Upper bounds for the constants $C_{k}$

With Theorem 4 established, let us now revisit Corollary 1 and carefully
analyse the expressions for $C_{k}$ provided in (1.13) and (1.14). Recall that
we have to differentiate between the case of even and odd period length
$\ell$. To ease the analysis, it will be useful to make an assumption on the
size of $\max_{j}a_{j}$ given a quadratic irrational
$\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$. Presuming a priori that we
cannot do better for general $\ell$ than for the $\ell=1$ case (see [2,
Theorem 6]), we assume throughout this section that $\max_{j}a_{j}\geq 6$.

We will show the following.

######  Theorem 5.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ for some even period length
$\ell$, and assume $a_{k}=\max_{j}a_{j}\geq 6$. Then the limit
$C_{k}=\lim_{m\to\infty}P_{\ell m+k}(\alpha)$ obeys the bound

$C_{k}^{\frac{c-2}{c}}\leq\frac{\pi}{2a_{k}}e^{1+f(a_{k})}\left(200e^{2.4}c^{2}\right)^{\frac{1}{c}}\cdot
2^{\frac{1}{q_{\ell}}}\cdot\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{\frac{1}{a_{k}}}.$

where $q_{\ell}=q_{\ell}(\alpha_{\sigma_{k}}),$ $c$ is defined in (1.3) and we
recall from (4.4) that

$f(a_{k})\leq\frac{13.7}{a_{k}}+0.1,\quad a_{k}\geq 6.$

Throughout this section, when there is no danger of confusion, we shall write
for abbreviation $q_{\ell}=q_{\ell}(\alpha_{\sigma_{k}})$ and
$q_{\ell+1}=q_{\ell+1}(\alpha_{\sigma_{k}}).$

For the special case $q_{\ell}=1$, we have an improved bound; note that this
only occurs if $\ell=2$ and $\alpha=[0;\overline{a_{1},a_{2}}]$ with either
$a_{1}=1$ or $a_{2}=1$.

######  Corollary 4.

Let $\alpha=[0;\overline{a_{1},a_{2}}]$, and assume
$a_{k}=\max\\{a_{1},a_{2}\\}\geq 6$ and $\min\\{a_{1},a_{2}\\}=1$. Then the
limit $C_{k}=\lim_{m\to\infty}P_{2m+k}(\alpha)$ obeys the bound

$C_{k}^{\frac{c-2}{c}}\leq\frac{\pi}{a_{k}}e^{1+g(a_{k})}\left(6.2(a_{k}+2)^{4}\right)^{\frac{1}{a_{k}+2}},$

where we recall from (4.6) that

$g(a_{k})\leq\frac{3.3}{a_{k}}+0.1,\quad a_{k}\geq 6.$

For the odd period case, we will establish the following.

######  Theorem 6.

Let $\alpha=[0;\overline{a_{1},\ldots,a_{\ell}}]$ for some odd period length
$\ell$, and assume $a_{k}=\max_{j}a_{j}\geq 6$. Then the limit
$C_{k}=\lim_{m\to\infty}P_{\ell m+k}(\alpha)$ obeys the bound

$C_{k}\leq\frac{\pi}{2a_{k}}e^{1+f(a_{k})}\left(40c^{\frac{3}{2}}\right)^{\frac{1}{c}}\cdot
2^{\frac{1}{q_{\ell}}}\cdot a_{k}^{\frac{5}{2a_{k}}},$

where $f$ is given in (4.4).

###  5.1. Bounding $C_{k}$ when $\ell\equiv 0\pmod{2}$

Considering first even period lengths $\ell$, we recall from (1.13) that

(5.1) $\displaystyle C_{k}^{c-2}$ $\displaystyle=$
$\displaystyle\frac{1+a^{2}}{c!}\left|\frac{G(1)^{c-1}G(1+2a^{2})}{G(1)G(3)G(5)\cdots
G(2c-1)}\right|,$

with $a$ and $c$ as given in (1.3), and the function $G$ defined in (4.1). The
term $|G(1)|$ in (5.1) is bounded above by $1$, and the term $G(1+2a^{2})$ can
be bounded by the upper bound in Theorem 4. Keeping the expression for
$C_{k}^{c-2}$ in mind, we will rather give a bound for $(1+a^{2})G(1+2a^{2})$.
By Theorem 4 we have

(5.2) $(1+a^{2})G(1+2a^{2})\leq\frac{14A}{9}\cdot
e^{f(a_{k})}\cdot\frac{(1+a^{2})}{(1+2a^{2})}\leq\frac{4}{5}Ae^{f(a_{k})},$

where for the last inequality we have used that $a\geq 5$ whenever $a_{k}\geq
6$.

Now let us find a lower bound on

(5.3) $G(1)G(3)\cdots G(2c-1)=\prod_{s=0}^{c-1}G(2s+1).$

In view of Theorem 4, some of the factors of (5.3) will be bounded using (4.3)
while others will be bounded using (4.5). Since $2a_{k}<A<2(a_{k}+2)$, the
integers $j$ satisfying $j<\frac{A}{2}$ are $j=1,2,\ldots,a_{k}$ and possibly
also $a_{k}+1$. Thus the factors of (5.3) with $0\leq
s\leq\left\lfloor\frac{a_{k}}{2}\right\rfloor-1$ will be bounded using (4.5)
and those with $s\geq\lfloor\frac{a_{k}}{2}\rfloor+1$ will be bounded using
(4.3). We get

(5.4)
$\prod_{s=0}^{\lfloor\frac{a_{k}}{2}\rfloor-1}G(2s+1)\geq\left(\frac{2}{\pi}e^{-f(a_{k})}\right)^{\lfloor\frac{a_{k}}{2}\rfloor}$

and

(5.5)
$\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}G(2s+1)\geq\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}\frac{2}{\pi}e^{-f(a_{k})}\left(1-\frac{2}{3Am_{s}}\right)\left(1-\frac{1}{Am_{s}}\right)^{2}\cdot\frac{\mathrm{dist}(2s+1,\mathcal{U})}{2s+1},$

with $\mathcal{U}=(u_{k}(t))_{t=1}^{\infty}$ given in (1.10) and
$m_{s}\in\mathbb{N}$ as defined in Theorem 4.

The factor $G(a_{k}+1)$ appears in (5.3) only when $a_{k}$ is even and
$s=a_{k}/2$. For this factor, it is not clear which of the two bounds (4.3)
and (4.5) apply. However, under the restriction that $a_{k}\geq 6$, we clearly
have $\mathrm{dist}(a_{k}+1,\mathcal{U})>1$, and by combining the bounds (4.3)
and (4.5) (keeping all terms that are below $1$), we get the universal bound

$G(2\lfloor\tfrac{a_{k}}{2}\rfloor+1)\geq\frac{2}{\pi}e^{-f(a_{k})}\left(1-\frac{2}{3A}\right)\left(1-\frac{1}{A}\right)^{2}\frac{1}{2\lfloor\tfrac{a_{k}}{2}\rfloor+1},$

which holds regardless of the parity of $a_{k}$, and of whether $a_{k}+1<A/2$
or $a_{k}+1\geq A/2$. Combining this bound with (5.4) and (5.5), we finally
get

(5.6)
$\prod_{s=0}^{c-1}G(2s+1)\geq\left(\frac{2}{\pi}e^{-f(a_{k})}\right)^{c}\cdot\frac{\Pi_{1}\cdot\Pi_{2}}{\Pi_{3}},$

where

(5.7)
$\Pi_{1}=\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\left(1-\frac{2}{3Am_{s}}\right)\left(1-\frac{1}{Am_{s}}\right)^{2},$
(5.8)
$\Pi_{2}\quad=\prod_{s={\lfloor\frac{a_{k}}{2}\rfloor}+1}^{c-1}\mathrm{dist}(2s+1,\mathcal{U})\,,$

and

(5.9) $\Pi_{3}=\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}(2s+1).$

We proceed by bounding the three product terms $\Pi_{1}$, $\Pi_{2}$ and
$\Pi_{3}$ separately. In the following, we will make use of the inequalities

$\displaystyle\sqrt{2\pi
n}\,\left(\frac{n}{e}\right)^{n}\,\leq\,n!\,\leq\,e\sqrt{n}\left(\frac{n}{e}\right)^{n},$
(5.10) $\displaystyle\sqrt{2\pi
n}\,\left(\frac{2n}{e}\right)^{n}\,\leq\,(2n)!!\,\leq\,e\sqrt{n}\left(\frac{2n}{e}\right)^{n},$
$\displaystyle\frac{\sqrt{4\pi}}{e}\,\left(\frac{2n}{e}\right)^{n}\,\leq\,(2n-1)!!\,\leq\,\frac{e}{\sqrt{\pi}}\left(\frac{2n}{e}\right)^{n}$

which are valid for all $n\in\mathbb{N}$.

