# Extended Loney trigonometric series

Robert Reynolds

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Robert Reynolds. Extended Loney trigonometric series. 2023. hal-04258377

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Extended Loney trigonometric series

Robert Reynolds

ABSTRACT. In this paper we use an interesting exercise of combining finite trigonometric series to a contour integral method to produce finite series involving special functions. The evaluations derived and conclusions reached are in terms of the fundamental constants and special functions. The finite series derived are in terms of the Hurwitz-Lerch zeta function and its composite functions, like the gamma function, polylogarithm function and zeta function. We complete this short study with a table of summarized results featuring finite and infinite products involving special functions and fundamental constants.

# 1. History

Sydney Luxton Loney, M.A. (March 16, 1860 - May 16, 1939) held the position of a Mathematics Professor at the Royal Holloway College in Egham, Surrey. He was also a fellow at Sidney Sussex College in Cambridge, England. Loney is credited with writing several mathematics textbooks, some of which have seen multiple reprints. He is recognized as one of the early mentors who had a significant impact on the mathematical genius Srinivasa Ramanujan. Loney received his education at Maidstone Grammar School in Tonbridge and pursued further studies at Sidney Sussex College, Cambridge. In 1882, he completed his undergraduate degree with honours, earning a B.A. and securing the position of 3rd Wrangler. His textbooks on Plane Trigonometry and Coordinate Geometry enjoy great popularity among senior high school students in India who are preparing for engineering entrance exams like JEE Advanced.

In chapter (4) of [15], finite series involving the Bessel functions $J_ν(z)$, $I_ν(z)$, Neumann function $Y_ν(z)$, Macdonald function $J_ν(z)$, Legendre polynomials $P_n(x)$, Laguerre polynomials $L_n^α(x)$, Hermite polynomials $H_n(x)$, Gegenbauer polynomials $C_n^λ(x)$ and Jacobi polynomials $P_n^{(α,β)}(x)$. In this paper we derive a finite series of the Hurwitz-Lerch zeta function in terms of the Hurwitz-Lerch zeta function which expands upon current literature of finite series involving special functions. Special functions are useful when solving problems that are not easily expressed in terms of elementary functions like polynomials, trigonometric functions, or exponential functions. Applications of finite series of special functions have many applications namely; Fourier series, Bessel functions, Legendre Polynomials, Hermite
Polynomials, Gamma Function, Error Function, Hypergeometric Functions, Special Functions in Quantum Mechanics, Airy Functions and Whittaker Functions.

## 2. Introduction

An examination of Loney's book on Plane Trigonometry, 1893 reveals a large number of intriguing and obscure formulas. In my opinion one of the most remarkable formula is a finite series given on page 416, question (18), where this finite trigonometric series has an argument involving $3^k$. The trigonometric functions already defined make it possible to use angles to calculate unknown quantities in plane rectilinear figures. Angles can often be measured with less effort and greater accuracy than lengths. As the name indicates, trigonometry is concerned with the measurement or calculation of triangles into which every figure bounded by straight lines can be subdivided by diagonals. In this one always has in mind the use of known angles. In this paper we will apply a contour integral method to a finite trigonometric series which produces a finite series in terms of a special function. This finite series is then used to derive composite formulae in terms of fundamental constants and other special functions. This type of work is an extension on the work produced by the author.

In this work we apply the contour integral method [1], to the generalized form of the finite secant sum on page 416, question (18) in [12] to derive a form involving the finite sum of the cotangent cosine-cosecant functions, resulting in

$$ (2.1) \quad \frac{1}{2\pi i} \int_C \sum_{p=0}^{n-1} \frac{2 \times 3^p a^w w^{-k-1} \sin(2 \times 3^p(m+w))}{2 \cos(2 \times 3^p(m+w)) + 1} dw \\ = \frac{1}{2\pi i} \int_C \frac{1}{2} a^w w^{-k-1} (\cot(m+w) - 3^n \cot(3^n(m+w))) dw $$

