# Extended Levett trigonometric series

Robert Reynolds  Department of Mathematics and Statistics, York University,
Toronto, ON, Canada, M3J1P3<EMAIL_ADDRESS>
###### Abstract.

An extension of two finite trigonometric series is studied to derive closed
form formulae involving the Hurwitz-Lerch zeta function. The trigonometric
series involves angles with a geometric series involving the powers of 3.
These closed formulae are used to derive composite finite and infinite series
involving special functions, trigonometric functions and fundamental
constants. A short table summarizing some interesting results is produced.

###### Key words and phrases:

Levett, Hurwitz-Lerch zeta function, contour integral, Catalan constant,
Glaisher-Kinkelin constant, gamma function

###### 2020 Mathematics Subject Classification:

Primary 30E20, 33-01, 33-03, 33-04

##  1\. History

RAWDON LEVETT (February 2, 1837 - February 22, 1914.), as aptly noted in The
Times, [1], was a distinguished educator, ranking among the most esteemed
teachers of his era. He passed away at the age of seventy-nine in his
residence in Colwyn Bay in February. Following his graduation as the Eleventh
Wrangler in 1865, he briefly taught at Rossall before assuming the role of a
mathematics instructor at King Edward’s School in Birmingham. It was here that
he dedicated his career, ultimately becoming the Head of the Mathematics
Department. Despite being offered the position of Head Master, he declined due
to health concerns. Approximately two decades ago, he retired from his
teaching profession. With his passing, the last of a remarkable trio of
educators Vardy, Hunter Smith, and Levett faded away. These individuals not
only left their indelible mark on their respective schools (which had
previously nurtured prominent figures like Lightfoot, Westcott, and Benson)
but also contributed significantly to the honorable legacy they inherited. In
the 19th century, Rawdon Levett made a significant contribution to education,
particularly in the area of mathematical training. He was honoured for his
skill as a teacher and dedication to his trade. At Birmingham’s King Edward’s
School, Levett taught mathematics and had a lasting impression on his
students. His legacy includes his commitment to teaching and his influence on
education.

##  2\. Introduction

In 1892 Rawdon Levett, published his landmark book entitled ”The elements of
plane geometry” [2], which was not written as a text-book, but rather as a
detailed syllabus of proofs, which is divided into three sections. These
sections respectively address the concepts related to arithmetic, real
algebra, and complex quantities. This arrangement is quite natural and offers
the advantage of introducing new terminology and formulas associated with the
subject before students encounter the challenge of applying signs to represent
the direction and orientation of lines. Part I is made even simpler by
deferring the discussion of circular angle measurement. For various practical
applications, such as surveying and basic mechanics involving trigonometry,
the concise introduction provided in Part I can prove to be valuable to
interested readers.

The focus of this research is in chapter XIV dedicated to Factors with respect
to the fundamental theorem on trigonometrical factors. In this section of the
book Levett produces many exciting finite and infinite trigonometric series.
We focus on two obscure but interesting finite series involving secant-sine
and cosecant-cosine series. In this current paper we applied a contour
integral method to two obscure finite trigonometric series tabled in the book
by Levett [2]. The forms of these two series are of particular interest as the
angle in these series are of a geometric series form involving 3 raised to an
integer power. These types of formulae have been studied by the author in
previous work with different geometric angular forms. In this work we will
look at the following finite series forms; (i) Finite sum of Hurwitz-Lerch
zeta function, (ii) Finite sum of composite special functions, (iii) Finite
products of trigonometric functions, (iv) Finite products of quotient gamma
functions, (v) Quotient trigonometric functions with reciprocal angles and
(vi) Table of interesting results. Some of the finite and infinite products
involving the gamma function can be simplified using the Gauss’s
multiplication formula [3].

A special function is a mathematical function, typically named after an early
researcher who studied its properties, and it serves a specific purpose in the
area of mathematical physics or various branches of mathematics. Notable
instances of such functions comprise the gamma function, hypergeometric
function, Whittaker function, Hurwitz-Lerch zeta and Meijer G-function.
Applications of special functions are prevalent in various areas of
mathematics and related fields, such as engineering, quantum physics,
astronomy and combinatorics. Finite sums of trigonometric functions are listed
in [4], [5], chapter 4 in [6], and page 131 in [7]. Finite sums of special
functions are tabled in [5] and chapter 4 in [8].

In this work we apply the contour integral method [9], to the finite secant-
sine and cosecant-cosine sums given in Examples XXII, numbers (50), (48) and
(62) on page 335 in [2] respectively to derive their contour integral finite
trigonometric forms, resulting in

(2.1) $\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3^{-p}a^{w}w^{-k-1}\sin^{3}\left(3^{p}(m+w)\right)\sec\left(3^{p+1}(m+w)\right)dw\\\
=\frac{1}{2\pi
i}\int_{C}\frac{3}{8}a^{w}w^{-k-1}\left(3^{-n}\tan\left(3^{n}(m+w)\right)-\tan(m+w)\right)dw$

where $a,m,k\in\mathbb{C},Re(m+w)>0,n\in\mathbb{Z^{+}}$. Using equation (2.1)
the main Theorem to be derived and evaluated is given by

(2.2)
$\sum_{p=0}^{n-1}\left(i3^{p+1}\right)^{k}e^{im3^{p}}\left(\Phi\left(e^{2i3^{p+1}m},-k,\frac{1}{6}\left(1-i3^{-p}\log(a)\right)\right)\right.\\\
\left.+e^{4im3^{p}}\Phi\left(e^{2i3^{p+1}m},-k,\frac{1}{6}\left(5-i3^{-p}\log(a)\right)\right)\right)\\\
=i^{k}e^{im}\Phi\left(e^{2im},-k,\frac{1}{2}-\frac{1}{2}i\log(a)\right)-\left(i3^{n}\right)^{k}e^{im3^{n}}\Phi\left(e^{2i3^{n}m},-k,\frac{1}{2}-\frac{1}{2}i3^{-n}\log(a)\right)$

where the variables $k,a,m$ are general complex numbers and $n$ are positive
integers. The second contour integral form used is given by;

(2.3) $\frac{1}{2\pi i}\int_{C}\sum_{p=0}^{n-1}a^{w}w^{-k-1}\cos\left(2\times
3^{p}(m+w)\right)\csc\left(3^{p+1}(m+w)\right)dw\\\ =\frac{1}{2\pi
i}\int_{C}\frac{1}{2}a^{w}w^{-k-1}\csc(m+w)-\frac{1}{2}a^{w}w^{-k-1}\csc\left(3^{n}(m+w)\right)dw$

where $a,m,k\in\mathbb{C},Re(m+w)>0,n\in\mathbb{Z^{+}}$. Using equation (2.3)
the main Theorem to be derived and evaluated is given by

(2.4)
$\sum_{p=0}^{n-1}3^{-p}\left(\log^{k}(a)+2^{k}\left(i3^{p+1}\right)^{k}e^{2im3^{p}}\left(-3\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(2-i3^{-p}\log(a)\right)\right)\right.\right.\\\
\left.\left.+3e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(4-i3^{-p}\log(a)\right)\right)\right.\right.\\\
\left.\left.-2e^{4im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(6-i3^{-p}\log(a)\right)\right)\right)\right)\\\
=\frac{1}{2}3^{1-n}\left(\left(3^{n}-1\right)\log^{k}(a)+2^{k+1}\left(\left(i3^{n}\right)^{k}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,1-\frac{1}{2}i3^{-n}\log(a)\right)\right.\right.\\\
\left.\left.-i^{k}e^{2im}3^{n}\Phi\left(-e^{2im},-k,1-\frac{1}{2}i\log(a)\right)\right)\right)$

where the variables $k,a,m$ are general complex numbers and $n$ are positive
integers. The third contour integral form used is given by;

(2.5) $\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3^{-p}a^{w}w^{-k-1}\sin\left(3^{p}(m+w)\right)\left(2\cos\left(2\times
3^{p}(m+w)\right)\right.\\\
\left.+3^{p+1}-2\right)\sec\left(3^{p+1}(m+w)\right)dw\\\ =\frac{1}{2\pi
i}\int_{C}\frac{1}{2}3^{1-n}\left(3^{n}-1\right)a^{w}w^{-k-1}\tan\left(3^{n}(m+w)\right)dw$

where $a,m,k\in\mathbb{C},Re(m+w)>0,n\in\mathbb{Z^{+}}$. Using equation (2.3)
the main Theorem to be derived and evaluated is given by

(2.6)
$\sum_{p=0}^{n-1}3^{-p}\left(i3^{p+1}\right)^{k}e^{2im3^{p}}\left(3\left(3^{p}-1\right)\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(3^{-p}a+2\right)\right)\right.\\\
\left.-3\left(3^{p}-1\right)e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(3^{-p}a+4\right)\right)\right.\\\
\left.-2e^{4im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(3^{-p}a+6\right)\right)\right)\\\
=-3i\left(3^{n}-1\right)\left(i3^{n}\right)^{k-1}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,\frac{3^{-n}a}{2}+1\right)$

where the variables $k,a,m$ are general complex numbers and $n$ are positive
integers. The derivations follow the method used by us in [9]. This method
involves using a form of the generalized Cauchy’s integral formula given by

(2.7) $\frac{y^{k}}{\Gamma(k+1)}=\frac{1}{2\pi
i}\int_{C}\frac{e^{wy}}{w^{k+1}}dw,$

where $y,w\in\mathbb{C}$ and $C$ is in general an open contour in the complex
plane where the bilinear concomitant [9] is equal to zero at the end points of
the contour. This method involves using a form of equation (2.7) then
multiplies both sides by a function, then takes the double finite sum of both
sides. This yields a double finite sum in terms of a contour integral. Then we
multiply both sides of equation (2.7) by another function and take the
infinite sum of both sides such that the contour integral of both equations
are the same.

##  3\. The Hurwitz-Lerch zeta function

We use equation (1.11.3) in [10] where $\Phi(z,s,v)$ is the Lerch function
which is a generalization of the Hurwitz zeta $\zeta(s,v)$ and Polylogarithm
functions $Li_{n}(z)$. In number theory and complex analysis, the Lerch
function is a mathematical function that appears in many branches of
mathematics and physics. It is named after Czech mathematician Mathias Lerch,
who published a paper about the function in 1887. Numerous areas of
mathematics, including number theory (especially in the investigation of the
Riemann zeta function and its generalizations), complex analysis, and
theoretical physics, all have uses for it. It can be used to express a variety
of complex functions and series and is involved in numerous mathematical
identities. The Lerch function has a series representation given by

(3.1) $\Phi(z,s,v)=\sum_{n=0}^{\infty}(v+n)^{-s}z^{n}$

where $|z|<1,v\neq 0,-1,-2,-3,..,$ and is continued analytically by its
integral representation given by

(3.2)
$\Phi(z,s,v)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{t^{s-1}e^{-(v-1)t}}{e^{t}-z}dt$

where $Re(v)>0$, and either $|z|\leq 1,z\neq 1,Re(s)>0$, or $z=1,Re(s)>1$.

##  4\. Derivation of generalized trigonometric contour integral
representations

In this section we will derive the contour integral representations for
Theorems (6.1) and (6.2) respectively.