Starting with the first and simplest of the three, we observe that

$\Pi_{1}\geq\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\left(1-\frac{1}{Am_{s}}\right)^{3}.$

Recall that $m_{s}$ is the unique positive integer for which $|Am_{s}-(s+1)|$
is minimized. As $s$ runs through the values $\lfloor
a_{k}/2\rfloor,\ldots,c-1$, the integer $m_{s}$ runs through the values
$1,\ldots,q_{\ell}$, and each integer occurs at most $a_{k}+1$ times. It
follows that

$\Pi_{1}\geq\prod_{m=1}^{q_{\ell}}\left(1-\frac{1}{Am}\right)^{3(a_{k}+1)}.$

Using that

$x-x^{2}\,<\,\log(1+x)\,<\,x\qquad\text{ for all }x>-\frac{1}{2}$

it is straightforward to show that

$\prod_{m=1}^{q_{\ell}}\left(1-\frac{1}{Am}\right)\geq\exp\Big{(}\\!\\!-\frac{1}{A}\left(1+\log
q_{\ell}\right)-\frac{\pi^{2}}{6A^{2}}\Big{)}\geq\exp\Big{(}\\!\\!-\frac{1}{A}(1.14+\log
q_{\ell})\Big{)},$

where for the final inequality we have used that $a_{k}\geq 6$, and thus
$A>12$. Inserting this in the expression for $\Pi_{1}$ above, we get

(5.11)
$\Pi_{1}\geq\left(\frac{1}{e^{1.14}q_{\ell}}\right)^{\frac{3(a_{k}+1)}{A}}\geq\left(\frac{1}{e^{1.14}q_{\ell}}\right)^{\frac{3(a_{k}+1)}{2a_{k}}}\geq\frac{1}{e^{2}q_{\ell}^{2}}.$

Let us now consider $\Pi_{3}$, for which we need to find an upper bound. From
(5.10) it follows that

$\Pi_{3}=\frac{(2c-1)!!}{\left(2\lfloor\frac{a_{k}}{2}\rfloor-1\right)!!}\leq\frac{e^{2}}{2\pi}\left(\frac{2c}{e}\right)^{c}\left(\frac{2\lfloor\frac{a_{k}}{2}\rfloor}{e}\right)^{-\lfloor\frac{a_{k}}{2}\rfloor}.$

We observe that

(5.12)
$\left(\frac{2\lfloor\frac{a_{k}}{2}\rfloor}{e}\right)^{\lfloor\frac{a_{k}}{2}\rfloor}\geq\left(\frac{a_{k}-1}{e}\right)^{\frac{a_{k}-1}{2}}\geq\frac{1}{2}\left(\frac{e}{a_{k}}\right)^{\frac{1}{2}}\left(\frac{a_{k}}{e}\right)^{\frac{a_{k}}{2}},$

where in the last step we have used that
$(1-1/n)^{\frac{n}{2}}\geq\frac{1}{2}$ for $n\geq 2$. Inserting this in the
bound above we find that

(5.13)
$\Pi_{3}\leq\frac{e^{\frac{3}{2}}\sqrt{a_{k}}}{\pi}\left(\frac{e}{a_{k}}\right)^{\frac{a_{k}}{2}}\left(\frac{2c}{e}\right)^{c}.$

The estimation of the product $\Pi_{2}$ is far more intricate. When we analyse
the size of the factors comprising $\Pi_{2},$ the numbers

(5.14)
$R_{t}\,=\,\left\\{\frac{tp_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}\right\\}+t\left(\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}-\alpha_{\sigma_{k}}\right)-\frac{2bt}{q_{\ell}(\alpha_{\sigma_{k}})},\quad
t=1,\ldots,q_{\ell}-1$

appear naturally. The size of the product of $\|R_{t}\|$ and its relation with
$\Pi_{2}$ is stated in the following lemmas, the proofs of which are postponed
for Section 5.3.

######  Lemma 4.

When $\ell$ is even, the numbers $R_{t},t=1,\ldots,q_{\ell}-1$ given in (5.14)
satisfy

(5.15)
$\prod_{t=1}^{q_{\ell}-1}\|R_{t}\|\geq\frac{\sqrt{q_{\ell}}}{2(2e)^{q_{\ell}+1}}\cdot$

######  Lemma 5.

The product $\Pi_{2}$ defined in (5.8) is bounded below by

(5.16) $\quad\Pi_{2}\geq\frac{2}{5c}\cdot
2^{c-\lfloor\frac{a_{k}}{2}\rfloor}\cdot\left\lfloor\frac{a_{k}}{2}\right\rfloor!\cdot[(a_{k}-1)!]^{q_{\ell}-1}\cdot\prod_{t=1}^{q_{\ell}-1}\|R_{t}\|.$

We now find a bound for $\Pi_{2}$ using Lemmas 4 and 5. By (5.10) we have

$\Big{\lfloor}\frac{a_{k}}{2}\Big{\rfloor}!\geq\left(\frac{a_{k}}{2e}\right)^{\frac{a_{k}}{2}}\quad\text{
and
}\quad(a_{k}-1)!\geq\frac{2}{\sqrt{a_{k}}}\left(\frac{a_{k}}{e}\right)^{a_{k}}.$

The former is obvious when $a_{k}$ is even, and follows by an argument similar
to (5.12) when $a_{k}$ is odd and greater than $6$. Inserting these bounds in
(5.16) and employing Lemma 4, we get

(5.17) $\Pi_{2}\geq\frac{\sqrt{a_{k}q_{\ell}}}{20ec}\cdot
2^{c-a_{k}}\left(\frac{1}{e\sqrt{a_{k}}}\right)^{q_{\ell}}\cdot\left(\frac{a_{k}}{e}\right)^{a_{k}(q_{\ell}-\frac{1}{2})}.$

Combining the bounds established for $\Pi_{1}$, $\Pi_{2}$ and $\Pi_{3}$, we
conclude as follows.

######  Lemma 6.

Let $G$ be given in (4.1). Under the assumption that $a_{k}\geq 6$ we have

$\prod_{s=0}^{c-1}G(2s+1)\geq\left(\frac{2a_{k}}{\pi
e^{f(a_{k})}c}\right)^{c}\cdot\frac{2^{-a_{k}}\pi}{20e^{\frac{9}{2}}q_{\ell}^{\frac{3}{2}}c}\cdot\left(\frac{e}{a_{k}^{5/2}}\right)^{q_{\ell}}.$

###### Proof.

Inserting the bounds established for $\Pi_{1}$, $\Pi_{2}$ and $\Pi_{3}$ in
(5.11), (5.17) and (5.13) into (5.6), we get

$\displaystyle\prod_{s=0}^{c-1}G(2s+1)$ $\displaystyle\geq\left(\frac{2}{\pi
c}e^{1-f(a_{k})}\right)^{c}\cdot\frac{2^{-a_{k}}\pi}{20e^{\frac{9}{2}}q_{\ell}^{\frac{3}{2}}c}\left(\frac{1}{e\sqrt{a_{k}}}\right)^{q_{\ell}}\left(\frac{a_{k}}{e}\right)^{a_{k}q_{\ell}}$
$\displaystyle\geq\left(\frac{2a_{k}}{\pi
e^{f(a_{k})}c}\right)^{c}\cdot\frac{2^{-a_{k}}\pi}{20e^{\frac{9}{2}}q_{\ell}^{\frac{3}{2}}c}\left(\frac{e}{a^{\frac{5}{2}}}\right)^{q_{\ell}},$

where for the last inequality we have used that $c<(a_{k}+2)q_{\ell}$. ∎

We are now fully equipped to prove Theorem 5.