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

$$ (2.2) \quad \sum_{p=0}^{n-1} 3^p (i3^{p+1})^k e^{2im3^p} \left( \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{6} \left(2 - i3^{-p} \log(a)\right) \right) \right. \\ \qquad \left. \qquad + e^{2im3^p} \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{6} \left(4 - i3^{-p} \log(a)\right) \right) \right. \\ \qquad \left. \qquad - 2e^{4im3^p} \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{6} \left(6 - i3^{-p} \log(a)\right) \right) \right) \\ = i^k e^{2im} \Phi \left( e^{2im}, -k, 1 - \frac{1}{2}i \log(a) \right) - 3^n (i3^n)^k e^{2im3^n} \Phi \left( e^{2i3^n m}, -k, 1 - \frac{1}{2}i3^{-n} \log(a) \right) $$

where the variables $k, a, m$ are general complex numbers and $n$, a positive integer. The derivations follow the method used by us in [1]. This method involves using a form of the generalized Cauchy's integral formula given by

$$ (2.3) \quad \frac{y^k}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C \frac{e^{wy}}{w^{k+1}} dw, $$

where $y, w \in \mathbb{C}$ and $C$ is in general an open contour in the complex plane where the bilinear concomitant [1] is equal to zero at the end points of the contour. This
method involves using a form of equation (2.3) then multiplies both sides by a function, then takes the finite sum of both sides. This yields a finite sum in terms of a contour integral. Then we multiply both sides of equation (2.3) 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 [3] 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

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

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

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

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

#### 3.1. Derivation of the generalized contour integrals.
In this section we derive generalized trigonometric contour integrals in terms of the Hurwitz-Lerch zeta function. These contour integrals will be used to derive equation (2.2).

##### 3.1.1. Hurwitz-Lerch zeta function in terms of the cotangent contour integral representation.

##### 3.1.2. Derivation of the additional contour.
Using a generalization of Cauchy's integral formula (2.3), first replace $y \to \log(a)$ then multiply both sides by $i$ and simplify to get;

$$ (3.3) \qquad \frac{i \log^k(a)}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C ia^w w^{-k-1} dw $$

Using a generalization of Cauchy's integral formula (2.3), first replace $y \to \log(a) + 2ib(y+1)$ then multiply both sides by $-2ie^{2ibm(y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch zeta function and subtract equation (3.3) to get

$$ (3.4) \qquad \frac{i \left( \log^k(a) + 2^{k+1} (ib)^k e^{2ibm} \Phi \left( e^{2ibm}, -k, 1 - \frac{i \log(a)}{2b} \right) \right)}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C a^w w^{-k-1} e^{2ib(y+1)(m+w)} dw $$
$$ = \frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} a^w w^{-k-1} e^{2ib(y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C a^w w^{-k-1} \cot(b(m+w)) dw $$

from equations (1.232.2) and (1.411.1) in [2] where $Re(w+m) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. We apply Tonelli's theorem for sums and integrals, see page 177 in [7] as the summand and integral are of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

3.1.3. *Hurwitz-Lerch zeta function in terms of the cosine-cosecant contour integral representation.* We use the method in [1]. Using a generalization of Cauchy's integral formula (2.3) 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

$$ (3.5) \quad \frac{e^{-imx} (e^{2imx}(\log(a) + ix + y)^k + (\log(a) - ix + y)^k)}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \int_C 2w^{-k-1} e^{w(\log(a)+y)} \cos(x(m+w)) dw $$

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

$$ (3.6) \quad \frac{i2^k (ib)^k e^{im(b-x)} \left( \Phi \left( e^{2ibm}, -k, \frac{b-x-i \log(a)}{2b} \right) + e^{2imx} \Phi \left( e^{2ibm}, -k, \frac{b+x-i \log(a)}{2b} \right) \right)}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 2a^w w^{-k-1} e^{ibm(2y+1)+ibw(2y+1)} \cos(x(m+w)) dw \\ = \frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 2a^w w^{-k-1} e^{ibm(2y+1)+ibw(2y+1)} \cos(x(m+w)) dw \\ = \frac{1}{2\pi i} \int_C a^w w^{-k-1} \csc(b(m+w)) \cos(x(m+w)) dw $$

from equation (1.232.2) and (1.411.3) in [2] where $Re(w+m) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. We apply Tonelli's theorem for sums and integrals, see page 177 in [7] as the summand and integral are of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