###  4.1. The Hurwitz-Lerch zeta function representation for the secant-sine
function contour integral

We use the method in [9]. Using a generalization of Cauchy’s integral formula
(2.7) we first replace $y$ by $\log(a)+ix+y$ then multiply both sides by
$e^{mxi}$ then form a second equation by replacing $x$ by $-x$ and subtract
both equations to get

(4.1)
$\frac{e^{-imx}\left((\log(a)-ix+y)^{k}-e^{2imx}(\log(a)+ix+y)^{k}\right)}{\Gamma(k+1)}\\\
=-\frac{1}{2\pi i}\int_{C}2ia^{w}w^{-k-1}e^{wy}\sin(x(m+w))dw$

Next we replace $y$ by $ib(2y+1)$ and multiply both sides by
$(-1)^{y}e^{ibm(2y+1)}$ and take the infinite sum over $y\in[0,\infty)$ and
simplify in terms of the Hurwitz-Lerch zeta function to get

(4.2)
$\frac{i2^{k}(ib)^{k}e^{im(b-x)}\left(\Phi\left(-e^{2ibm},-k,\frac{b-x-i\log(a)}{2b}\right)-e^{2imx}\Phi\left(-e^{2ibm},-k,\frac{b+x-i\log(a)}{2b}\right)\right)}{\Gamma(k+1)}\\\
=-\frac{1}{2\pi
i}\sum_{y=0}^{\infty}\int_{C}2i(-1)^{y}a^{w}w^{-k-1}e^{ib(2y+1)(m+w)}\sin(x(m+w))dw\\\
=-\frac{1}{2\pi
i}\int_{C}\sum_{y=0}^{\infty}2i(-1)^{y}a^{w}w^{-k-1}e^{ib(2y+1)(m+w)}\sin(x(m+w))dw\\\
=\frac{1}{2\pi i}\int_{C}a^{w}w^{-k-1}\sec(b(m+w))\sin(x(m+w))dw$

from equation (1.232.2) and (1.411.1) in [4] where $Re(w+m)>0$ and
$Im\left(m+w\right)>0$ in order for the sums to converge. We apply Tonelli’s
theorem for sums and integrals, see page 177 in [11] as the summand and
integral are of bounded measure over the space $\mathbb{C}\times[0,\infty)$.

###  4.2. The Hurwitz-Lerch zeta function in terms of the tangent contour
integral representation

####  4.2.1. Derivation of the additional contour

Using a generalization of Cauchy’s integral formula (2.7), first replace
$y\to\log(a)$ then multiply both sides by $-i$ and simplify to get

(4.3) $-\frac{i\log^{k}(a)}{\Gamma(k+1)}=-\frac{1}{2\pi
i}\int_{C}ia^{w}w^{-k-1}dw$

Using a generalization of Cauchy’s integral formula (2.7), first replace
$y\to\log(a)+2ib(y+1)$ then multiply both sides by $-2i(-1)^{y}e^{2ibm(y+1)}$
and take the infinite sums over $y\in[0,\infty)$ and simplify in terms of the
Hurwitz-Lerch zeta function and subtract equation (4.3) to get

(4.4)
$\frac{i\left(\log^{k}(a)-2^{k+1}(ib)^{k}e^{2ibm}\Phi\left(-e^{2ibm},-k,1-\frac{i\log(a)}{2b}\right)\right)}{\Gamma(k+1)}\\\
=\frac{1}{2\pi
i}\sum_{y=0}^{\infty}\int_{C}(-1)^{y}a^{w}w^{-k-1}e^{2ib(y+1)(m+w)}dw\\\
=-\frac{1}{2\pi
i}\int_{C}\sum_{y=0}^{\infty}(-1)^{y}a^{w}w^{-k-1}e^{2ib(y+1)(m+w)}dw\\\
=\frac{1}{2\pi i}\int_{C}a^{w}w^{-k-1}\tan(b(m+w))dw$

from equation (1.232.1) in [4] where $Re\left(m+w\right)>0$ and
$Im\left(m+w\right)>0$ in order for the sums to converge. Apply Tonelli’s
theorem for multiple sums, see page 177 in [11] as the summands are of bounded
measure over the space $\mathbb{C}\times[0,\infty)$.

###  4.3. Derivation of the generalized cosecant-cosine contour integral

We use the method in [9]. Using a generalization of Cauchy’s integral formula
(2.7) we first replace $y$ by $\log(a)+ix+y$ then multiply both sides by
$e^{mxi}$ then form a second equation by replacing $x$ by $-x$ and add both
equations to get

(4.5)
$\frac{e^{-imx}\left(e^{2imx}(\log(a)+ix+y)^{k}+(\log(a)-ix+y)^{k}\right)}{\Gamma(k+1)}\\\
=\frac{1}{2\pi i}\int_{C}2w^{-k-1}e^{w(\log(a)+y)}\cos(x(m+w))dw$

Next we replace $y$ by $bI(2y+1)$ and multiply both sides by $-ie^{ibm(2y+1)}$
and take the infinite and finite sums over $y\in[0,\infty)$ and simplify in
terms of the Hurwitz-Lerch zeta function to get

(4.6)
$-\frac{i2^{k}(ib)^{k}e^{im(b-x)}\left(\Phi\left(e^{2ibm},-k,\frac{b-x-i\log(a)}{2b}\right)+e^{2imx}\Phi\left(e^{2ibm},-k,\frac{b+x-i\log(a)}{2b}\right)\right)}{\Gamma(k+1)}\\\
=\frac{1}{2\pi
i}\sum_{y=0}^{\infty}\int_{C}2a^{w}w^{-k-1}e^{ib(2y+1)(m+w)}\cos(x(m+w))dw\\\
=\frac{1}{2\pi
i}\int_{C}\sum_{y=0}^{\infty}2a^{w}w^{-k-1}e^{ib(2y+1)(m+w)}\cos(x(m+w))dw\\\
=\frac{1}{2\pi i}\int_{C}a^{w}w^{-k-1}\csc(b(m+w))\cos(x(m+w))dw$

from equation (1.232.3) and (1.411.3) in [4] where $Re(w+m)>0$ and
$Im\left(m+w\right)>0$ in order for the sums to converge. We apply Tonelli’s
theorem for sums and integrals, see page 177 in [11] as the summand and
integral are of bounded measure over the space $\mathbb{C}\times[0,\infty)$.

###  4.4. Derivation of the generalized cosecant contour integral

We use the method in [9]. Using equation (2.7) we first replace
$\log(a)+ib(2y+1)$ and multiply both sides by $-2ie^{ibm(2y+1)}$ then take the
infinite sum over $y\in[0,\infty)$ and simplify in terms of the Hurwitz-Lerch
zeta function to get

(4.7)
$-\frac{i2^{k+1}(ib)^{k}e^{ibm}\Phi\left(e^{2ibm},-k,\frac{b-i\log(a)}{2b}\right)}{\Gamma(k+1)}\\\
=\frac{1}{2\pi
i}\sum_{y=0}^{\infty}\int_{C}a^{w}w^{-k-1}e^{ib(2y+1)(m+w)}dw\\\
=\frac{1}{2\pi
i}\int_{C}\sum_{y=0}^{\infty}a^{w}w^{-k-1}e^{ib(2y+1)(m+w)}dw\\\
=\frac{1}{2\pi i}\int_{C}a^{w}w^{-k-1}\csc(b(m+w))dw$

from equation (1.232.3) in [4] where $Re(w+m)>0$ and $Im\left(m+w\right)>0$ in
order for the sums to converge. We apply Tonelli’s theorem for multiple sums,
see page 177 in [11] as the summands are of bounded measure over the space
$\mathbb{C}\times[0,\infty)$.

##  5\. Derivation of the Hurwitz-Lerch zeta function contour integrals

In this section we will derive the contour integrals by simple substitution in
the previous contour integral representations.

###  5.1. The secant-sine series

####  5.1.1. Left-hand side first contour integral

Use equation (4.2) and replace $b$ by $3^{p+1}$ and $x$ by $3^{p}$ then
multiply both sides by $3^{1-p}$ and take the finite sum over $p\in[0,n-1]$ to
get;

(5.1)
$\sum_{p=0}^{n-1}\frac{1}{\Gamma(k+1)}i2^{k}3^{1-p}\left(i3^{p+1}\right)^{k}e^{im\left(3^{p+1}-3^{p}\right)}\\\
\left(\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)-3^{p}+3^{p+1}\right)\right)\right.\\\
\left.-e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)+3^{p}+3^{p+1}\right)\right)\right)\\\
=\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3^{1-p}a^{w}w^{-k-1}\sin\left(3^{p}(m+w)\right)\sec\left(3^{p+1}(m+w)\right)dw$

####  5.1.2. Left-hand side second contour integral

Use equation (4.4) and replace $b$ by $3^{p+1}$ then multiply both sides by
$-3^{-p}$ and take the finite sum over $p\in[0,n-1]$ to get;

(5.2)
$-\sum_{p=0}^{n-1}\frac{i3^{-p}\left(\log^{k}(a)-2^{k+1}\left(i3^{p+1}\right)^{k}e^{2im3^{p+1}}\Phi\left(-e^{2i3^{p+1}m},-k,1-\frac{1}{2}i3^{-p-1}\log(a)\right)\right)}{\Gamma(k+1)}\\\
=-\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3^{-p}a^{w}w^{-k-1}\tan\left(3^{p+1}(m+w)\right)dw$

####  5.1.3. Right-hand side first contour integral

Use equation (4.4) and replace $b$ by $1$ then multiply both sides by $-3/2$
to get;

(5.3)
$-\frac{3i\left(\log^{k}(a)-i^{k}2^{k+1}e^{2im}\Phi\left(-e^{2im},-k,1-\frac{1}{2}i\log(a)\right)\right)}{2\Gamma(k+1)}\\\
=-\frac{1}{2\pi i}\int_{C}\frac{3}{2}a^{w}w^{-k-1}\tan(m+w)dw$

####  5.1.4. Right-hand side first contour integral

Use equation (4.4) and replace $b$ by $3^{n}$ then multiply both sides by
$\frac{3^{1-n}}{2}$ to get;

(5.4)
$\frac{i3^{1-n}\left(\log^{k}(a)-2^{k+1}\left(i3^{n}\right)^{k}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,1-\frac{1}{2}i3^{-n}\log(a)\right)\right)}{2\Gamma(k+1)}\\\
=\frac{1}{2\pi
i}\int_{C}\frac{1}{2}3^{1-n}a^{w}w^{-k-1}\tan\left(3^{n}(m+w)\right)dw$

###  5.2. The cosecant-cosine series

####  5.2.1. Left-hand side contour integral

Use equation (4.6) and replace $b$ by $3^{p+1}$ and $x$ by $2\times 3^{p}$ and
take the finite sum over $p\in[0,n-1]$ to get;

(5.5)
$-\sum_{p=0}^{n-1}\frac{1}{\Gamma(k+1)}i2^{k}\left(i3^{p+1}\right)^{k}e^{im\left(3^{p+1}-2\times
3^{p}\right)}\\\
\left(\Phi\left(e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)-2\times
3^{p}+3^{p+1}\right)\right)\right.\\\
\left.+e^{4im3^{p}}\Phi\left(e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)+2\times
3^{p}+3^{p+1}\right)\right)\right)\\\ =\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}a^{w}w^{-k-1}\cos\left(2\times
3^{p}(m+w)\right)\csc\left(3^{p+1}(m+w)\right)dw$