###### Proof of Theorem 5.

We recall from (5.1) and (5.2) that

$C_{k}^{c-2}=\frac{1+a^{2}}{c!}\left|\frac{G(1)^{c-1}G(1+2a^{2})}{\prod_{s=1}^{c-1}G(2s+1)}\right|\leq\frac{1}{c!}\cdot\frac{4Ae^{f(a_{k})}}{5}\cdot\frac{1}{\left|\prod_{s=1}^{c-1}G(2s+1)\right|}.$

Using Lemma 6 and the bound on $c!$ in (5.10), we thus get

$C_{k}^{c-2}\leq\frac{16Ae^{\frac{9}{2}}q_{\ell}^{\frac{3}{2}}\sqrt{c}e^{f(a_{k})}}{\pi\sqrt{2\pi}}\left(\frac{\pi
e^{1+f(a_{k})}}{2a_{k}}\right)^{c}\cdot
2^{a_{k}}\cdot\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{q_{\ell}}$

Recall that $A<2(a_{k}+2)$ and $q_{\ell}<c/a_{k}$. This allows us to bound the
first term on the right hand side by

$\frac{32e^{\frac{9}{2}}}{\pi\sqrt{2\pi}}\,\frac{(a_{k}+2)}{a_{k}^{3/2}}e^{f(a_{k})}c^{2}\leq\frac{32e^{\frac{9}{2}}}{\pi\sqrt{2\pi}}\,\frac{8}{6^{3/2}}e^{f(6)}c^{2}\leq
200e^{2.4}c^{2}.$

Inserting this in the expression above, we get

$C_{k}^{c-2}\leq\left(\frac{\pi
e^{1+f(a_{k})}}{2a_{k}}\right)^{c}\cdot(200e^{2.4}c^{2})\cdot
2^{a_{k}}\cdot\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{q_{\ell}}.$

Raising both sides to the power $1/c$ and using again that $c>a_{k}q_{\ell}$
completes the proof of Theorem 5. ∎

###### Proof of Corollary 4.

When $q_{\ell}=1$ we have $c=a_{k}+2$, and revisiting Lemma 6 we see that we
can obtain an improved bound on $\prod G(2s+1)$ in this case. From Corollary 2
and Remark 6 it follows that

$\prod_{s=0}^{c-1}G(2s+1)\geq\left(\frac{2}{\pi}e^{-g(a_{k})}\right)^{c}\cdot\frac{\Pi_{1}\cdot\Pi_{2}}{\Pi_{3}},$

with $\Pi_{1}$, $\Pi_{2}$ and $\Pi_{3}$ defined as in (5.7)–(5.9),
respectively, and $g(a_{k})$ given in (4.6). We keep the bounds for $\Pi_{1}$
and $\Pi_{3}$ established in (5.11) and (5.13), and note that the bound on
$\Pi_{2}$ in (5.16) simplifies to

$\Pi_{2}\geq\frac{2}{5c}\cdot
2^{c-\lfloor\frac{a_{k}}{2}\rfloor}\cdot\Big{\lfloor}\frac{a_{k}}{2}\Big{\rfloor}!\geq\frac{8}{5c}\left(\frac{a_{k}}{e}\right)^{\frac{a_{k}}{2}}.$

Inserting all three bounds above, we get

(5.18)
$\prod_{s=0}^{c-1}G(2s+1)\geq\frac{8\pi}{5e^{\frac{7}{2}}c\sqrt{a_{k}}}\left(\frac{e^{1-g(a_{k})}}{\pi
c}\right)^{c}\left(\frac{a_{k}}{e}\right)^{a_{k}}\geq\frac{8\pi}{5e^{\frac{3}{2}}c^{\frac{7}{2}}}\left(\frac{e^{-g(a_{k})}a_{k}}{\pi
c}\right)^{c},$

where we have used that $c=a_{k}+2>a_{k}$.

Again we have that

$C_{k}^{c-2}=\frac{1+a^{2}}{c!}\left|\frac{G(1)^{c-1}G(1+2a^{2})}{\prod_{s=1}^{c-1}G(2s+1)}\right|\leq\frac{1}{c!}\cdot\frac{4Ae^{f(a_{k})}}{5}\cdot\frac{1}{\left|\prod_{s=1}^{c-1}G(2s+1)\right|}.$

Inserting $A<2(a_{k}+2)=2c$ and the improved bound (5.18) on $\prod G(2s+1)$,
we get

$C_{k}^{c-2}\leq\frac{1}{c!}\cdot\frac{e^{f(a_{k})+\frac{3}{2}}c^{\frac{9}{2}}}{\pi}\left(\frac{\pi
e^{g(a_{k})}c}{a_{k}}\right)^{c}\leq\frac{e^{f(a_{k})+\frac{3}{2}}}{\pi\sqrt{2\pi}}\cdot
c^{4}\cdot\left(\frac{\pi e^{1+g(a_{k})}}{a_{k}}\right)^{c},$

where for the last inequality we have used the lower bound on $c!$ in (5.10).
The proof is completed by bounding $e^{f(a_{k})}$ by $e^{f(6)}$ and raising
both sides to the power $1/c$. ∎

###  5.2. Bounding $C_{k}$ when $\ell\equiv 1\pmod{2}$

Now let us consider the case of odd period lengths $\ell$. By (1.14), the
constant $C_{k}$ is then given by

(5.19)
$C_{k}^{c}=\frac{G(1)^{c}}{\prod\limits_{s=1}^{c}\left|s-a\right|}\prod_{s=0}^{c-1}|G(2a-2s-1)|^{-1},$

where $a$ and $c$ are given in (1.3) and $G$ is the function defined in (4.1).
Our goal is again to derive a bound for $C_{k}$ in the case when $k$ is the
index such that $\max_{j}a_{j}=a_{k}$.

The assumptions that $\ell$ is odd and $a_{k}\geq 6$ necessarily imply that
$c\geq 13;$ we make use of this inequality in the estimates that follow. The
definitions of $a$ and $c$ in (1.3) imply that

$\frac{1}{a-c}\,=\,\frac{\sqrt{c^{2}+4}+c}{2}<c+\frac{1}{13}\qquad\text{ and
}\qquad|a-s|\,>\,c-s,\quad s=1,\ldots,c-1$

and therefore

(5.20)
$\prod\limits_{s=1}^{c}\left|s-a\right|^{-1}\leq\frac{c+\frac{1}{13}}{(c-1)!}=\frac{c(c+\frac{1}{13})}{c!}\stackrel{{\scriptstyle\eqref{factorials}}}{{\leq}}\frac{\sqrt{c}(c+\frac{1}{13})}{\sqrt{2\pi}}\Big{(}\frac{e}{c}\Big{)}^{c}.$

We now seek a lower bound for the product

(5.21)
$\prod\limits_{s=0}^{c-1}|G(2a-2s-1)|=\prod_{s=0}^{c-1}|G(2s+1+2(a-c))|=\prod_{s=0}^{c-1}|G(2s+1-2b)|.$

We argue as in the case of even $\ell$, and use Theorem 4 to bound the terms
of this product. We use (4.5) to bound the factors of (5.21) with $0\leq
s\leq\lfloor\frac{a_{k}}{2}\rfloor-1$, which gives

$\prod_{s=0}^{\lfloor\frac{a_{k}}{2}\rfloor-1}|G(2s+1-2b)|\geq\left(\frac{2}{\pi
e^{f(a_{k})}}\right)^{\lfloor\frac{a_{k}}{2}\rfloor}.$

For the factors of (5.21) corresponding to
$\lfloor\frac{a_{k}}{2}\rfloor+1\leq s\leq c-1,$ the bound (4.3) gives

$\displaystyle\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}|G(2s+1-2b)|\geq$
$\displaystyle\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}\frac{2}{\pi
e^{f(a_{k})}}\\!\\!\left(1-\frac{2}{3Am_{s}}\right)\\!\\!\left(1-\frac{1}{Am_{s}}\right)^{2}\frac{\mathrm{dist}\big{(}2s+1-2b,\mathcal{U}\big{)}}{2s+1-2b},$

where the integers $m_{s}$ are defined in Theorem 4. For the factor
$|G(2\lfloor\tfrac{a_{k}}{2}\rfloor+1-2b)|$ we use the bound