3.1.4. *Left-hand side first contour integral.* Use equation (3.4) and replace $b$ by $3^{p+1}$ and take the finite sum over $p \in [0, n-1]$ and simplify to get;

$$ (3.7) \quad \frac{i3^p \left( \log^k(a) + 2^{k+1} (i3^{p+1})^k e^{2im3^{p+1}} \Phi \left( e^{2i3^{p+1}m}, -k, 1 - \frac{1}{2}i3^{-p-1} \log(a) \right) \right)}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \int_C \sum_{p=0}^{n-1} 3^p a^w w^{-k-1} \cot(3^{p+1}(m+w)) dw $$
3.1.5. Left-hand side second contour integral. Use equation (3.6) and replace b by $3^{p+1}$, x by $3^p$ and take the finite sum over $p \in [0, n-1]$ and simplify to get;

$$ (3.8) \quad - \sum_{p=0}^{n-1} \frac{1}{\Gamma(k+1)} i^{2k} 3^p (i3^{p+1})^k e^{im(3^{p+1}-3^p)} \\ \qquad \left( \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{2}3^{-p-1} \left(-i\log(a) - 3^p + 3^{p+1}\right) \right) \right. \\ \qquad \left. + e^{2im3^p} \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{2}3^{-p-1} \left(-i\log(a) + 3^p + 3^{p+1}\right) \right) \right) \\ \qquad = \frac{1}{2\pi i} \int_C \sum_{p=0}^{n-1} 3^p a^w w^{-k-1} \cos(3^p(m+w)) \csc(3^{p+1}(m+w)) dw $$

3.1.6. Right-hand side first contour integral. Use equation (3.4) and replace b by 1 and simplify to get;

$$ (3.9) \quad - \frac{i \left( \log^k(a) + i^k 2^{k+1} e^{2im\Phi} \left( e^{2im}, -k, 1 - \frac{1}{2}i \log(a) \right) \right)}{2\Gamma(k+1)} \\ = \frac{1}{2\pi i} \int_C \frac{1}{2} a^w w^{-k-1} \cot(m+w) dw $$

3.1.7. Right-hand side second contour integral. Use equation (3.4) and replace b by $3^n$ and simplify to get;

$$ (3.10) \quad \frac{i3^n \left( \log^k(a) + 2^{k+1} (i3^n)^k e^{2im3^n} \Phi \left( e^{2i3^n m}, -k, 1 - \frac{1}{2}i3^{-n} \log(a) \right) \right)}{2\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \int_C \frac{1}{2} 3^n a^w w^{-k-1} \cot(3^n(m+w)) dw $$

## 4. Evaluations involving the finite sum and products of special functions

In this section we will derive our main theorem and evaluate this theorem to produce composite finite sums and products involving special constants and special functions.

THEOREM 4.1. For all $k, a, m \in \mathbb{C}$ then,

$$ (4.1) \quad \sum_{p=0}^{n-1} 3^p (i3^{p+1})^k e^{2im3^p} \left( \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{6} \left(2 - i3^{-p} \log(a)\right) \right) \right. \\ \qquad \left. + e^{2im3^p} \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{6} \left(4 - i3^{-p} \log(a)\right) \right) \right. \\ \qquad \left. - 2e^{4im3^p} \Phi \left( e^{2i3^{p+1}m}, -k, \frac{1}{6} \left(6 - i3^{-p} \log(a)\right) \right) \right) \\ = i^k e^{2im} \Phi \left( e^{2im}, -k, 1 - \frac{1}{2}i \log(a) \right) - 3^n (i3^n)^k e^{2im3^n} \Phi \left( e^{2i3^n m}, -k, 1 - \frac{1}{2}i3^{-n} \log(a) \right) $$
PROOF. With respect to equation (2.1) and observing the addition of the right-hand sides of relations (3.7) and (3.8), and the addition of relations (3.9) and (3.10) are identical; hence, the left-hand sides of the same are identical too. Simplifying with the Gamma function yields the desired conclusion. □

EXAMPLE 4.2. The degenerate case.

$$ (4.2) \qquad \sum_{p=0}^{n-1} \frac{3^p \sin(2m3^p)}{2 \cos(2m3^p) + 1} = \frac{1}{4} (\cot(m) - 3^n \cot(m3^n)) $$

PROOF. Use equation (4.1) and set $k = 0$ and simplify using entry (2) in the Table below equation (64:12:7) on page 692 in [4] to yield the stated result. □

EXAMPLE 4.3. Finite product involving quotient gamma functions.

$$ (4.3) \qquad \prod_{p=0}^{n-1} \left( \frac{\Gamma(3^{-1-p}a)^2 \Gamma(-1+3^{-1-p}a)}{3^{-3-p}a (3^p-a) (2 \times 3^p-a) \Gamma(-3+3^{-p}a)} \right)^{3^p} \\ = \frac{3^{-2+2 \times 3^n+an} \left( \frac{\Gamma(1+3^{-n}a)}{a} \right)^{3^n}}{\Gamma(a)} $$