####  5.2.2. Right-hand side first contour integral

Use equation (4.7) and replace $b$ by $1$ then multiply both sides by $1/2$ to
get;

(5.6)
$-\frac{i^{k+1}2^{k}e^{im}\Phi\left(e^{2im},-k,\frac{1}{2}(1-i\log(a))\right)}{\Gamma(k+1)}=\frac{1}{2\pi
i}\int_{C}\frac{1}{2}a^{w}w^{-k-1}\csc(m+w)dw$

####  5.2.3. Right-hand side second contour integral

Use equation (4.7) and replace $b$ by $3^{n}$ then multiply both sides by
$-1/2$ to get;

(5.7)
$\frac{i2^{k}\left(i3^{n}\right)^{k}e^{im3^{n}}\Phi\left(e^{2i3^{n}m},-k,\frac{1}{2}3^{-n}\left(3^{n}-i\log(a)\right)\right)}{\Gamma(k+1)}\\\
=-\frac{1}{2\pi
i}\int_{C}\frac{1}{2}a^{w}w^{-k-1}\csc\left(3^{n}(m+w)\right)dw$

###  5.3. The sine-secant series

####  5.3.1. Left-hand side first contour integral

Use equation (4.2) and replace $b$ by $3^{p+1}$ and $x$ by $3^{p}$ and
multiply both sides by 3 and simplify to get;

(5.8)
$\sum_{p=0}^{n-1}\frac{1}{k!}3i2^{k}\left(i3^{p+1}\right)^{k}e^{im\left(3^{p+1}-3^{p}\right)}\\\
\left(\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)-3^{p}+3^{p+1}\right)\right)\right.\\\
\left.-e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)+3^{p}+3^{p+1}\right)\right)\right)\\\
=\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3a^{w}w^{-k-1}\sin\left(3^{p}(m+w)\right)\sec\left(3^{p+1}(m+w)\right)dw$

####  5.3.2. Left-hand side second contour integral

Use equation (4.2) and replace $b$ by $3^{p+1}$ and $x$ by $3^{p}$ and
multiply both sides by $-3^{1-p}$ and simplify to get;

(5.9)
$-\sum_{p=0}^{n-1}\frac{1}{k!}i2^{k}3^{1-p}\left(i3^{p+1}\right)^{k}e^{im\left(3^{p+1}-3^{p}\right)}\\\
\left(\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)-3^{p}+3^{p+1}\right)\right)\right.\\\
\left.=-e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{2}3^{-p-1}\left(-i\log(a)+3^{p}+3^{p+1}\right)\right)\right)\\\
=-\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3^{1-p}a^{w}w^{-k-1}\sin\left(3^{p}(m+w)\right)\sec\left(3^{p+1}(m+w)\right)dw$

####  5.3.3. Left-hand side third contour integral

Use equation (4.4) and replace $b$ by $3^{p+1}$ and multiply both sides by
$3^{-p}$ and simplify to get;

(5.10)
$\sum_{p=0}^{n-1}\frac{i3^{-p}\left(\log^{k}(a)-2^{k+1}\left(i3^{p+1}\right)^{k}e^{2im3^{p+1}}\Phi\left(-e^{2i3^{p+1}m},-k,1-\frac{1}{2}i3^{-p-1}\log(a)\right)\right)}{k!}\\\
=\frac{1}{2\pi
i}\int_{C}\sum_{p=0}^{n-1}3^{-p}a^{w}w^{-k-1}\tan\left(3^{p+1}(m+w)\right)dw$

####  5.3.4. Right-hand side first contour integral

Use equation (4.4) and replace $b$ by $3^{n}$ and multiply both sides by $3/2$
and simplify to get;

(5.11)
$\frac{3i\left(\log^{k}(a)-2^{k+1}\left(i3^{n}\right)^{k}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,1-\frac{1}{2}i3^{-n}\log(a)\right)\right)}{2k!}\\\
=\frac{1}{2\pi
i}\int_{C}\frac{3}{2}a^{w}w^{-k-1}\tan\left(3^{n}(m+w)\right)dw$

####  5.3.5. Right-hand side second contour integral

Use equation (4.4) and replace $b$ by $3^{n}$ and multiply both sides by
$-\frac{3^{1-n}}{2}$ and simplify to get;

(5.12)
$-\frac{i3^{1-n}\left(\log^{k}(a)-2^{k+1}\left(i3^{n}\right)^{k}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,1-\frac{1}{2}i3^{-n}\log(a)\right)\right)}{2k!}\\\
=-\frac{1}{2\pi
i}\int_{C}\frac{1}{2}3^{1-n}a^{w}w^{-k-1}\tan\left(3^{n}(m+w)\right)dw$

##  6\. Derivation of the main theorems in terms of the Hurwitz-Lerch zeta
function

In this section we will derive and evaluate formulae involving the finite sum
of the Hurwitz-Lerch zeta function in terms other special functions,
trigonometric functions and fundamental constants.

######  Theorem 6.1.

Main theorem: secant-sine series. For all $k,a,m\in\mathbb{C}$ then,

(6.1)
$\sum_{p=0}^{n-1}3^{-p}\left(\log^{k}(a)+2^{k}\left(i3^{p+1}\right)^{k}e^{2im3^{p}}\left(-3\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(2-i3^{-p}\log(a)\right)\right)\right.\right.\\\
\left.\left.+3e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(4-i3^{-p}\log(a)\right)\right)\right.\right.\\\
\left.\left.-2e^{4im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(6-i3^{-p}\log(a)\right)\right)\right)\right)\\\
=\frac{1}{2}3^{1-n}\left(\left(3^{n}-1\right)\log^{k}(a)+2^{k+1}\left(\left(i3^{n}\right)^{k}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,1-\frac{1}{2}i3^{-n}\log(a)\right)\right.\right.\\\
\left.\left.-i^{k}e^{2im}3^{n}\Phi\left(-e^{2im},-k,1-\frac{1}{2}i\log(a)\right)\right)\right)$

###### Proof.

Observe that the addition of the right-hand sides of equations (5.1) and
(5.2), is equal to the addition of the right-hand sides of equations (5.3) and
(5.4) so we may equate the left-hand sides and simplify relative to equation
(2.1) and the Gamma function to yield the stated result. ∎

######  Theorem 6.2.

Main theorem: cosecant-cosine series. For all $k,a,m\in\mathbb{C}$ then,

(6.2)
$\sum_{p=0}^{n-1}\left(i3^{p+1}\right)^{k}e^{im3^{p}}\left(\Phi\left(e^{2i3^{p+1}m},-k,\frac{1}{6}\left(1-i3^{-p}\log(a)\right)\right)\right.\\\
\left.+e^{4im3^{p}}\Phi\left(e^{2i3^{p+1}m},-k,\frac{1}{6}\left(5-i3^{-p}\log(a)\right)\right)\right)\\\
=i^{k}e^{im}\Phi\left(e^{2im},-k,\frac{1}{2}-\frac{1}{2}i\log(a)\right)-\left(i3^{n}\right)^{k}e^{im3^{n}}\Phi\left(e^{2i3^{n}m},-k,\frac{1}{2}-\frac{1}{2}i3^{-n}\log(a)\right)$

###### Proof.

Observe that the addition of the right-hand side of equation (5.5), is equal
to the addition of the right-hand sides of equations (5.6) and (5.7) so we may
equate the left-hand sides and simplify relative to equation (2.1) and the
Gamma function to yield the stated result. ∎

######  Theorem 6.3.

Main theorem: secant-sine series. For all $k,a,m\in\mathbb{C}$ then,

(6.3)
$\sum_{p=0}^{n-1}3^{-p}\left(i3^{p+1}\right)^{k}e^{2im3^{p}}\left(3\left(3^{p}-1\right)\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(3^{-p}a+2\right)\right)\right.\\\
\left.-3\left(3^{p}-1\right)e^{2im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(3^{-p}a+4\right)\right)\right.\\\
\left.-2e^{4im3^{p}}\Phi\left(-e^{2i3^{p+1}m},-k,\frac{1}{6}\left(3^{-p}a+6\right)\right)\right)\\\
=-3i\left(3^{n}-1\right)\left(i3^{n}\right)^{k-1}e^{2im3^{n}}\Phi\left(-e^{2i3^{n}m},-k,\frac{3^{-n}a}{2}+1\right)$

###### Proof.

Observe that the addition of the right-hand side of equations (5.8), (5.9) and
(5.10) are equal to the right-hand sides of equation (5.11) so we may equate
the left-hand sides replace $a$ by $e^{ai}$ and simplify relative to equation
(2.1) and the Gamma function to yield the stated result. ∎

######  Example 6.4.

The degenerate case.

(6.4)
$\sum_{p=0}^{n-1}3^{-p}\sin^{3}\left(m3^{p}\right)\sec\left(m3^{p+1}\right)=\frac{3}{8}\left(3^{-n}\tan\left(m3^{n}\right)-\tan(m)\right)$

###### Proof.

Use equation (6.1) and set $k=0$ and simplify using entry (2) in the Table
below equation (64:12:7) on page 692 in [12] to yield the stated result. ∎

######  Example 6.5.

The degenerate case.

(6.5)
$\sum_{p=0}^{n-1}\cos\left(2m3^{p}\right)\csc\left(m3^{p+1}\right)=\frac{1}{2}\left(\csc(m)-\csc\left(m3^{n}\right)\right)$

###### Proof.

Use equation (6.2) and set $k=0$ and simplify using entry (2) in the Table
below equation (64:12:7) on page 692 in [12] to yield the stated result. ∎

######  Example 6.6.

The degenerate case.

(6.6)
$\sum_{p=0}^{n-1}\frac{3^{-p}\left(-2i\sin\left(2m3^{p}\right)-2\cos\left(2m3^{p}\right)-i\left(3^{p+1}-4\right)\tan\left(m3^{p}\right)+1\right)}{2\cos\left(2m3^{p}\right)-1}\\\
=\frac{3}{2}\left(3^{-n}-1\right)\left(1+i\tan\left(m3^{n}\right)\right)$

###### Proof.

Use equation (6.3) and set $k=0$ and simplify using entry (2) in the Table
below equation (64:12:7) on page 692 in [12] to yield the stated result. ∎

##  7\. Table of results: Part I

In this section we evaluate Theorem (6.1) for various values of the parameters
involved to derive related closed form formulae in terms of special functions
and fundamental constants.

######  Example 7.1.

Finite product of quotient gamma functions.