$|G(2\lfloor\tfrac{a_{k}}{2}\rfloor+1-2b)|\geq\frac{2}{\pi}e^{-f(a_{k})}\left(1-\frac{2}{3A}\right)\left(1-\frac{1}{A}\right)^{2}\frac{1}{2\lfloor\tfrac{a_{k}}{2}\rfloor+1-2b}\cdot$

Combining the estimates above, we obtain

(5.22) $\prod_{s=0}^{c-1}|G(2s+1-2b)|\geq\left(\frac{2}{\pi
e^{f(a_{k})}}\right)^{c}\cdot\frac{\Pi_{1}\cdot\Pi_{2}^{\prime}}{\Pi_{3}^{\prime}},$

where $\Pi_{1}$ is the product defined in (5.7), while

(5.23)
$\Pi_{2}^{\prime}=\\!\\!\\!\prod_{s={\lfloor\frac{a_{k}}{2}\rfloor}+1}^{c-1}\\!\\!\mathrm{dist}(2s+1-2b,\,\mathcal{U})\quad\text{
and
}\quad\Pi_{3}^{\prime}=\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}(2s+1-2b).$

Let us first find a bound for $\Pi_{3}^{\prime}$ by comparing it with
$\Pi_{3}$ in (5.9). Note that the bound (5.13) on $\Pi_{3}$ does not depend on
the parity of $\ell$. Since

$-2b=\frac{4}{\sqrt{c^{2}+4}+c}<\frac{2}{c},$

we have

$\displaystyle\Pi_{3}^{\prime}$ $\displaystyle\leq$
$\displaystyle\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\Big{(}2s+1+\frac{2}{c}\Big{)}\,\,=\,\,\Pi_{3}\cdot\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\Big{(}1+\frac{2}{(2s+1)c}\Big{)}.$

We find that

$\displaystyle\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\Big{(}1+\frac{2}{(2s+1)c}\Big{)}$
$\displaystyle=\exp\Bigg{(}\sum_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\log\Big{(}1+\frac{2}{(2s+1)c}\Big{)}\Bigg{)}$
$\displaystyle<\exp\Bigg{(}\frac{2}{c}\sum_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\frac{1}{2s+1}\Bigg{)}<\exp\Big{(}\frac{\log(c-1)}{c}\Big{)}<e^{\frac{1}{5}},$

so in view of (5.13) we deduce that

(5.24)
$\Pi_{3}^{\prime}\leq\frac{e^{\frac{17}{10}}\sqrt{a_{k}}}{\pi}\left(\frac{e}{a_{k}}\right)^{\frac{a_{k}}{2}}\left(\frac{2c}{e}\right)^{c}.$

As for the even period case, the estimation of $\Pi_{2}^{\prime}$ is more
elaborate. When analysing the factors in $\Pi_{2}^{\prime}$, the numbers
$R_{t}+b$ appear naturally, where $R_{t}$ is defined as in (5.14). The size of
products over $\|R_{t}+b\|$ and its relation to $\Pi_{2}^{\prime}$ is stated
in the lemmas below. Note that these are analogues of Lemmas 4 and 5 for the
even period case. The proofs are postponed to Section 5.3.

######  Lemma 7.

When $\ell$ is odd, the numbers $R_{t},t=1,\ldots,q_{\ell}-1$ defined in
(5.14) satisfy

(5.25)
$\prod_{t=1}^{q_{\ell}-1}\|R_{t}+b\|\geq\frac{4\pi}{e^{3}(2e)^{q_{\ell}}}\,\cdot$

######  Lemma 8.

The product $\Pi_{2}^{\prime}$ defined in (5.23) is bounded below by

(5.26) $\Pi_{2}^{\prime}\geq
2^{c-\lfloor\frac{a_{k}}{2}\rfloor-1}\cdot\left\lfloor\frac{a_{k}}{2}\right\rfloor!\cdot[(a_{k}-1)!]^{q_{\ell}-1}\cdot\prod_{t=1}^{q_{\ell}-1}\|R_{t}+b\|$

Let us now bound $\Pi_{2}^{\prime}$ using Lemmas 7 and 8. We bound the
factorials in (5.26) using (5.10), and combined with Lemma 7 this gives

(5.27) $\Pi_{2}^{\prime}\geq\frac{3\sqrt{a_{k}}}{20}\cdot
2^{c-a_{k}}\left(\frac{1}{e\sqrt{a_{k}}}\right)^{q_{\ell}}\cdot\Big{(}\frac{a_{k}}{e}\Big{)}^{a_{k}(q_{\ell}-\frac{1}{2})}\cdot$

Combining the bounds for $\Pi_{1}$, $\Pi_{2}^{\prime}$ and $\Pi_{3}^{\prime}$
established in (5.11), (5.24) and (5.27), we get the following analogue of
Lemma 6. The proof is omitted.

######  Lemma 9.

Let $G$ be defined in (4.1). Under the assumption that $a_{k}\geq 6$ we have

$\prod_{s=0}^{c-1}|G(2s+1-2b)|\geq\left(\frac{2a_{k}}{\pi
e^{f(a_{k})}c}\right)^{c}\cdot\frac{3\pi\cdot
2^{-a_{k}}}{20e^{19/5}q_{\ell}^{2}}\cdot\left(\frac{e}{a_{k}^{5/2}}\right)^{q_{\ell}}.$

With Lemma 9 established, we are equipped to prove Theorem 6.

###### Proof of Theorem 6.

Recall from (5.19) that

$C_{k}^{c}\leq\left(\prod_{s=1}^{c}|s-a|\prod_{s=0}^{c-1}|G(2a-2s-1)|\right)^{-1},$

where we have used that $|G(1)|\leq 1$. Using (5.20) and Lemma 9, we get

$C_{k}^{c}\leq 40c^{\frac{3}{2}}\left(\frac{\pi
e^{1+f(a_{k})}}{2a_{k}}\right)^{c}\cdot 2^{a_{k}}\cdot
q_{\ell}^{2}\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{q_{\ell}},$

where we have replaced $(c+1/13)$ in (5.20) by the upper bound $27c/26$.
Raising both sides to the power $1/c$ and recalling that $c>a_{k}q_{\ell}$, we
get

$\displaystyle C_{k}$
$\displaystyle\leq\frac{\pi}{2a_{k}}e^{1+f(a_{k})}\left(40c^{\frac{3}{2}}\right)^{\frac{1}{c}}\cdot
2^{\frac{1}{q_{\ell}}}\cdot
q_{\ell}^{\frac{2}{a_{k}q_{\ell}}}\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{\frac{1}{a_{k}}}$
$\displaystyle\leq\frac{\pi}{2a_{k}}e^{1+f(a_{k})}\left(40c^{\frac{3}{2}}\right)^{\frac{1}{c}}\cdot
2^{\frac{1}{q_{\ell}}}\cdot a_{k}^{\frac{5}{2a_{k}}},$

where in the final step we have used that $q_{\ell}^{2/q_{\ell}}<e$. ∎

###  5.3. Estimation of $\Pi_{2}$ and $\Pi_{2}^{\prime}$

Let us now turn to the proofs of Lemmas 4 – 5 and 7 – 8, which lay the
foundation for the estimates of $\Pi_{2}$ and $\Pi_{2}^{\prime}$ above.

####  5.3.1. Proofs of Lemmas 4 and 7

We consider first Lemmas 4 and 7, which provide bounds on products of factors
$\|R_{t}\|$ and $\|R_{t}+b\|$, with $R_{t}$ given in (5.14). Recall that these
are analogous statements relevant to the cases of even and odd period lengths
$\ell$, respectively.

We treat first the proof of Lemma 4, and begin by examining the size of each
of the three terms appearing in the expression for $R_{t}$ in (5.14).