PROOF. Use equation (4.1) 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 [4]. 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 [4]. Finally take the exponential function of both sides and simplify both sides to yield the stated result. □

EXAMPLE 4.4. Finite product of quotient gamma functions.

$$ (4.4) \qquad \prod_{p=0}^{n-1} \left( \frac{3^{-\frac{1}{4}} a^{3^{-p}} - p + \frac{5}{2}}{\Gamma(\frac{1}{4}3^{-p-1}a)^2 \Gamma(\frac{1}{4}3^{-p-1}a-1) \Gamma(\frac{1}{12}(3^{-p}a+2)) \Gamma(\frac{1}{12}(3^{-p}a+10))} \right)^{3^p} \\ = - \frac{(3^{-n/2})^{3^n} e^{-i\pi 3^n} (2\pi)^{\frac{1}{2}(1-3^n)} a^{3^n-1} \Gamma(\frac{a}{4}+1) \left( \frac{\Gamma(\frac{1}{4}(3^{-n}a+2))}{\Gamma(\frac{3-n}{4}+1)} \right)^{3^n}}{\Gamma(\frac{a+2}{4})} $$

PROOF. Use equation (4.1) 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 [4]. 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 [4]. Finally take the exponential function of both sides and simplify both sides to yield the stated result. □

EXAMPLE 4.5. Finite product in terms of quotient sine functions.

$$ (4.5) \qquad \prod_{p=0}^{n-1} \frac{1}{1 + 2 \cos (2 \times 3^p m)} = \frac{\sin(m)}{\sin(3^m m)} $$
PROOF. Use equation (4.1) and set $a = 1$ and simplify in terms of the Polylogarithm function using equation (64:12:2) in [4]. Next simplify the Polylogarithm function in terms of the Hurwitz zeta function using equation (6) in [19]. 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 [4]. Next take the exponential function of both sides and simplify. $\square$

EXAMPLE 4.6. Finite product in terms of quotient sine functions.

$$ (4.6) \qquad \prod_{p=0}^{n-1} \frac{1}{2 \cos (2m3^{-p-1}) + 1} = \csc(m) \sin (m3^{-n}) $$

PROOF. Use equation (4.5) and simplify using the method in equations (91.8.7) and (91.8.9) in [14]. $\square$

EXAMPLE 4.7. Finite product in terms of quotient sine functions.

$$ (4.7) \qquad \prod_{p=0}^{n-1} \frac{1 + 2 \cos (2 \times 3^p r)}{1 + 2 \cos (2 \times 3^p m)} = \frac{\sin (3^n r) \sin (m)}{\sin (3^n m) \sin (r)} $$

PROOF. Use equation (4.1) and form a second equation by replacing $m \to r$. Next take the difference 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 [4]. Next take the exponential function of both sides and simplify to yield the stated result. $\square$

EXAMPLE 4.8. Finite product in terms of quotient sine functions.

$$ (4.8) \qquad \prod_{p=0}^{n-1} \frac{1 + 2 \cos (2 \times 3^{-1+p}x)}{1 + 2 \cos (2 \times 3^p x)} = \frac{\sin(x) \sin (3^{-1+n}x)}{\sin (3^n x) \sin (\frac{x}{3})} $$

PROOF. Use equation (4.1) and set $k = 1, a = 1, m = x$ and apply the method in section (8) in [10]. $\square$

EXAMPLE 4.9. Finite series in terms of Catalan's constant $C$.

$$ (4.9) \qquad \begin{aligned} & \sum_{p=0}^{n-1} 3^{-p} \left( -\psi^{(1)}\left(\frac{1}{12}(2-3^{-p})\right) + 2e^{2i\pi 3^p} \left( \psi^{(1)}\left(\frac{1}{12}(6-3^{-p})\right) - \psi^{(1)}\left(1-\frac{3^{-p-1}}{4}\right) \right) \right. \\ & \left. \qquad + \psi^{(1)}\left(\frac{1}{12}(8-3^{-p})\right) + e^{i\pi 3^p} \left( \psi^{(1)}\left(\frac{1}{12}(10-3^{-p})\right) - \psi^{(1)}\left(\frac{1}{12}(4-3^{-p})\right) \right) \right) \\ & = 9 \left( 3^{-n} \left( \psi^{(1)}\left(\frac{1}{2}-\frac{3^{-n}}{4}\right) - \psi^{(1)}\left(1-\frac{3^{-n}}{4}\right) \right) - 16C \right) \end{aligned} $$