(7.1)
$\prod_{p=0}^{n-1}\left(\frac{\Gamma\left(3^{-p-1}a+\frac{1}{2}\right)}{\Gamma\left(3^{-p-1}a+1\right)}\right)^{2\times
3^{-p-1}}\\\
\left(\frac{9^{p+1}\Gamma\left(3^{-p-1}a+\frac{1}{6}\right)\Gamma\left(3^{-p-1}a+\frac{5}{6}\right)}{\left(3^{p}-a\right)\left(2\times
3^{p}-a\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-2\right)\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-1\right)\right)}\right)^{3^{-p}}\\\
=\frac{3^{\frac{3}{4}\left(1-3^{-n}\right)}\Gamma\left(a+\frac{1}{2}\right)\left(\frac{\Gamma\left(3^{-n}a+1\right)}{\Gamma\left(3^{-n}a+\frac{1}{2}\right)}\right)^{3^{-n}}}{\Gamma(a+1)}$

###### Proof.

Use equation (6.1) and set $m=0$ and simplify in terms of the Hurwitz zeta
function using entry (4) in the Table below equation (64:12:7) on page 692 in
[12]. Next take the first partial derivative with respect to $k$ and set $k=0$
and simplify in terms of the log-gamma function using equation (64:10:2) in
[12]. Finally take the exponential function of both sides and simplify both
sides to yield the stated result. ∎

######  Example 7.2.

Infinite product of quotient gamma functions.

(7.2)
$\prod_{p=0}^{\infty}\left(\frac{\Gamma\left(3^{-p-1}a+\frac{1}{2}\right)}{\Gamma\left(3^{-p-1}a+1\right)}\right)^{2\times
3^{-p-1}}\\\
\left(\frac{9^{p+1}\Gamma\left(3^{-p-1}a+\frac{1}{6}\right)\Gamma\left(3^{-p-1}a+\frac{5}{6}\right)}{\left(3^{p}-a\right)\left(2\times
3^{p}-a\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-2\right)\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-1\right)\right)}\right)^{3^{-p}}\\\
=\frac{3^{3/4}\Gamma\left(a+\frac{1}{2}\right)}{\Gamma(a+1)}$

###### Proof.

Use equation (7.1) and take the limit as $n\to\infty$ of the right-hand side
and simplify to yield the stated result. ∎

######  Example 7.3.

Finite product of root of quotient cosine functions.

(7.3)
$\prod_{p=0}^{n-1}\left(\frac{\left(\frac{\cos\left(3^{p}m\right)}{\cos\left(3^{p}r\right)}\right)^{16}\left(1-2\cos\left(2\times
3^{p}r\right)\right)^{2}}{\left(1-2\cos\left(2\times
3^{p}m\right)\right)^{2}}\right)^{3^{-2p}}=\left(\frac{\cos(m)}{\cos(r)}\right)^{18}\left(\frac{\cos\left(3^{n}r\right)}{\cos\left(3^{n}m\right)}\right)^{2\times
3^{2-2n}}$

###### Proof.

Use equation (6.1) and form a second equation by replacing $m\to r$. Next take
the differnce of the two equations and simplify. Then set $k=-1,a=1$ and
simplify in terms of the logarithm function using entry (5) in the Table below
equation (64:12:7) on page 692 in [12]. Next take the exponential function of
both sides and simplify to yield the stated result. ∎

######  Example 7.4.

Infinite product of root of quotient cosine functions.

(7.4)
$\prod_{p=0}^{\infty}\left(\frac{\left(\frac{\cos\left(3^{p}m\right)}{\cos\left(3^{p}r\right)}\right)^{16}\left(1-2\cos\left(2\times
3^{p}r\right)\right)^{2}}{\left(1-2\cos\left(2\times
3^{p}m\right)\right)^{2}}\right)^{\frac{3^{-2p}}{18}}=\frac{\cos(m)}{\cos(r)}$

###### Proof.

Use equation (7.3) and analyze the right-hand side as $n\to\infty$ using
Figure 2. ∎

Figure 1. Plot of
$f(m,r,n)=\left(\sec\left(m3^{n}\right)\cos\left(3^{n}r\right)\right)^{2\times
3^{2-2n}}$, $m,r\in\mathbb{R}$.

##  8\. Quotient cosine functions with reciprocal angles: the Secant-sine
series

Figure 2. Plot of $f(m)=\cos(m)\sec\left(\frac{1}{m}\right)$,
$m\in\mathbb{R}$. Figure 3. Plot of
$f(m)=Re\left(\cos(m)\sec\left(\frac{1}{m}\right)\right)$, $m\in\mathbb{C}$.
Figure 4. Plot of $f(m)=Im\left(\cos(m)\sec\left(\frac{1}{m}\right)\right)$,
$m\in\mathbb{C}$. Figure 5. Plot of
$f(m)=Abs\left(\cos(m)\sec\left(\frac{1}{m}\right)\right)$, $m\in\mathbb{C}$.

###  8.1. Elliptic functions

A traditional area of mathematics called elliptic functions [13] is used often
in physics and engineering applications, either directly or as a building
block for more complex ones. However, it cannot be denied that they are not as
well recognised as they ought to be. It is typically feasible to discover an
even simpler (and not always sufficient) skeleton of the theory in
trigonometry in many instances where the elliptic functions might be usefully
applied, or at least examined as a potentially effective first approximation.
Here we present a simple infinite product in terms of the ratio involving
elliptic functions. In this example we look at an extension of the ratio of
Elliptic functions in terms of the infinite product involving the ratio of the
cosine function using equation (7.4) and equation (29’) in [14], where
$-1<Re(k)<Re(a)<Re(b)<1$.

######  Example 8.1.

Infinite product of sine and cosine functions in terms of Elliptic functions.

(8.1)
$\prod_{p=0}^{\infty}\left(\frac{\cos^{16}\left(3^{p}k\sin(a)\right)\left(1-2\cos\left(2\times
3^{p}k\sin(b)\right)\right)^{2}\sec^{16}\left(3^{p}k\sin(b)\right)}{\left(1-2\cos\left(2\times
3^{p}k\sin(a)\right)\right)^{2}}\right)^{\frac{1}{2}3^{-2-2p}}\\\
=\frac{\cos(k\sin(a))}{\cos(k\sin(b))}=\frac{\sqrt{1-(k\sin(a))^{2}}}{\sqrt{1-(k\sin(b))^{2}}}$

######  Example 8.2.

Functional equation in terms of the Hurwitz-Lerch zeta function.

(8.2)
$\Phi(z,s,a)=3^{-2s-1}\left(3^{s}\left(3\Phi\left(z^{3},s,\frac{a}{3}\right)+z\left(3\Phi\left(z^{3},s,\frac{a+1}{3}\right)+2z\Phi\left(z^{3},s,\frac{a+2}{3}\right)\right)\right)\right.\\\
\left.+z^{2}\left(\Phi\left(z^{9},s,\frac{a+2}{9}\right)+z^{6}\Phi\left(z^{9},s,\frac{a+8}{9}\right)+z^{3}\Phi\left(z^{9},s,\frac{a+5}{9}\right)\right)\right)$

###### Proof.

Use equation (6.1) and set $n=2,m=\frac{\log(-z)}{2i},a=e^{ai},k=-s$ and
simplify. ∎

##  9\. Special cases of the Hurwitz-Lerch zeta function

In this section we will evaluate equation (6.1) and derive formulae in terms
of fundamental constants; namely Catalan’s constant $C$ given in equations
(20) and (21) in [15], Glaisher’s constant $A$, given in equation (18) in
[15], Apery’s constant $\zeta(3)$, given in equation (19).

######  Example 9.1.

Finite series in terms of Catalan’s constant $C$.

(9.1)
$\sum_{p=0}^{n-1}27^{-p}\left(3\psi^{(1)}\left(\frac{1}{12}\left(2+3^{-p}\right)\right)-3\psi^{(1)}\left(\frac{1}{12}\left(4+3^{-p}\right)\right)+2\psi^{(1)}\left(\frac{1}{12}\left(6+3^{-p}\right)\right)\right.\\\
\left.-3\psi^{(1)}\left(\frac{1}{12}\left(8+3^{-p}\right)\right)+3\psi^{(1)}\left(\frac{1}{12}\left(10+3^{-p}\right)\right)-2\left(8\times
9^{p+1}+\psi^{(1)}\left(\frac{1}{12}\left(12+3^{-p}\right)\right)\right)\right)\\\
=27^{1-n}\left(8\times
9^{n}\left(-2C3^{n}+3^{n}+1\right)+\psi^{(1)}\left(1+\frac{3^{-n}}{4}\right)-\psi^{(1)}\left(\frac{1}{4}\left(2+3^{-n}\right)\right)\right)$

###### Proof.

Use equation (6.1) and set $m=0,k=-2,a=e^{i}$ and simplify. ∎

######  Example 9.2.

Difference of Hurwitz-Lerch zeta functions in terms of Glaisher’s constant
$A$.

(9.2)
$\Phi^{\prime}\left(-1,-1,\frac{1}{3}\right)-\Phi^{\prime}\left(-1,-1,\frac{2}{3}\right)=\log\left(\frac{2^{2/9}\sqrt[12]{3}\sqrt[6]{e}}{A^{2}}\right)$

###### Proof.

Use equation (6.1) take the first partial derivative with respect to $k$ and
set $m=0,k=a=1$ and simplify. ∎

######  Example 9.3.

Difference of Hurwitz-Lerch zeta functions in terms of Apery’s constant
$\zeta](3)$.

(9.3)
$\Phi^{\prime}\left(-1,-2,\frac{2}{3}\right)-\Phi^{\prime}\left(-1,-2,\frac{1}{3}\right)=\frac{14\zeta(3)}{9\pi^{2}}$

###### Proof.

Use equation (6.1) take the first partial derivative with respect to $k$ and
set $m=0,k=2,a=1$ and simplify. ∎

######  Example 9.4.

Finite series in terms of Catalan’s constant $C$.

(9.4)
$\sum_{p=0}^{n-1}\left(18\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(2-3^{-p}\right)\right)-18\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(4-3^{-p}\right)\right)\right.\\\
\left.+12\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(6-3^{-p}\right)\right)+3^{-p}\log\left(i3^{p+1}\right)\right)\\\
=\frac{-6\pi\Phi^{\prime}\left(-1,-1,1-\frac{3^{-n}}{2}\right)+6C+\frac{3}{2}\left(\pi-\pi
3^{-n}\right)\log\left(i3^{n}\right)}{\pi}$

###### Proof.

Use equation (6.1) take the first partial derivative with respect to $k$ and
set $m=0,k=1,a=e^{-i}$ and simplify. ∎

######  Example 9.5.

Finite product of quotient gamma functions.

(9.5)
$\prod_{p=0}^{n-1}\left(\frac{3^{a3^{-p}+2p}\Gamma\left(3^{-p-1}a-1\right)}{\left(a^{2}-a3^{p+1}+2\times
9^{p}\right)\Gamma\left(3^{-p-1}a+1\right)^{2/3}\Gamma\left(3^{-p}a-3\right)}\right)^{3^{-p}}\\\
=\frac{3^{\frac{1}{8}9^{-n}\left(9a\left(9^{n}-1\right)+8\
3^{n}\left(-n+3^{n+1}-3\right)\right)}a^{3^{-n}}\Gamma\left(3^{-n}a\right)^{3^{-n}}}{\Gamma(a+1)}$

###### Proof.

Use equation (6.1) and set $m=\pi/2$ and simplify in terms of the Hurwitz zeta
function using entry (4) in the Table below equation (64:12:7) on page 692 in
[12]. Next take the first partial derivative with respect to $k$ and set $k=0$
and simplify in terms of the log-gamma function using equation (64:10:2) in
[12]. Finally take the exponential function of both sides and simplify both
sides to yield the stated result. ∎

######  Example 9.6.

Infinite product of quotient gamma functions.