######  Lemma 10.

When $\ell$ is even, we have

$-\frac{1}{q_{\ell}q_{\ell+1}}\,<\,\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-\frac{2b}{q_{\ell}}\,<\,-\frac{a_{k}}{a_{k}+1}\,\frac{1}{q_{\ell}q_{\ell+1}}\,\,\cdot$

###### Proof.

First we observe that setting $m=1$ and $k=0$ in (1.6) yields
$p_{\ell}-q_{\ell}\alpha_{\sigma_{k}}=b$, and therefore

(5.28)
$\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)\,-\,\frac{2b}{q_{\ell}}\,\,=\,\,-\frac{b}{q_{\ell}}\,\cdot$

When $\ell$ is even, the definition of $b$ in (1.3) gives

$\displaystyle b$ $\displaystyle<$
$\displaystyle\frac{1}{c-1}\,=\,\frac{1}{q_{\ell+1}+p_{\ell}-1}\,<\,\frac{1}{q_{\ell+1}}$

and

$\displaystyle b\,$ $\displaystyle>$
$\displaystyle\,\frac{1}{c}\,\,=\,\,\frac{1}{q_{\ell+1}+p_{\ell}}\,\,=\,\,\frac{1}{q_{\ell+1}}\left(1+\frac{p_{\ell}}{q_{\ell+1}}\right)^{-1}$
$\displaystyle\,>\,$
$\displaystyle\frac{1}{q_{\ell+1}}\left(1+\frac{q_{\ell}}{q_{\ell+1}}\right)^{-1}\,\,>\,\,\frac{1}{q_{\ell+1}}\left(1+\frac{q_{\ell}}{a_{k}q_{\ell}}\right)^{-1}\,=\,\frac{1}{q_{\ell+1}}\,\,\frac{a_{k}}{a_{k}+1}\,\,,$

whence the claim follows. ∎

######  Lemma 11.

Let $\ell$ be even. For each $t=1,2,\ldots,q_{\ell}-1$ there exists a unique
integer $i=i_{t}\in\\{1,2,\ldots,q_{\ell}-1\\}$ such that

$\frac{i}{q_{\ell}}-\frac{1}{q_{\ell+1}}\,\,\leq\,\,\,\left\\{\frac{tp_{\ell}}{q_{\ell}}\right\\}+t\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-\frac{2bt}{q_{\ell}}\,\,\,\leq\,\,\frac{i}{q_{\ell}}-\frac{6}{7q_{\ell}q_{\ell+1}}\,\,\cdot$

Moreover, the correspondence $t\mapsto i_{t}$ is one-to-one.

###### Proof.

In view of the condition $a_{k}\geq 6$, by Lemma 10 we get

$-\frac{1}{q_{\ell+1}}\,<\,t\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-\frac{2bt}{q_{\ell}}<\,-\frac{6}{7q_{\ell}q_{\ell+1}},\quad
t=1,2,\ldots,q_{\ell}-1.$

Also since $(p_{\ell},q_{\ell})=1$, for each $t=1,2,\ldots,q_{\ell}-1$ there
exists a unique integer $1\leq i<q_{\ell}$ such that
$\left\\{{tp_{\ell}/q_{\ell}}\right\\}={i/q_{\ell}},$ and the correspondence
$t\mapsto i$ is one-to-one. The result follows. ∎

Lemma 11 gives the following estimates for the values of
$\|R_{t}\|,\,t=1,2,\ldots,q_{\ell}-1.$

######  Corollary 5.

Let $\ell$ be even and $R_{t}$ be as in (5.14). Then for every
$t=1,2,\ldots,q_{\ell}-1$ there exists an integer
$j_{t}\in\left\\{1,2,\ldots,\lfloor\tfrac{1}{2}q_{\ell}\rfloor\right\\}$ such
that

(5.29)
$\|R_{t}\|\,\geq\,\frac{j_{t}}{q_{\ell}}-\frac{1}{q_{\ell+1}}\quad\text{ or
}\quad\|R_{t}\|\,\geq\,\frac{j_{t}}{q_{\ell}}+\frac{6}{7q_{\ell}q_{\ell+1}}\,\cdot$

The first inequality in (5.29) holds when the residue of
$tp_{\ell}(\alpha_{\tau_{k}})$ modulo $q_{\ell}(\alpha_{\sigma_{k}})$ is one
of $1,2,\ldots,\lfloor\tfrac{1}{2}q_{\ell}(\alpha_{\tau_{k}})\rfloor$ while
the second holds when the residue of $tp_{\ell}(\alpha_{\tau_{k}})$ modulo
$q_{\ell}(\alpha_{\sigma_{k}})$ is one of
$\lfloor\tfrac{1}{2}q_{\ell}(\alpha_{\tau_{k}})\rfloor+1,\ldots,q_{\ell}(\alpha_{\tau_{k}})-1.$
Moreover, there exists at most one integer $t$ such that
$j_{t}=\lfloor\tfrac{1}{2}q_{\ell}(\alpha_{\tau_{k}})\rfloor$, and the
correspondence $t\mapsto j_{t}$ is “at most two-to-one”.

With Corollary 5 established, we are equipped to complete the proof of Lemma
4.

###### Proof of Lemma 4.

We consider first the case $q_{\ell}\geq 4$ and for abbreviation set
$Q=\lfloor\frac{1}{2}q_{\ell}\rfloor$. From Corollary 5 it follows that

$\displaystyle\prod_{t=1}^{q_{\ell}-1}\|R_{t}\|$ $\displaystyle\geq$
$\displaystyle\left(\frac{1}{2}-\frac{1}{q_{\ell+1}}\right)\prod_{j=1}^{Q-1}\\!\\!\left(\frac{j}{q_{\ell}}-\frac{1}{q_{\ell+1}}\right)\left(\frac{j}{q_{\ell}}+\frac{6}{7q_{\ell}q_{\ell+1}}\right)$
$\displaystyle\geq$
$\displaystyle\left(\frac{1}{2}-\frac{1}{6}\right)\frac{1}{q_{\ell}}\left(1-\frac{q_{\ell}}{q_{\ell+1}}\right)\prod_{j=2}^{Q-1}\left(\frac{j}{q_{\ell}}-\frac{1}{q_{\ell+1}}\right)\prod_{j=1}^{Q-1}\frac{j}{q_{\ell}}$
$\displaystyle\geq$
$\displaystyle\frac{5}{18q_{\ell}}\prod_{j=1}^{Q-2}\left(\frac{j}{q_{\ell}}+\frac{1}{2q_{\ell}}\right)\prod_{j=1}^{Q-1}\frac{j}{q_{\ell}}\qquad\left(\text{because
}1-\frac{q_{\ell}}{q_{\ell+1}}>\frac{5}{6}\right).$

We may therefore continue to find

$\displaystyle\prod_{t=1}^{q_{\ell}-1}\|R_{t}\|$ $\displaystyle\geq$
$\displaystyle\frac{5}{18}\frac{(2Q-3)!!}{(2q_{\ell})^{Q-1}}\frac{(Q-1)!}{(q_{\ell})^{Q-1}}=\frac{5}{18}\frac{(2Q-2)!}{(2q_{\ell})^{2(Q-1)}}$
$\displaystyle\stackrel{{\scriptstyle\eqref{factorials}}}{{\geq}}$
$\displaystyle\frac{5}{18}\sqrt{2\pi\cdot
2(Q-1)}\left(\frac{Q-1}{q_{\ell}e}\right)^{2(Q-1)}$ $\displaystyle\geq$
$\displaystyle\frac{5}{18}\sqrt{2\pi(q_{\ell}-3)}\left(\frac{q_{\ell}-3}{2q_{\ell}e}\right)^{q_{\ell}-2}$
$\displaystyle\geq$
$\displaystyle\frac{5}{18}\sqrt{2\pi(q_{\ell}-3)}\left(\frac{1}{2e}\right)^{q_{\ell}-2}\left(1-\frac{3}{q_{\ell}}\right)^{q_{\ell}-3}\left(1-\frac{3}{q_{\ell}}\right)$
$\displaystyle\geq$ $\displaystyle\frac{5}{18}\,\frac{\sqrt{2\pi
q_{\ell}}}{(2e)^{q_{\ell}+1}}\cdot$

In the last step we have used the bound
$\tfrac{q_{\ell}-3}{q_{\ell}}\geq\tfrac{1}{4}$ as well as the inequality
$\left(1+\frac{r}{n}\right)^{n}<e^{r}$ for $n>r$, which implies that
$\left(1-\frac{r}{n}\right)^{n-r}=\left(\frac{n}{n-r}\right)^{-(n-r)}>e^{-r}$
whenever $n>r$.