PROOF. Use equation (4.1) and set $m = \pi/2, k = -2, a = e^{-i}$ and simplify using equation (21) in [20]. $\square$

EXAMPLE 4.10. Finite series of the cosine function in terms of the Glaisher-Kinkelin constant $A$.

$$ (4.10) \qquad \sum_{p=0}^{n-1} \frac{9^p \log (i 3^{p+1}) (\cos (\pi 3^p) + 2)}{(2 \cos (\pi 3^p) + 1)^2} $$
$$ = \frac{1}{96} \left( -3(9^n - 1) \left( 6\Phi' \left(-1, -1, \frac{1}{3}\right) - 6\Phi' \left(-1, -1, \frac{2}{3}\right) + 1 \right) + 3^{2n} \log \left( 16e^6 A^{4 \times 3^{2-2n}-36} \right) + 4 \times 3^{2n+1} \log(i3^n) \csc^2 \left( \frac{\pi 3^n}{2} \right) - 6i\pi - 6 - 4 \log(2) \right) $$

PROOF. Use equation (4.1) and take the first derivative with respect to $k$ and set $m = \pi/2, k = 1, a = 1$ and simplify using equation (18) in [20]. □

EXAMPLE 4.11. Finite series of the cosine function in terms of the Catalan's constant $C$.

$$ (4.11) \quad \sum_{p=0}^{n-1} 9^p \left( \begin{aligned} & -6\Phi' \left(-1, -1, \frac{1}{6}(2-3^{-p})\right) + 6\Phi' \left(-1, -1, \frac{1}{6}(4-3^{-p})\right) \\ & +12\Phi' \left(-1, -1, \frac{1}{6}(6-3^{-p})\right) + \frac{3^{-p}(1-4 \times 3^p) \log(i3^{p+1})}{(2 \cos(\pi 3^p) + 1)^2} \end{aligned} \right) \\ = 2 \times 9^n \Phi' \left(-1, -1, 1 - \frac{3^{-n}}{2}\right) - \frac{2C}{\pi} + \frac{3^n(3^n-1) \log(i3^n)}{\cos(\pi 3^n) - 1} $$

PROOF. Use equation (4.1) and take the first derivative with respect to $k$ and set $m = \pi/2, k = 1, a = e^{-i}$ and simplify using equation (20) in [20]. □

EXAMPLE 4.12. Functional equation for the Hurwitz-Lerch zeta function.

$$ (4.12) \quad \Phi(z, s, a) = 3^{-s} \left( z \left( \Phi \left( z^3, s, \frac{a+1}{3} \right) - 2z\Phi \left( z^3, s, \frac{a+2}{3} \right) \right) + \Phi \left( z^3, s, \frac{a}{3} \right) + 3z^2\Phi \left( z^3, s, \frac{a+2}{3} \right) \right) $$

PROOF. Use equation (4.1) and set $n = 1, m = \frac{\log(z)}{2i}, a = e^{ai}, k = -s$ and simplify. □

EXAMPLE 4.13. Functional equation involving the exponential function.

$$ (4.13) \quad \Phi \left( e^t, s, \frac{a}{2} \right) - e^{t/2} \Phi \left( e^t, s, \frac{a+1}{2} \right) = 3^{-s} e^{-\frac{at}{2}} \left( e^{\frac{at}{2}} \Phi \left( e^{3t}, s, \frac{a}{6} \right) \right. \\ \left. - e^{\frac{1}{2}(a+1)t} \Phi \left( e^{3t}, s, \frac{a+1}{6} \right) + e^{\frac{1}{2}(a+2)t} \Phi \left( e^{3t}, s, \frac{a+2}{6} \right) - e^{\frac{1}{2}(a+3)t} \Phi \left( e^{3t}, s, \frac{a+3}{6} \right) \right. \\ \left. + e^{\frac{1}{2}(a+4)t} \Phi \left( e^{3t}, s, \frac{a+4}{6} \right) - e^{\frac{1}{2}(a+5)t} \Phi \left( e^{3t}, s, \frac{a+5}{6} \right) \right) $$