(9.6)
$\prod_{p=0}^{\infty}\left(\frac{3^{a3^{-p}+2p}\Gamma\left(3^{-p-1}a-1\right)}{\left(a^{2}-a3^{p+1}+2\times
9^{p}\right)\Gamma\left(3^{-p-1}a+1\right)^{2/3}\Gamma\left(3^{-p}a-3\right)}\right)^{3^{-p}}=\frac{3^{\frac{9a}{8}+3}}{\Gamma(a+1)}$

###### Proof.

Use equation (9.5) and take the limit of the right-hand side as $n\to\infty$
and simplify. ∎

######  Example 9.7.

Finite product involving the cosine function.

(9.7)
$\prod_{p=0}^{n-1}\left(\frac{\left(e^{-2m3^{p+1}}+1\right)^{2/3}\cosh^{2}\left(m3^{p}\right)}{2\cosh\left(2m3^{p}\right)-1}\right)^{3^{-2p}}=\frac{2^{\frac{3}{4}\left(1+3^{1-2n}\right)}\cosh^{3}(m)}{e^{3m}\left(1+e^{-2\times
3^{n}m}\right)^{3^{1-2n}}}$

###### Proof.

Use equation (6.1) and set $a=1$ and simplify in terms of the Polylogarithm
function using equation (64:12:2) in [12]. Next simplify the Polylogarithm
function in terms of the Hurwitz zeta function using equation (6) in [16].
Next take the limit of both sides as $k\to-1$ and simplify in terms of the
log-gamma function using equation (64:10:2) in [12]. Next take the exponential
function of both sides and simplify. ∎

######  Example 9.8.

Infinite product involving the cosine function.

(9.8)
$\prod_{p=0}^{\infty}\left(\frac{\left(e^{-2m3^{p+1}}+1\right)^{2/3}\cosh^{2}\left(m3^{p}\right)}{2\cosh\left(2m3^{p}\right)-1}\right)^{3^{-2p-1}}=\frac{e^{-2m}+1}{2^{3/4}}$

###### Proof.

Use equation (9.7) and take the limit as $n\to\infty$ and simplify. ∎

##  10\. Table of results: Part II

In this section we evaluate Theorem (6.2) for various values of the parameters
involved to derive related closed form formulae in terms of special functions
and fundamental constants.

######  Example 10.1.

Finite product of gamma functions.

(10.1)
$\prod_{p=0}^{n-1}\frac{\Gamma\left(\frac{1}{6}\left(3^{-p}a+1\right)\right)\Gamma\left(\frac{1}{6}\left(3^{-p}a+5\right)\right)}{2\pi}=\frac{3^{-\frac{1}{4}a3^{1-n}\left(3^{n}-1\right)}\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{1}{2}\left(3^{-n}a+1\right)\right)}$

###### Proof.

Use equation (6.2) and set $m=0$ and simplify in terms of the Hurwitz zeta
function using entry (4) in the Table below equation (64:12:7) on page 692 in
[12]. Next take the first partial derivative with respect to $k$ and set $k=0$
and simplify in terms of the log-gamma function using equation (64:10:2) in
[12]. Finally take the exponential function of both sides and simplify both
sides to yield the stated result. ∎

######  Example 10.2.

Infinite product of gamma functions.

(10.2)
$\prod_{p=0}^{\infty}\frac{\Gamma\left(\frac{1}{6}\left(3^{-p}a+1\right)\right)\Gamma\left(\frac{1}{6}\left(3^{-p}a+5\right)\right)}{2\pi}=\frac{3^{-3a/4}\Gamma\left(\frac{a+1}{2}\right)}{\sqrt{\pi}}$

###### Proof.

Use equation (10.1) and take the limit as $n\to\infty$ and simplify. ∎

######  Example 10.3.

Finite product of quotient hyperbolic tangent functions.

(10.3) $\prod_{p=0}^{n-1}\left(\frac{\left(1+2\cosh\left(2\times
3^{p}m\right)\right)\left(-1+2\cosh\left(2\times
3^{p}r\right)\right)}{\left(-1+2\cosh\left(2\times
3^{p}m\right)\right)\left(1+2\cosh\left(2\times
3^{p}r\right)\right)}\right)^{3^{-p}}\left(\frac{\tanh\left(3^{p}r\right)}{\tanh\left(3^{p}m\right)}\right)^{2\times
3^{-p}}\\\
=\left(\frac{\tanh\left(3^{n}m\right)}{\tanh\left(3^{n}r\right)}\right)^{3^{1-n}}\left(\frac{\tanh(r)}{\tanh(m)}\right)^{3}$

###### Proof.

Use equation (6.2) and form a second equation by replacing $m\to r$. Next take
the differnce of the two equations and simplify. Then set $k=-1,a=1$ and
simplify in terms of the logarithm function using entry (5) in the Table below
equation (64:12:7) on page 692 in [12]. Next take the exponential function of
both sides and simplify to yield the stated result. ∎

######  Example 10.4.

Infinite product of quotient hyperbolic tangent functions.

(10.4) $\prod_{p=0}^{\infty}\left(\frac{\left(1+2\cosh\left(2\times
3^{p}m\right)\right)\left(-1+2\cosh\left(2\times
3^{p}r\right)\right)}{\left(-1+2\cosh\left(2\times
3^{p}m\right)\right)\left(1+2\cosh\left(2\times
3^{p}r\right)\right)}\right)^{3^{-p}}\left(\frac{\tanh\left(3^{p}r\right)}{\tanh\left(3^{p}m\right)}\right)^{2\times
3^{-p}}\\\ =\left(\frac{\tanh(r)}{\tanh(m)}\right)^{3}$

###### Proof.

Use equation (10.3) and take the limit as $n\to\infty$ of the right-hand side
and simplify. ∎

##  11\. Quotient cosine functions with reciprocal angles: the cosecant-cosine
series

In this section we look at plots involving the ratio of the hyperbolic tangent
function and reciprocal angles.

Figure 6. Plot of $f(m)=\tanh^{3}(r)\coth^{3}\left(\frac{1}{r}\right)$,
$m\in\mathbb{R}$. Figure 7. Plot of
$f(r)=Re\left(\tanh^{3}(r)\coth^{3}\left(\frac{1}{r}\right)\right)$,
$r\in\mathbb{C}$. Figure 8. Plot of
$f(r)=Im\left(\tanh^{3}(r)\coth^{3}\left(\frac{1}{r}\right)\right)$,
$r\in\mathbb{C}$. Figure 9. Plot of
$f(r)=Abs\left(\tanh^{3}(r)\coth^{3}\left(\frac{1}{r}\right)\right)$,
$r\in\mathbb{C}$.

######  Example 11.1.

Functional equation for the Hurwitz-Lerch zeta function.

(11.1)
$\Phi(z,s,a)=3^{-s}\left(\Phi\left(z^{3},s,\frac{a}{3}\right)+z\left(\Phi\left(z^{3},s,\frac{a+1}{3}\right)+z\Phi\left(z^{3},s,\frac{a+2}{3}\right)\right)\right)$

###### Proof.

Use equation (6.2) and set $n=1,m=\frac{\log(z)}{2i},a=e^{ai},k=-s$ and
simplify. ∎

######  Example 11.2.

Finite series in terms of Catalan’s constant C.

(11.2)
$\sum_{p=0}^{n-1}9^{-p}\left(\psi^{(1)}\left(\frac{1}{12}\left(2+3^{-p}\right)\right)+\psi^{(1)}\left(\frac{1}{12}\left(10+3^{-p}\right)\right)\right)\\\
=9\left(-8C-9^{-n}\psi^{(1)}\left(\frac{1}{4}\left(2+3^{-n}\right)\right)+\pi^{2}\right)$

###### Proof.

Use equation (6.2) and take the first partial derivative with respect to $k$
and set $k=-2,m=0,a=e^{i/2}$ and simplify in terms of the Polygamma function
using equation (64:12:2) in [12]. ∎

######  Example 11.3.

Finite series in terms of Catalan’s constant C.

(11.3)
$\sum_{p=0}^{n-1}9^{-p}\left(\Phi^{\prime}\left(1,2,\frac{1}{12}\left(2-3^{-p}\right)\right)+\Phi^{\prime}\left(1,2,\frac{1}{6}\left(5-\frac{3^{-p}}{2}\right)\right)\right.\\\
\left.-\log\left(i3^{p+1}\right)\left(\psi^{(1)}\left(\frac{1}{12}\left(2-3^{-p}\right)\right)+\psi^{(1)}\left(\frac{1}{12}\left(10-3^{-p}\right)\right)\right)\right)\\\
=9\left(-9^{-n}\Phi^{\prime}\left(1,2,\frac{1}{2}-\frac{3^{-n}}{4}\right)+\Phi^{\prime}\left(1,2,\frac{1}{4}\right)-\frac{1}{2}i\pi\left(8C+\pi^{2}\right)\right.\\\
\left.+9^{-n}\log\left(i3^{n}\right)\psi^{(1)}\left(\frac{1}{2}-\frac{3^{-n}}{4}\right)\right)$

###### Proof.

Use equation (6.2) and take the first partial derivative with respect to $k$
and set $k=-2,m=0,a=e^{-i/2}$ and simplify in terms of the Polygamma function
using equation (64:12:2) in [12].

∎

######  Example 11.4.

Finite product of quotient gamma functions.

(11.4)
$\prod_{p=0}^{n-1}\left(\frac{\Gamma\left(\frac{1}{12}\left(3^{-p}a+7\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+11\right)\right)}{\Gamma\left(\frac{1}{4}3^{-p-1}\left(a+3^{p}\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+5\right)\right)}\right)^{(-1)^{p}}\\\
=\frac{3^{\frac{1}{4}\left((-1)^{n}-1\right)}\Gamma\left(\frac{a+3}{4}\right)\left(\frac{\Gamma\left(\frac{1}{4}\left(3^{-n}a+1\right)\right)}{\Gamma\left(\frac{1}{4}\left(3^{-n}a+3\right)\right)}\right)^{(-1)^{n}}}{\Gamma\left(\frac{a+1}{4}\right)}$

###### Proof.

Use equation (6.2) and set $m=\pi/2$ and simplify in terms of the Hurwitz zeta
function using entry (4) in the Table below equation (64:12:7) on page 692 in
[12]. Next take the first partial derivative with respect to $k$ and set $k=0$
and simplify in terms of the log-gamma function using equation (64:10:2) in
[12]. Finally take the exponential function of both sides and simplify both
sides to yield the stated result. ∎

######  Example 11.5.

An upper-bound for an infinite product of the ratio of gamma functions.