We have now shown that (5.15) holds for $q_{\ell}\geq 4$, and proceed by
considering smaller values of $q_{\ell}$. When $q_{\ell}=1$, the product in
question is empty and hence by definition equals $1$. When $q_{\ell}=2$, the
product consists only of the factor $\|R_{1}\|$; by (5.29) and the assumption
$a_{k}\geq 6$ we have $\|R_{1}\|\geq 1/3,$ so (5.15) is still valid. Finally,
when $q_{\ell}=3$ again (5.29) implies $\|R_{1}\|\|R_{2}\|\geq 1/12$ which
proves that (5.15) is true also in this case. This completes the proof of
Lemma 4. ∎

We now turn to Lemma 7. We will not write out the proof of this result in full
detail, as it is very similar to that of Lemma 4. We simply point out that it
is a consequence of the following results, analogous to Lemmas 10–11 and
Corollary 5 above.

######  Lemma 12.

When $\ell$ is odd, we have

(5.30)
$\frac{a_{k}}{a_{k}+1}\frac{1}{q_{\ell}q_{\ell+1}}\,<\,\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-\frac{2b}{q_{\ell}}\,<\,\frac{1}{q_{\ell}q_{\ell+1}}\,\cdot$

######  Lemma 13.

Let $\ell$ be odd. For each $t=1,2,\ldots,q_{\ell}-1$ there exists a unique
integer $i=i_{t}\in\\{1,2,\ldots,q_{\ell}-1\\}$ such that

$\frac{i}{q_{\ell}}-\frac{1}{q_{\ell+1}}\,\,\leq\,\,\,\|R_{t}+b\|\,\,\,\leq\,\,\frac{i}{q_{\ell}}+\frac{1}{7q_{\ell+1}}\,\,\cdot$

Moreover, the correspondence $t\mapsto i_{t}$ is one-to-one.

######  Corollary 6.

Let $\ell$ be odd and $R_{t}$ be as in (5.14). Then for every
$t=1,2,\ldots,q_{\ell}-1$ there exists an integer
$j_{t}\in\left\\{1,2,\ldots,\lfloor\tfrac{1}{2}q_{\ell}\rfloor\right\\}$ such
that

(5.31)
$\|R_{t}+b\|\,\geq\,\frac{j_{t}}{q_{\ell}}-\frac{1}{q_{\ell+1}}\quad\text{ or
}\quad\|R_{t}+b\|\,\geq\,\frac{j_{t}}{q_{\ell}}-\frac{1}{7q_{\ell+1}}\,\,\cdot$

The first inequality in (5.31) holds when the residue of
$tp_{\ell}(\alpha_{\tau_{k}})$ modulo $q_{\ell}(\alpha_{\sigma_{k}})$ is one
of $1,2,\ldots,\lfloor\tfrac{1}{2}q_{\ell}(\alpha_{\tau_{k}})\rfloor$ while
the second holds when the residue of $tp_{\ell}(\alpha_{\tau_{k}})$ modulo
$q_{\ell}(\alpha_{\sigma_{k}})$ is one of
$\lfloor\tfrac{1}{2}q_{\ell}(\alpha_{\tau_{k}})\rfloor+1,\ldots,q_{\ell}(\alpha_{\tau_{k}})-1.$
Moreover, there exists at most one integer $t$ such that
$j_{t}=\lfloor\tfrac{1}{2}q_{\ell}(\alpha_{\tau_{k}})\rfloor$, and the
correspondence $t\mapsto j_{t}$ is “at most two-to-one”.

####  5.3.2. Proofs of Lemmas 5 and 8

Now let us treat Lemmas 5 and 8. Recall that these provide lower bounds on the
factors $\Pi_{2}$ and $\Pi_{2}^{\prime}$, given in (5.8) and (5.23). As in the
previous subsection, we will provide a full proof of Lemma 5, and then simply
sketch the proof of Lemma 8.

###### Proof of Lemma 5.

Recall from (5.8) that $\Pi_{2}$ is defined as

$\Pi_{2}=\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor}^{c-1}\text{dist}\left(2s+1,\mathcal{U}\right),\qquad
u_{k}(t)=\frac{2t}{|e_{k}c_{k}|}-2\\{t\alpha_{\sigma_{k}}\\}+1.$

For each $s\geq 0$ let $t_{s}\geq 1$ be the unique positive integer such that

(5.32) $\mathrm{dist}(2s+1,\mathcal{U})=|2s+1-u_{k}(t_{s})|.$

Then

(5.33) $\displaystyle\Pi_{2}$ $\displaystyle=$
$\displaystyle\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}\\!\\!|2s+1-u_{k}(t_{s})|\,=\,2^{c-\lfloor\frac{a_{k}}{2}\rfloor-1}\,\prod_{t=1}^{\infty}\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=t}\\!\\!\tfrac{1}{2}|u_{k}(t)-(2s+1)|\,\,.$

We now analyze the factors appearing in (5.33). For any $t\geq 1$ and
$s=0,1,\ldots,c-1$,

(5.34) $\displaystyle\frac{1}{2}\left(u_{k}(t)-(2s+1)\right)$ $\displaystyle=$
$\displaystyle\frac{t}{|c_{k}e_{k}|}-\\{t\alpha_{\sigma_{k}}\\}-s$
$\displaystyle\stackrel{{\scriptstyle\eqref{ckekformula}}}{{=}}$
$\displaystyle
ta_{k}+t\left(\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\right)-\\{t\alpha_{\sigma_{k}}\\}-s$
$\displaystyle=$
$\displaystyle\underbrace{ta_{k}+\left\lfloor\frac{tp_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}\right\rfloor+\lfloor
t\alpha_{\sigma_{k}}\rfloor-s}_{\text{$\in\mathbb{Z}$ }}\,\,+\,\,R_{t},$

where the terms $R_{t}$ are as in (5.14), with this definition extended to all
values $t\geq 1.$ When $t=q_{\ell}$ and $s=c-1$ the right hand side of (5.34)
is equal to

$a_{k}q_{\ell}+p_{\ell}+\lfloor
q_{\ell}\alpha_{\sigma_{k}}\rfloor-(c-1)+q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-2b=\\\
=a_{k}q_{\ell}+q_{\ell-1}+\left\lfloor
q_{\ell}\left(\alpha_{\sigma_{k}}-\frac{p_{\ell}}{q_{\ell}}\right)+p_{\ell}\right\rfloor-(c-1)+q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-2b=\\\
=q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-2b\,\stackrel{{\scriptstyle\eqref{ltherror}}}{{=}}-b\in\left(-\frac{1}{q_{\ell+1}},0\right).$

This observation implies that $t_{c-1}=q_{\ell}$. Since $t_{s}$ is increasing
in $s$, $t=q_{\ell}$ is the maximum value of $t$ for which the product in
(5.33) is non-empty, and we may write

(5.35)
$\Pi_{2}=2^{c-\lfloor\frac{a_{k}}{2}\rfloor-1}\,\prod_{t=1}^{q_{\ell}}\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=t}\tfrac{1}{2}|u_{k}(t)-(2s+1)|.$

We now want to find lower bounds for the factors in the innermost product in
(5.35) for each $1\leq t\leq q_{\ell}.$ When $t=1$, the values of $s$
appearing in the product

$\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=1}\\!\tfrac{1}{2}|u_{k}(1)-(2s+1)|$

are the integers $s\geq 1+\lfloor\frac{a_{k}}{2}\rfloor$ for which
$2s+1<\tfrac{1}{2}\left(u_{k}(1)+u_{k}(2)\right),$ or equivalently

$1+\left\lfloor\frac{a_{k}}{2}\right\rfloor\,\leq\,s\,<\,\frac{3a_{k}}{2}+\frac{3}{2}\left(\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\right)-\frac{\\{\alpha_{\sigma_{k}}\\}+\\{2\alpha_{\sigma_{k}}\\}}{2}\,\cdot$

The number of integers $s$ in any interval of the form $[n,\kappa)$, where $n$
is an integer, is $1+\lfloor\kappa\rfloor-n$. In our case we have
$n=1+\lfloor\frac{a_{k}}{2}\rfloor$ and

$\displaystyle\kappa$ $\displaystyle=$
$\displaystyle\frac{3a_{k}}{2}+\frac{3}{2}\left(\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\right)-\frac{\\{\alpha_{\sigma_{k}}\\}+\\{2\alpha_{\sigma_{k}}\\}}{2}$
$\displaystyle\geq$
$\displaystyle\frac{3a_{k}}{2}+\frac{3}{2}\left(\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\right)-\frac{3}{2}{\\{\alpha_{\sigma_{k}}\\}}$
$\displaystyle=$
$\displaystyle\frac{3a_{k}}{2}+\frac{3}{2}\Big{(}\underbrace{\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}-\alpha_{\sigma_{k}}}_{\frac{1}{2q_{\ell}q_{\ell+1}}\leq\ldots\leq\frac{1}{q_{\ell}q_{\ell+1}}}+\underbrace{\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}}_{>0}\Big{)},$

so there are at least $a_{k}$ values of $s$ for which $t_{s}=1$. Consider now
the corresponding factors in the innermost product in (5.35). From (5.34) and
Lemma 11 it follows that the minimal factor is equal to $\|R_{1}\|$, whereas
the remaining factors can be bounded below by the values $1,2,\ldots,a_{k}-1.$
Therefore

(5.36)
$\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=1}\\!\\!\tfrac{1}{2}|u_{k}(1)-(2s+1)|\,\geq\,\|R_{1}\|\cdot(a_{k}-1)!$

We then examine the innermost product of (5.35) for $1<t<q_{\ell}$. The
factors appearing are those corresponding to integers $s$ such that

$\tfrac{1}{2}\left(u_{k}(t-1)+u_{k}(t)\right)\,<\,2s+1\,<\,\tfrac{1}{2}\left(u_{k}(t)+u_{k}(t+1)\right),$

or equivalently

$\frac{2t-1}{|c_{k}e_{k}|}-\\{(t-1)\alpha_{\sigma_{k}}\\}-\\{t\alpha_{\sigma_{k}}\\}+1<2s+1<\frac{2t+1}{|c_{k}e_{k}|}-\\{t\alpha_{\sigma_{k}}\\}-\\{(t+1)\alpha_{\sigma_{k}}\\}+1.$

Observe that the number of integers in an interval $(\alpha,\beta)$ is at
least $\lfloor\beta-\alpha\rfloor.$ Here the length of the interval of
possible values of $s$ is

$\displaystyle\left(\frac{t+\frac{1}{2}}{|c_{k}e_{k}|}-\frac{\\{t\alpha_{\sigma_{k}}\\}-\\{(t+1)\alpha_{\sigma_{k}}\\}}{2}\right)$
$\displaystyle-$
$\displaystyle\left(\frac{t-\frac{1}{2}}{|c_{k}e_{k}|}-\frac{\\{t\alpha_{\sigma_{k}}\\}-\\{(t-1)\alpha_{\sigma_{k}}\\}}{2}\right)=$
$\displaystyle=$
$\displaystyle\frac{1}{|c_{k}e_{k}|}+\frac{\\{(t-1)\alpha_{\sigma_{k}}\\}+\\{(t+1)\alpha_{\sigma_{k}}\\}}{2}$
$\displaystyle\geq$ $\displaystyle
a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}-\alpha_{\sigma_{k}}$
$\displaystyle\geq$ $\displaystyle a_{k},$

so there exist at least $a_{k}$ values of $s$. The minimal factor
$\frac{1}{2}|u_{k}(t)-(2s+1)|$ is equal to $\|R_{t}\|$, while the remaining
ones can be bounded from below by $1,2,\ldots,a_{k}-1.$ We thus have

(5.37)
$\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=t}\\!\\!\tfrac{1}{2}|u_{k}(t)-(2s+1)|\,\geq\,\|R_{t}\|\cdot(a_{k}-1)!,\,\qquad
t=2,\ldots,q_{\ell}-1.$

Finally we deal with the innermost product in (5.35) when $t=q_{\ell}$. The
values of $s$ appearing in the product are those for which

$\displaystyle 2c-1\,\geq\,2s+1$ $\displaystyle>$
$\displaystyle\tfrac{1}{2}\left(u_{k}(q_{\ell})+u_{k}(q_{\ell}-1)\right)$
$\displaystyle=$ $\displaystyle
2(c-2b)-\left(a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\,\right)$
$\displaystyle\qquad\qquad-\\{q_{\ell}\alpha_{\sigma_{k}}\\}-\\{(q_{\ell}-1)\alpha_{\sigma_{k}}\\}+1,$

which is equivalent to

$\displaystyle c-1\,\,\geq\,\,s$ $\displaystyle>$ $\displaystyle
c-\frac{1}{2}\left(a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\,\right)$
$\displaystyle\qquad-\frac{\\{q_{\ell}\alpha_{\sigma_{k}}\\}+\\{(q_{\ell}-1)\alpha_{\sigma_{k}}\\}}{2}-2b\,.$

For any $\kappa>1$, the interval $(c-\kappa,c-1]$ contains precisely
$\lfloor\kappa\rfloor$ integers. Here we need to apply this observation with

$\displaystyle\kappa$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\,\right)+\frac{\\{q_{\ell}\alpha_{\sigma_{k}}\\}+\\{(q_{\ell}-1)\alpha_{\sigma_{k}}\\}}{2}+2b$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\left(a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\,\right)+\frac{\left(1-\frac{1}{q_{\ell+1}}\right)+\left(1-\frac{1}{q_{\ell+1}}-\alpha_{\sigma_{k}}\right)}{2}+2b$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\left(a_{k}+\frac{p_{\ell}(\alpha_{\sigma_{k}})}{q_{\ell}(\alpha_{\sigma_{k}})}-\alpha_{\sigma_{k}}+\frac{p_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}-\frac{2b}{q_{\ell}(\alpha_{\tau_{k}})}\,+2-\frac{2}{q_{\ell+1}}+4b\right)$

hence there must be at least $1+\left\lfloor\dfrac{a_{k}}{2}\right\rfloor$
such factors.

When $t=q_{\ell}$ and $s=c-1$, we saw already that

$\displaystyle\tfrac{1}{2}|u_{k}(q_{\ell})-2c+1|$
$\displaystyle=|q_{\ell}a_{k}+p_{\ell}(\alpha_{\tau_{k}})+p_{\ell}(\alpha_{\sigma_{k}})-c+R_{q_{\ell}}|=|R_{q_{\ell}}|=b.$

For the remaining values of $s$ with $t_{s}=q_{\ell}$, we bound the factors
$\tfrac{1}{2}|u_{k}(q_{\ell})-(2s+1)|$ from below by

$r\,-\,\frac{1}{q_{\ell+1}},\qquad
r=1,2,\ldots,\left\lfloor\frac{a_{k}}{2}\right\rfloor.$

Consequently,

(5.38) $\displaystyle\mathop{\prod_{\lfloor a_{k}/2\rfloor+1\leq
s<c-1}}_{t_{s}=q_{\ell}}\tfrac{1}{2}|u_{k}(q_{\ell})-(2s-1)|$
$\displaystyle\geq$ $\displaystyle b\prod_{r=1}^{\lfloor
a_{k}/2\rfloor}\\!\\!\left(r-\frac{1}{q_{\ell+1}}\right)$ $\displaystyle=$
$\displaystyle b\left\lfloor\frac{a_{k}}{2}\right\rfloor!\prod_{r=1}^{\lfloor
a_{k}/2\rfloor}\\!\\!\left(1-\frac{1}{rq_{\ell+1}}\right)$ $\displaystyle\geq$
$\displaystyle\frac{1}{c}\left\lfloor\frac{a_{k}}{2}\right\rfloor!\Big{(}1-\sum_{r=1}^{\lfloor
a_{k}/2\rfloor}\frac{1}{rq_{\ell+1}}\Big{)}$ $\displaystyle\geq$
$\displaystyle\frac{4}{5c}\left\lfloor\frac{a_{k}}{2}\right\rfloor!\quad.$