(11.5)
$\prod_{p=0}^{\infty}\left(\frac{\Gamma\left(\frac{1}{12}\left(3^{-p}a+7\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+11\right)\right)}{\Gamma\left(\frac{1}{4}3^{-p-1}\left(a+3^{p}\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+5\right)\right)}\right)^{(-1)^{p}}<\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{a+3}{4}\right)}{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{a+1}{4}\right)}$

###### Proof.

Use equation (11.4) and apply the limit as $n\to\infty$ to the right-hand
side. In this analysis we see that the three exponential terms which have
bounds as $n\to\infty$ namely
$\frac{\Gamma\left(\frac{a+3}{4}\right)3^{\frac{1}{4}\left(-1+e^{2i\times[0,\pi]}\right)}}{\Gamma\left(\frac{a+1}{4}\right)}\left(\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\right)^{e^{2i\times[0,\pi]}}$.
We can see the bound on the right-hand side using Figure 11. over finite
values of $n$. ∎

Figure 10. Plot of
$\frac{3^{\frac{1}{4}\left((-1)^{n}-1\right)}\Gamma\left(\frac{a+3}{4}\right)\left(\frac{\Gamma\left(\frac{1}{4}\left(3^{-n}a+1\right)\right)}{\Gamma\left(\frac{1}{4}\left(3^{-n}a+3\right)\right)}\right)^{(-1)^{n}}}{\Gamma\left(\frac{a+1}{4}\right)}$,
$a\in\mathbb{R}$.

######  Example 11.6.

Finite product involving the cosine function.

(11.6)
$\prod_{p=0}^{n}\left(\frac{\left(\sqrt{3}+i\tanh\left(\frac{3^{p}x}{2}\right)\right)\left(\sqrt{3}-3i\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)}{\left(\sqrt{3}-3i\tanh\left(\frac{3^{p}x}{2}\right)\right)\left(\sqrt{3}+i\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)}\right)^{3^{-p-1}}\\\
\left(\tanh\left(\frac{1}{2}3^{p+1}x\right)\coth\left(\frac{3^{p}x}{2}\right)\right)^{2\times
3^{-p-1}}\\\ \left(\sinh\left(2\times
3^{-p-1}\left(\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{3^{p}x}{2}\right)\right)\right)\right.\right.\right.\\\
\left.\left.\left.-\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)\right)\right)\right)\right.\\\
\left.+\cosh\left(2\times
3^{-p-1}\left(\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{3^{p}x}{2}\right)\right)\right)\right.\right.\right.\\\
\left.\left.\left.-\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)\right)\right)\right)\right)\\\
=\frac{\left(2\cos\left(\frac{x}{3}\right)+1\right)\left(\tan\left(\frac{1}{2}3^{n-1}x\right)\cot\left(\frac{3^{n}x}{2}\right)\right)^{3^{-n}}}{2\cos\left(\frac{x}{3}\right)-1}$

###### Proof.

Use equation (6.2) and set $k=1,a=1,m=x$ and apply the method in section (8)
in [17]. ∎

######  Example 11.7.

Infinite product involving the cosine function.

(11.7)
$\prod_{p=0}^{\infty}\left(\frac{\left(\sqrt{3}+i\tanh\left(\frac{3^{p}x}{2}\right)\right)\left(\sqrt{3}-3i\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)}{\left(\sqrt{3}-3i\tanh\left(\frac{3^{p}x}{2}\right)\right)\left(\sqrt{3}+i\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)}\right)^{3^{-p-1}}\\\
\left(\tanh\left(\frac{1}{2}3^{p+1}x\right)\coth\left(\frac{3^{p}x}{2}\right)\right)^{2\times
3^{-p-1}}\\\ \left(\sinh\left(2\times
3^{-p-1}\left(\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{3^{p}x}{2}\right)\right)\right)\right.\right.\right.\\\
\left.\left.\left.-\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)\right)\right)\right)\right.\\\
\left.+\cosh\left(2\times
3^{-p-1}\left(\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{3^{p}x}{2}\right)\right)\right)\right.\right.\right.\\\
\left.\left.\left.-\tanh^{-1}\left(\frac{1}{2}\left(1-i\sqrt{3}\tanh\left(\frac{1}{2}3^{p+1}x\right)\right)\right)\right)\right)\right)\\\
=\frac{2\cosh(x)+1}{2\cosh(x)-1}$

###### Proof.

Use equation (11.6) set $x=3xi$ and analyze the right-hand side as
$n\to\infty$ using Figure 12. ∎

Figure 11. Plot of
$\left(\tan\left(\frac{1}{2}3^{n-1}x\right)\cot\left(\frac{3^{n}x}{2}\right)\right)^{3^{-n}}$,
$x\in\mathbb{R}$.

######  Example 11.8.

Finite product involving the Hurwitz-Lerch zeta function.

(11.8)
$\prod_{p=0}^{n-1}\exp\left(3^{-p}e^{-5m3^{p}}\left(e^{4m3^{p}}\Phi\left(e^{-2\times
3^{p+1}m},1,\frac{1}{6}\left(1+3^{-p}\right)\right)\right.\right.\\\
\left.\left.+\Phi\left(e^{-2\times
3^{p+1}m},1,\frac{1}{6}\left(5+3^{-p}\right)\right)\right)\right)\\\
=2^{-3e^{m}}(\coth(m)+1)^{3e^{m}}\exp\left(-3^{1-n}e^{-m3^{n}}\Phi\left(e^{-2\times
3^{n}m},1,\frac{1}{2}\left(1+3^{-n}\right)\right)\right)$

###### Proof.

Use equation (6.2) and set $a=e^{i}$ and simplify in terms of the
Polylogarithm function using equation (64:12:2) in [12]. Next simplify the
Polylogarithm function in terms of the Hurwitz zeta function using equation
(6) in [16]. Next take the limit of both sides as $k\to-1$ and simplify in
terms of the log-gamma function using equation (64:10:2) in [12]. Next take
the exponential function of both sides and simplify. ∎

######  Example 11.9.

Finite product involving the Hurwitz-Lerch zeta function.

(11.9)
$\prod_{p=0}^{n-1}\exp\left(3^{-p}e^{im3^{p}}\left(\Phi\left(e^{2i3^{p+1}m},1,\frac{1}{6}\left(1+3^{n-p}\right)\right)+e^{4im3^{p}}\Phi\left(e^{2i3^{p+1}m},1,\frac{1}{6}\left(5+3^{n-p}\right)\right)\right)\right)\\\
=\left(1-e^{2im3^{n}}\right)^{3^{1-n}e^{-im3^{n}}}e^{3e^{im}\Phi\left(e^{2im},1,\frac{1}{2}\left(1+3^{n}\right)\right)}$

###### Proof.

Use equation (6.2) and set $a=e^{I3^{n}}$ and simplify in terms of the
Polylogarithm function using equation (64:12:2) in [12]. Next simplify the
Polylogarithm function in terms of the Hurwitz zeta function using equation
(6) in [16]. Next take the limit of both sides as $k\to-1$ and simplify in
terms of the log-gamma function using equation (64:10:2) in [12]. Next take
the exponential function of both sides and simplify. ∎

######  Example 11.10.

Finite series involving the Glaisher-Kinkelin constant $A$.

(11.10) $\sum_{p=0}^{n-1}\frac{\csc^{2}\left(\frac{\pi
3^{p}}{2}\right)}{8\left(2\cos\left(\pi
3^{p}\right)+1\right)^{2}}\left(2\log\left(i3^{p+1}\right)\left(3^{p}\left(5\cos\left(\frac{\pi
3^{p}}{2}\right)+\cos\left(\frac{5\pi 3^{p}}{2}\right)\right)\right.\right.\\\
\left.\left.-i\left(\sin\left(\frac{\pi 3^{p}}{2}\right)+\sin\left(\frac{5\pi
3^{p}}{2}\right)\right)\right)-2\times 3^{p+1}e^{-\frac{5}{2}i\pi
3^{p}}\left(-1+e^{i\pi 3^{p+1}}\right)^{2}\right.\\\
\left.\left(\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(3^{-p}+1\right)\right)+e^{2i\pi
3^{p}}\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(3^{-p}+5\right)\right)\right)\right)\\\
=-3^{n}e^{\frac{1}{2}i\pi
3^{n}}\Phi^{\prime}\left(-1,-1,\frac{1}{2}\left(3^{-n}+1\right)\right)+i\log\left(\frac{A^{3}}{\sqrt[3]{2}\sqrt[4]{e}}\right)\\\
+\frac{\pi\cos\left(\pi
3^{n}\right)+4\log\left(i3^{n}\right)\left(3^{n}\cos\left(\frac{\pi
3^{n}}{2}\right)-i\sin\left(\frac{\pi
3^{n}}{2}\right)\right)-\pi}{8\left(\cos\left(\pi 3^{n}\right)-1\right)}$

###### Proof.

Use equation (6.2) and take the first derivative with respect to $k$ and set
$k=1,m=\pi/2,a=e^{i}$ and simplify using equation (18) in [15]. ∎

######  Example 11.11.

Finite series involving Apéry’s constant $\zeta(3)$.

(11.11) $\sum_{p=0}^{n-1}\csc^{3}\left(\frac{1}{2}\pi
3^{p+1}\right)\left(4\times
9^{p+1}\Phi^{\prime}\left(-1,-2,\frac{1}{6}\left(3^{-p}+1\right)\right)\right.\\\
\left.+4\times
9^{p+1}\Phi^{\prime}\left(-1,-2,\frac{1}{6}\left(3^{-p}+5\right)\right)+\left(5\times
9^{p}-1\right)\log\left(i3^{p+1}\right)\right)\\\ =4\times
9^{n}\csc^{3}\left(\frac{\pi
3^{n}}{2}\right)\Phi^{\prime}\left(-1,-2,\frac{1}{2}\left(3^{-n}+1\right)\right)+\frac{1}{2}\left(9^{n}-1\right)\log\left(i3^{n}\right)\csc^{3}\left(\frac{\pi
3^{n}}{2}\right)\\\ -\frac{7\zeta(3)}{\pi^{2}}$

###### Proof.

Use equation (6.2) and take the first derivative with respect to $k$ and set
$k=2,m=\pi/2,a=0$ and simplify using equation (19) in [15]. ∎

######  Example 11.12.

Finite series involving Catalan’s constant $C$.

(11.12) $\sum_{p=0}^{n-1}3^{p}e^{-\frac{5}{2}i\pi
3^{p}}\csc^{2}\left(\frac{1}{2}\pi
3^{p+1}\right)\left(-12\left(\Phi^{\prime}\left(-1,-1,\frac{1}{6}\right)+\Phi^{\prime}\left(-1,-1,\frac{5}{6}\right)\right)\right.\\\
\left.+e^{\frac{5}{2}i\pi
3^{p}}\log\left(i3^{p+1}\right)\left(5\cos\left(\frac{\pi
3^{p}}{2}\right)+\cos\left(\frac{5\pi 3^{p}}{2}\right)\right)\right)\\\
=\frac{2\left(4C\left(3^{n}e^{\frac{1}{2}i\pi 3^{n}}\sin^{2}\left(\frac{\pi
3^{n}}{2}\right)-i\right)+\pi 3^{n}\log\left(i3^{n}\right)\cos\left(\frac{\pi
3^{n}}{2}\right)\right)}{\pi\left(\cos\left(\pi 3^{n}\right)-1\right)}$

###### Proof.

Use equation (6.2) and take the first derivative with respect to $k$ and set
$k=1,m=\pi/2,a=0$ and simplify using equation (20) in [15]. ∎

##  12\. Table of results: Part III

In this section we evaluate Theorem (6.3) for various values of the parameters
involved to derive related closed form formulae in terms of special functions
and fundamental constants.