On combining (5.35) with (5.36), (5.37) and (5.38), we obtain inequality
(5.16). This completes the proof of Lemma 5. ∎

Let us now give an outline of the proof of Lemma 8. Recall that this result
gives a bound on $\Pi_{2}^{\prime}$ defined in (5.23), relevant to the case of
odd period lengths $\ell$. Using the same arguments that gave the bound for
$\Pi_{2}$ above, we find that

$\Pi_{2}^{\prime}=2^{c-\lfloor\frac{a_{k}}{2}\rfloor-1}\,\prod_{t=1}^{q_{\ell}}\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=t}\tfrac{1}{2}|u_{k}(t)-(2s+1-2b)|,$

where $t_{s}$ is defined in (5.32). We now proceed as before, and find lower
bounds on the factors in the innermost product for each $1\leq t\leq
q_{\ell}$. According to (5.34), the factors appearing in the product are

(5.39) $\displaystyle\frac{1}{2}\left(u_{k}(t)-(2s+1-2b)\right)$
$\displaystyle=$
$\displaystyle\underbrace{ta_{k}+\left\lfloor\frac{tp_{\ell}(\alpha_{\tau_{k}})}{q_{\ell}(\alpha_{\tau_{k}})}\right\rfloor+\lfloor
t\alpha_{\sigma_{k}}\rfloor-s}_{\text{$\in\mathbb{Z}$ }}\,\,+\,\,R_{t}+b,$

with $R_{t}$ given in (5.14).

When $\ell$ is odd, $t=q_{\ell}$ and $s=c-1$, the right hand side of (5.39) is
equal to

(5.40) $a_{k}q_{\ell}+p_{\ell}+\lfloor
q_{\ell}\alpha_{\sigma_{k}}\rfloor-c+1+q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-b=\\\
=a_{k}q_{\ell}+q_{\ell-1}+\left\lfloor
q_{\ell}\left(\alpha_{\sigma_{k}}-\frac{p_{\ell}}{q_{\ell}}\right)+p_{\ell}\right\rfloor-c+1+q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-b\\\
=q_{\ell+1}+p_{\ell}-(q_{\ell+1}+p_{\ell})+1+q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-b\\\
=1+q_{\ell}\left(\frac{p_{\ell}}{q_{\ell}}-\alpha_{\sigma_{k}}\right)-b\,\stackrel{{\scriptstyle\eqref{ltherror}}}{{=}}1.\qquad\qquad\quad$

It follows that

$\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=q_{\ell}}\tfrac{1}{2}|u_{k}(q_{\ell})-(2s+1-2b)|\,\,\geq\,\,\left\lfloor\frac{a_{k}}{2}\right\rfloor!\quad.$

For $t=1,\ldots,q_{\ell}-1$, we argue as for the even period case in Lemma 5,
and obtain

$\mathop{\prod_{s=\lfloor\frac{a_{k}}{2}\rfloor+1}^{c-1}}_{t_{s}=t}\\!\\!\tfrac{1}{2}|u_{k}(t)-(2s+1-2b)|\,\geq\,\|R_{t}+b\|\cdot(a_{k}-1)!,\,\qquad
t=1,\ldots,q_{\ell}-1.$

Combining these two estimates, we deduce that

$\displaystyle\Pi_{2}^{\prime}$ $\displaystyle\geq$ $\displaystyle
2^{c-\lfloor\frac{a_{k}}{2}\rfloor-1}\cdot[(a_{k}-1)!]^{q_{\ell}-1}\cdot\left\lfloor\frac{a_{k}}{2}\right\rfloor!\cdot\prod_{t=1}^{q_{\ell}-1}\|R_{t}+b\|,$

which confirms Lemma 8.

##  6\. Proof of Theorem 3

With Theorems 5 and 6 established, let us now prove Theorem 3. Recall that by
Theorem 2, Theorem 3 is verified if we can show that

$C_{k}=\lim_{m\to\infty}P_{q_{m\ell+k}}(\alpha)<1$

for every $\alpha=[0;\overline{a_{1},a_{2},\ldots,a_{\ell}}]$ with
$a_{k}=\max_{j}a_{j}\geq 23$.

Assume first that the period length $\ell$ is odd. Then $q_{\ell}>1$, and from
Theorem 6 it follows that

$C_{k}\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}\left(40c^{\frac{3}{2}}\right)^{\frac{1}{c}}a_{k}^{\frac{5}{2a_{k}}}.$

The factor $(40c^{3/2})^{1/c}$ is decreasing in $c$ (within the range of
relevant values of $c$), so we may safely use the fact that $c>a_{k}q_{\ell}$
to obtain

$\displaystyle C_{k}$
$\displaystyle\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}\left(40a_{k}^{\frac{3}{2}}q_{\ell}^{\frac{3}{2}}\right)^{\frac{1}{a_{k}q_{\ell}}}a_{k}^{\frac{5}{2a_{k}}}$
$\displaystyle\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}\left(40a_{k}^{\frac{3}{2}}\right)^{\frac{1}{a_{k}q_{\ell}}}2^{\frac{1}{a_{k}}}a_{k}^{\frac{5}{2a_{k}}},$

where for the last inequality we have used that $q_{\ell}^{3/2q_{\ell}}<2$.
This expression is decreasing in $q_{\ell}$, so we insert $q_{\ell}=2$ to
obtain

$C_{k}\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}\left(160a_{k}^{\frac{13}{2}}\right)^{\frac{1}{2a_{k}}}.$

The right hand side is a decreasing function of $a_{k}$, and it can be easily
verified that $C_{k}\leq 1$ whenever $a_{k}\geq 22$. This proves Theorem 3 for
odd period lengths $\ell$.

Now assume that the period length $\ell$ is even, and consider first the case
$q_{\ell}>1$. By Theorem 5 we then have

$C_{k}^{\frac{c-2}{c}}\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}(200e^{2.4}c^{2})^{\frac{1}{c}}\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{\frac{1}{a_{k}}}.$

The factor $(200e^{2.4}c^{2})^{1/c}$ is again decreasing in $c$, so we replace
$c$ by $a_{k}q_{\ell}$ to obtain

$\displaystyle C_{k}^{\frac{c-2}{c}}$
$\displaystyle\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}(200e^{2.4}a_{k}^{2}q_{\ell}^{2})^{\frac{1}{a_{k}q_{\ell}}}\left(\frac{a_{k}^{\frac{5}{2}}}{e}\right)^{\frac{1}{a_{k}}}$
$\displaystyle\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}(200e^{2.4}a_{k}^{2})^{\frac{1}{a_{k}q_{\ell}}}\cdot
a_{k}^{\frac{5}{2a_{k}}},$

where we have used that $q_{\ell}^{2/q_{\ell}}<e$. This expression is
decreasing in $q_{\ell}$, so we insert $q_{\ell}=2$ to obtain

$C_{k}^{\frac{c-2}{c}}\leq\frac{\pi}{\sqrt{2}a_{k}}e^{1+f(a_{k})}\left(200e^{2.4}a_{k}^{7}\right)^{\frac{1}{2a_{k}}}.$

One can again verify that the right hand side is below one whenever $a_{k}\geq
23$. This proves Theorem 3 for even $\ell$ in the case $q_{\ell}>1$.

Finally, we consider the case $q_{\ell}=1$. Recall that this can only occur if
$\ell=2$ and $\alpha=[0;\overline{a_{1},a_{2}}]$, with either $a_{1}=1$ or
$a_{2}=1$. By Corollary 4, we then have the bound

$C_{k}^{\frac{c-2}{c}}\leq\frac{\pi}{a_{k}}e^{1+g(a_{k})}\left(6.2(a_{k}+2)^{4}\right)^{\frac{1}{a_{k}+2}},$

where $g(a_{k})\leq 3.3/a_{k}+0.1$ and one can again verify that the right
hand side is below one whenever $a_{k}\geq 21$. This verifies Theorem 3 when
$q_{\ell}=1$, and completes the proof.

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