######  Example 12.1.

Finite product of quotient gamma functions.

(12.1) $\prod_{p=0}^{n-1}\left(1-\frac{2\times 3^{p+1}}{a}\right)^{2\times
3^{-p-1}}\left(\frac{\Gamma\left(\frac{1}{12}\left(3^{-p}a-6\right)\right)}{\Gamma\left(\frac{1}{4}3^{-p-1}a\right)}\right)^{2\times
3^{-p-1}}\\\
\left(\frac{3^{-\frac{1}{4}a3^{-p}-2p+\frac{3}{2}}\left(a^{2}-4a3^{p+1}+32\times
9^{p}\right)\Gamma\left(\frac{3^{-p}a}{4}-3\right)}{\Gamma\left(\frac{1}{4}3^{-p-1}a-1\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+2\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+10\right)\right)}\right)^{1-3^{-p}}\\\
=\frac{2^{\frac{3}{2}\left(2n+3^{1-n}-3\right)}3^{\frac{1}{4}\left(-2n-3^{1-n}+3\right)}\pi^{\frac{3}{2}\left(1-3^{-n}\right)-n}\left(1-\frac{2\times
3^{n}}{a}\right)^{-3^{-n}}}{a}\\\ \left(a-2\times
3^{n}\right)\left(\frac{\Gamma\left(\frac{1}{4}\left(3^{-n}a-2\right)\right)}{\Gamma\left(\frac{3^{-n}a}{4}\right)}\right)^{1-3^{-n}}$

###### Proof.

Use equation (6.3) and set $m=0$ and simplify in terms of the Hurwitz zeta
function using entry (4) in the Table below equation (64:12:7) on page 692 in
[12]. Next take the first partial derivative with respect to $k$ and set $k=0$
and simplify in terms of the log-gamma function using equation (64:10:2) in
[12]. Finally take the exponential function of both sides and simplify both
sides to yield the stated result. ∎

######  Example 12.2.

Finite product of quotient gamma functions.

(12.2)
$\prod_{p=0}^{n-1}\frac{\left(\left(a3^{-p}-4\right)\left(a3^{-p}-2\right)\left(a3^{-p-1}\Gamma\left(\frac{1}{2}3^{-p-1}a\right)\right)^{2/3}\Gamma\left(\frac{1}{6}\left(3^{-p}a-4\right)\right)\Gamma\left(\frac{1}{6}\left(3^{-p}a-2\right)\right)\right)^{1-3^{-p}}}{\Gamma\left(\frac{1}{2}3^{-p-1}a\right)^{2/3}}\\\
=2^{\frac{3}{2}\left(2n+3^{1-n}-3\right)}\pi^{\frac{3}{2}\left(3^{-n}-1\right)+n}3^{\frac{1}{16}9^{-n}\left(4\times
3^{n}\left(2\left(5\times
3^{n}-2\right)n-9\left(3^{n}-1\right)\right)-3a\left(-4\times
3^{n}+9^{n}+3\right)\right)-\frac{1}{3}n(n+1)}a^{2n/3}\\\
\left(a\Gamma\left(\frac{3^{-n}a}{2}\right)\right)^{3^{-n}-1}$

###### Proof.

Use equation (6.1) and set $m=\pi/2$ and simplify in terms of the Hurwitz zeta
function using entry (4) in the Table below equation (64:12:7) on page 692 in
[12]. Next take the first partial derivative with respect to $k$ and set $k=0$
and simplify in terms of the log-gamma function using equation (64:10:2) in
[12]. Finally take the exponential function of both sides and simplify both
sides to yield the stated result. ∎

######  Example 12.3.

Functional equation for the Hurwtiz-Lerch zeta function.

(12.3)
$\Phi(z,s,a)=3^{-2s-1}\left(3^{s}\left(3\Phi\left(z^{3},s,\frac{a}{3}\right)+z\left(3\Phi\left(z^{3},s,\frac{a+1}{3}\right)-z\Phi\left(z^{3},s,\frac{a+2}{3}\right)\right)\right)\right.\\\
\left.+4z^{2}\left(\Phi\left(z^{9},s,\frac{a+2}{9}\right)+z^{6}\Phi\left(z^{9},s,\frac{a+8}{9}\right)+z^{3}\Phi\left(z^{9},s,\frac{a+5}{9}\right)\right)\right)$

###### Proof.

Use equation (6.3) and set $n=3,m=\frac{\log(-z)}{6i},a=6(ai-1),k=-s$ and
simplify. ∎

######  Example 12.4.

Finite product of quotient cosine functions.

(12.4) $\prod_{p=0}^{n-1}\left(\frac{\left(-1+2\cosh\left(2\times
3^{p}m\right)\right)\left(\frac{\cosh\left(3^{p}r\right)}{\cosh\left(3^{p}m\right)}\right)^{2}}{-1+2\cosh\left(2\times
3^{p}r\right)}\right)^{3^{1-2p}\left(-1+3^{p}\right)}\left(\frac{\cosh\left(3^{1+p}m\right)}{\cosh\left(3^{1+p}r\right)}\right)^{2\times
3^{-2p}}\\\
=\left(\frac{\cosh\left(3^{n}m\right)}{\cosh\left(3^{n}r\right)}\right)^{9^{1-n}\left(-1+3^{n}\right)}$

###### Proof.

Use equation (6.3) and form a second equation by replacing $m\to r$. Next take
the differnce of the two equations and simplify. Then set $k=-1,a=1$ and
simplify in terms of the logarithm function using entry (5) in the Table below
equation (64:12:7) on page 692 in [12]. Next take the exponential function of
both sides and simplify to yield the stated result. ∎

######  Example 12.5.

Finite product of quotient cosine functions.

(12.5) $\prod_{p=0}^{n-1}\left(\frac{1-2\cosh\left(2\times
3^{p}x\right)}{1-2\cosh\left(2\times
3^{-1+p}x\right)}\right)^{\frac{1}{2}9^{-p}\left(-1+3^{1+p}\right)}\left(\frac{\cosh\left(3^{-1+p}x\right)}{\cosh\left(3^{p}x\right)}\right)^{9^{-p}\left(-4+3^{1+p}\right)}\\\
=\left(\frac{\cosh\left(3^{n}x\right)}{\cosh\left(3^{-1+n}x\right)}\right)^{\frac{1}{2}9^{1-n}\left(-1+3^{n}\right)}$

###### Proof.

Use equation (6.3) and set $k=1,a=0,m=x$ and apply the method in section (8)
in [17]. ∎

###  12.1. Finite product involving polynomials

Finite products involving polynomials are used in the study of the Thue-Morse
constant [18, 19], where this constant is not a badly approximable number. In
his renowned 1936 publication, Alan Turing explored the concept of the
computability of real numbers. In simpler terms, a real number, denoted as
$\alpha$, is deemed computable when there exists a Turing machine capable of
generating a rational approximation to $\alpha$, which is accurate within a
margin of $2^{-i}$, where $i$ represents the input value. The finite product
involving polynomials is used to study the transcendence of the Thue–Morse
Number on page 388 in [19], where a proof using a theorem on analytic
functions was employed.

######  Example 12.6.

Finite product involving polynomial functions.

(12.6)
$\prod_{p=0}^{n-1}\left(z^{3^{p}}+1\right)^{3^{-2p-1}}\left(\frac{\sqrt{\left(z^{3^{p-1}}-1\right)z^{3^{p-1}}+1}}{z^{3^{p-1}}+1}\right)^{3^{-2p}\left(3^{p}-1\right)}\\\
=\left(z^{3^{n-1}}+1\right)^{\frac{1}{2}3^{1-2n}\left(3^{n}-1\right)}\\\
=\sum_{p=0}^{\infty}\binom{\frac{1}{2}3^{1-2n}\left(-1+3^{n}\right)}{p}\left(z^{3^{n-1}}\right)^{\frac{1}{2}3^{1-2n}\left(3^{n}-1\right)-p}$

###### Proof.

Use equation (6.3) and set $a=0,m=\log(z)/(6i)$ and simplify in terms of the
Polylogarithm function using equation (64:12:2) in [12]. Next simplify the
Polylogarithm function in terms of the Hurwitz zeta function using equation
(6) in [16]. Next take the limit of both sides as $k\to-1$ and simplify in
terms of the log-gamma function using equation (64:10:2) in [12]. Next take
the exponential function of both sides and simplify. ∎

Figure 12. Plot of
$f(z)=Re\left(\left(z^{3^{n-1}}+1\right)^{\frac{1}{2}3^{1-2n}\left(3^{n}-1\right)}\right)$,
$z\in\mathbb{C}$. Figure 13. Plot of
$f(z)=Im\left(\left(z^{3^{n-1}}+1\right)^{\frac{1}{2}3^{1-2n}\left(3^{n}-1\right)}\right)$,
$z\in\mathbb{C}$. Figure 14. Plot of
$f(z)=Abs\left(\left(z^{3^{n-1}}+1\right)^{\frac{1}{2}3^{1-2n}\left(3^{n}-1\right)}\right)$,
$z\in\mathbb{C}$.

######  Example 12.7.

Infinite product involving polynomial functions.

(12.7)
$\prod_{p=0}^{\infty}\left(z^{3^{p}}+1\right)^{3^{-2p-1}}\left(\frac{\sqrt{\left(z^{3^{p-1}}-1\right)z^{3^{p-1}}+1}}{z^{3^{p-1}}+1}\right)^{3^{-2p}\left(3^{p}-1\right)}=1$

###### Proof.

Use equation (12.6) and take the limit of the right-hand side as $n\to\infty$
and simplify. ∎

######  Example 12.8.

Finite sum of the digamma function in terms of Catalan’s constant $C$.

(12.8)
$\sum_{p=0}^{n-1}27^{-p}\left(\psi^{(1)}\left(1-\frac{1}{4}3^{n-p-1}\right)+\frac{3}{2}\left(3^{p}-1\right)\left(\psi^{(1)}\left(\frac{5}{6}-\frac{1}{4}3^{n-p-1}\right)-\psi^{(1)}\left(\frac{1}{12}\left(4-3^{n-p}\right)\right)\right)\right.\\\
\left.-\psi^{(1)}\left(\frac{1}{12}\left(6-3^{n-p}\right)\right)+\frac{3}{2}\left(3^{p}-1\right)\left(\psi^{(1)}\left(\frac{1}{12}\left(2-3^{n-p}\right)\right)-\psi^{(1)}\left(\frac{1}{12}\left(8-3^{n-p}\right)\right)\right)\right)\\\
=-8C3^{3-3n}\left(3^{n}-1\right)$

###### Proof.

Use equation (6.3) and set $k=-2,m=0,a=-3^{n}$ and simply using equation (21)
in [15]. ∎

######  Example 12.9.

Finite series in terms of the Glaisher-Kinkelin constant $A$.

(12.9)
$\sum_{p=0}^{n-1}\left(6\left(3^{p}-1\right)\left(\Phi^{\prime}\left(-1,-1,\frac{1}{3}\right)-\Phi^{\prime}\left(-1,-1,\frac{2}{3}\right)\right)+3^{p}\log\left(i3^{p+1}\right)\right)\\\
=\frac{1}{6}\left(\left(-2n+3^{n}-1\right)\log\left(\frac{16e^{3}}{A^{36}}\right)+3\left(3^{n}-1\right)\log\left(i3^{n}\right)\right)$

###### Proof.

Use equation (6.3) and take the first derivative with respect to $k$ and set
$k=1,m=0,a=0$ and simplify using equation (18) in [15]. ∎

######  Example 12.10.

Finite series in terms of the Apery’s constant $\zeta(3)$.

(12.10)
$\sum_{p=0}^{n-1}3^{p}\left(3^{p}-1\right)\left(\Phi^{\prime}\left(-1,-2,\frac{1}{3}\right)-\Phi^{\prime}\left(-1,-2,\frac{2}{3}\right)\right)=-\frac{7\left(-4\times
3^{n}+9^{n}+3\right)\zeta(3)}{36\pi^{2}}$

###### Proof.

Use equation (6.3) and take the first derivative with respect to $k$ and set
$k=2,m=0,a=0$ and simplify using equation (19) in [15]. ∎

######  Example 12.11.

Finite sum of the digamma function in terms of Catalan’s constant $C$.

(12.11)
$\sum_{p=0}^{n-1}\left(12\Phi^{\prime}\left(-1,-1,1-\frac{1}{2}3^{n-p-1}\right)-18\left(3^{p}-1\right)\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(2-3^{n-p}\right)\right)\right.\\\
\left.+18\left(3^{p}-1\right)\Phi^{\prime}\left(-1,-1,\frac{1}{6}\left(4-3^{n-p}\right)\right)+3^{-p}\left(3^{n}-3^{2p+1}\right)\log\left(i3^{p+1}\right)\right)\\\
=\frac{6C\left(3^{n}-1\right)}{\pi}$

###### Proof.

Use equation (6.3) and take the first derivative with respect to $k$ and set
$k=1,m=0,a=-3^{n}$ and simplify using equation (20) in [15]. ∎

$\prod\limits_{p=0}^{n-1}\left(\frac{9^{p+1}\Gamma\left(3^{-p-1}a+\frac{1}{6}\right)\Gamma\left(3^{-p-1}a+\frac{5}{6}\right)\left(\frac{\Gamma\left(3^{-p-1}a+\frac{1}{2}\right)}{\Gamma\left(3^{-p-1}a+1\right)}\right)^{2/3}}{\left(3^{p}-a\right)\left(2\times 3^{p}-a\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-2\right)\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-1\right)\right)}\right)^{3^{-p}}$ | $\frac{3^{\frac{3}{4}\left(1-3^{-n}\right)}\Gamma\left(a+\frac{1}{2}\right)\left(\frac{\Gamma\left(3^{-n}a+1\right)}{\Gamma\left(3^{-n}a+\frac{1}{2}\right)}\right)^{3^{-n}}}{\Gamma(a+1)}$
---|---
$\prod\limits_{p=0}^{\infty}\left(\frac{9^{p+1}\Gamma\left(3^{-p-1}a+\frac{1}{6}\right)\Gamma\left(3^{-p-1}a+\frac{5}{6}\right)\left(\frac{\Gamma\left(3^{-p-1}a+\frac{1}{2}\right)}{\Gamma\left(3^{-p-1}a+1\right)}\right)^{2/3}}{\left(3^{p}-a\right)\left(2\times 3^{p}-a\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-2\right)\right)\Gamma\left(\frac{1}{3}\left(3^{-p}a-1\right)\right)}\right)^{3^{-p}}$ | $\frac{3^{3/4}\Gamma\left(a+\frac{1}{2}\right)}{\Gamma(a+1)}$
$\prod\limits_{p=0}^{n-1}\left(\frac{\left(\frac{\cos\left(3^{p}m\right)}{\cos\left(3^{p}r\right)}\right)^{16}\left(1-2\cos\left(2\times 3^{p}r\right)\right)^{2}}{\left(1-2\cos\left(2\times 3^{p}m\right)\right)^{2}}\right)^{3^{-2p}}$ | $\left(\frac{\cos(m)}{\cos(r)}\right)^{18}\left(\frac{\cos\left(3^{n}r\right)}{\cos\left(3^{n}m\right)}\right)^{2\times 3^{2-2n}}$
$\prod\limits_{p=0}^{\infty}\left(\frac{\left(\frac{\cos\left(3^{p}m\right)}{\cos\left(3^{p}r\right)}\right)^{16}\left(1-2\cos\left(2\times 3^{p}r\right)\right)^{2}}{\left(1-2\cos\left(2\times 3^{p}m\right)\right)^{2}}\right)^{\frac{3^{-2p}}{18}}$ | $\frac{\cos(m)}{\cos(r)}$
$\prod\limits_{p=0}^{\infty}\left(\frac{3^{a3^{-p}+2p}\Gamma\left(3^{-p-1}a-1\right)}{\left(a^{2}-a3^{p+1}+2\times 9^{p}\right)\Gamma\left(3^{-p-1}a+1\right)^{2/3}\Gamma\left(3^{-p}a-3\right)}\right)^{3^{-p}}$ | $\frac{3^{\frac{9a}{8}+3}}{\Gamma(a+1)}$
$\prod\limits_{p=0}^{n-1}\frac{\Gamma\left(\frac{1}{6}\left(3^{-p}a+1\right)\right)\Gamma\left(\frac{1}{6}\left(3^{-p}a+5\right)\right)}{2\pi}$ | $\frac{3^{-\frac{1}{4}a3^{1-n}\left(3^{n}-1\right)}\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{1}{2}\left(3^{-n}a+1\right)\right)}$
$\prod\limits_{p=0}^{\infty}\frac{\Gamma\left(\frac{1}{6}\left(3^{-p}a+1\right)\right)\Gamma\left(\frac{1}{6}\left(3^{-p}a+5\right)\right)}{2\pi}$ | $\frac{3^{-3a/4}\Gamma\left(\frac{a+1}{2}\right)}{\sqrt{\pi}}$
$\prod\limits_{p=0}^{\infty}\left(\frac{\left(1+2\cosh\left(2\times 3^{p}m\right)\right)\left(-1+2\cosh\left(2\times 3^{p}r\right)\right)}{\left(-1+2\cosh\left(2\times 3^{p}m\right)\right)\left(1+2\cosh\left(2\times 3^{p}r\right)\right)}\right)^{3^{-p}}\left(\frac{\tanh\left(3^{p}r\right)}{\tanh\left(3^{p}m\right)}\right)^{2\times 3^{-p}}$ | $\left(\frac{\tanh(r)}{\tanh(m)}\right)^{3}$
$\prod\limits_{p=0}^{n-1}\left(\frac{\Gamma\left(\frac{1}{12}\left(3^{-p}a+7\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+11\right)\right)}{\Gamma\left(\frac{1}{4}3^{-p-1}\left(a+3^{p}\right)\right)\Gamma\left(\frac{1}{12}\left(3^{-p}a+5\right)\right)}\right)^{(-1)^{p}}$ | $\frac{3^{\frac{1}{4}\left((-1)^{n}-1\right)}\Gamma\left(\frac{a+3}{4}\right)\left(\frac{\Gamma\left(\frac{1}{4}\left(3^{-n}a+1\right)\right)}{\Gamma\left(\frac{1}{4}\left(3^{-n}a+3\right)\right)}\right)^{(-1)^{n}}}{\Gamma\left(\frac{a+1}{4}\right)}$
$\sum\limits_{p=0}^{n-1}9^{-p}\left(\psi^{(1)}\left(\frac{1}{12}\left(2+3^{-p}\right)\right)+\psi^{(1)}\left(\frac{1}{12}\left(10+3^{-p}\right)\right)\right)$ | $9\left(-8C-9^{-n}\psi^{(1)}\left(\frac{1}{4}\left(2+3^{-n}\right)\right)+\pi^{2}\right)$
$\prod\limits_{p=0}^{n-1}\left(\frac{\left(e^{-2m3^{p+1}}+1\right)^{2/3}\cosh^{2}\left(m3^{p}\right)}{2\cosh\left(2m3^{p}\right)-1}\right)^{3^{-2p}}$ | $\frac{2^{\frac{3}{4}\left(1+3^{1-2n}\right)}\cosh^{3}(m)}{e^{3m}\left(1+e^{-23^{n}m}\right)^{3^{1-2n}}}$
$\prod\limits_{p=0}^{\infty}\left(\frac{\left(e^{-2m3^{p+1}}+1\right)^{2/3}\cosh^{2}\left(m3^{p}\right)}{2\cosh\left(2m3^{p}\right)-1}\right)^{3^{-2p-1}}$ | $\frac{e^{-2m}+1}{2^{3/4}}$
$\prod\limits_{p=0}^{n-1}\left(\frac{\left(-1+2\cosh\left(2\times 3^{p}m\right)\right)\left(\frac{\cosh\left(3^{p}r\right)}{\cosh\left(3^{p}m\right)}\right)^{2}}{-1+2\cosh\left(2\times 3^{p}r\right)}\right)^{3^{1-2p}\left(-1+3^{p}\right)}\left(\frac{\cosh\left(3^{1+p}m\right)}{\cosh\left(3^{1+p}r\right)}\right)^{2\times 3^{-2p}}$ | $\left(\frac{\cosh\left(3^{n}m\right)}{\cosh\left(3^{n}r\right)}\right)^{9^{1-n}\left(-1+3^{n}\right)}$
$\prod\limits_{p=0}^{n-1}\left(\frac{1-2\cosh\left(2\times 3^{p}x\right)}{1-2\cosh\left(2\times 3^{-1+p}x\right)}\right)^{\frac{1}{2}9^{-p}\left(-1+3^{1+p}\right)}\left(\frac{\cosh\left(3^{-1+p}x\right)}{\cosh\left(3^{p}x\right)}\right)^{9^{-p}\left(-4+3^{1+p}\right)}$ | $\left(\frac{\cosh\left(3^{n}x\right)}{\cosh\left(3^{-1+n}x\right)}\right)^{\frac{1}{2}9^{1-n}\left(-1+3^{n}\right)}$
$\prod_{p=0}^{n-1}\left(z^{3^{p}}+1\right)^{3^{-2p-1}}\left(\frac{\sqrt{\left(z^{3^{p-1}}-1\right)z^{3^{p-1}}+1}}{z^{3^{p-1}}+1}\right)^{3^{-2p}\left(3^{p}-1\right)}$ | $\left(z^{3^{n-1}}+1\right)^{\frac{1}{2}3^{1-2n}\left(3^{n}-1\right)}$
Table 1. Table of sums and products

##  13\. Conclusion

In this article, we’ve introduced an approach for obtaining formulae for
finite and infinite sums and products that incorporate the Hurwitz-Lerch zeta
function. We’ve also explored intriguing specific instances through a
combination of contour integration and established algebraic methods. Our
intention is to employ these techniques in forthcoming research to derive
additional formulae for other sums and products involving various special
functions.

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