PROOF. Use equation (4.12), set $z = -1$ and apply equation (6.2.1.7) in [15] and simplify. □

## 5. Summation of the Hurwitz-Lerch zeta function using the functional equation.

In this section we consider an alternate form, and express the Hurwitz-Lerch zeta function as a finite sum over the Hurwitz-Lerch zeta function and derive special cases in terms of fundamental constants using equation [16].
EXAMPLE 5.1.

$$
\begin{equation} \tag{5.1}
\begin{aligned}
& \sum_{k=1}^{q} 3^{-s} z^{k-1} \left( 3z^{2q} \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{a+k-1}{q} + 2 \right) \right) + \Phi \left( z^{3q}, s, \frac{a+k-1}{3q} \right) \right. \\
& \qquad \left. + z^q \left( \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{a+k-1}{q} + 1 \right) \right) - 2z^q \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{a+k-1}{q} + 2 \right) \right) \right) \right) \\
& = q^s \Phi(z, s, a)
\end{aligned}
\end{equation}
$$

PROOF. Use equation (4.12) and apply equation [16] and simplify.

<figure><img src="image_2.png" /></figure>

EXAMPLE 5.2. Finite sum of the Hurwitz zeta function over even and odd intervals.

$$
\begin{align*}
& \sum_{k=1}^{2q+1} (-1)^k 6^{-s} \left(q + \frac{1}{2}\right)^{-s} \left( -\zeta\left(s, \frac{a+k-1}{12q+6}\right) + \zeta\left(s, \frac{a+k+6q+2}{12q+6}\right) \right. \\
& \qquad \left. + (-1)^{2q} \left( \zeta\left(s, \frac{a+k+2q}{12q+6}\right) - \zeta\left(s, \frac{a+k+8q+3}{12q+6}\right) \right) \right. \\
& \qquad \left. + (-1)^{4q} \left( \zeta\left(s, \frac{a+k+10q+4}{12q+6}\right) - \zeta\left(s, \frac{a+k+4q+1}{12q+6}\right) \right) \right) \\
& = -\sum_{k=1}^{2q} (-1)^k 3^{-s} q^{-s} \left( \zeta\left(s, \frac{a+k-1}{6q}\right) + (-1)^{2q} \left( \zeta\left(s, \frac{a+k+2q-1}{6q}\right) \right. \right. \\
& \qquad \left. \left. + (-1)^{2q} \zeta\left(s, \frac{a+k+4q-1}{6q}\right) \right) \right)
\end{align*}
$$

PROOF. Use equation (5.1) and set $z = -1$ and simplify into even and odd summation formulae with respect to the first parameter in the Hurwitz-Lerch zeta function. Then simplify and equate the right-hand sides and simplify. $\square$

<figure><img src="image_3.png" /></figure>

EXAMPLE 5.3. Finite product of the gamma function over even and odd inter-
vals.

$$
\begin{equation}
\begin{split}
& \frac{1}{\sqrt{3+6q}} \prod_{k=1}^{2q+1} \left( \frac{\Gamma\left(\frac{a+k+2q}{6+12q}\right) \Gamma\left(\frac{2+a+k+6q}{6+12q}\right) \Gamma\left(\frac{4+a+k+10q}{6+12q}\right)}{\Gamma\left(\frac{-1+a+k}{6+12q}\right) \Gamma\left(\frac{1+a+k+4q}{6+12q}\right) \Gamma\left(\frac{3+a+k+8q}{6+12q}\right)} \right)^{(-1)^k} \\
& = \prod_{k=1}^{2q} \left(2\pi\right)^{\frac{(-1)^k}{2}} q^{-\frac{(-1)^k (-1+a+k-q)}{2q}} \Gamma\left(\frac{-1+a+k}{2q}\right)^{(-1)^k}
\end{split}
\end{equation}
$$

PROOF. Use equation (5.2) and take the first derivative with respect to $s$ and set $s=0$ and simplify using equation (25.11.17) in [5]. Next take the exponential function of both sides and simplify. $\square$

<figure><img src="image_4.png" /></figure>

EXAMPLE 5.4. Ratio of the finite product of the gamma function over even and odd intervals.

$$
(5.4) \quad \sqrt{6q+3} = \frac{\prod_{k=1}^{2q+1} \left( \frac{\Gamma\left(\frac{a+k+2q}{12q+6}\right) \Gamma\left(\frac{a+k+6q+2}{12q+6}\right) \Gamma\left(\frac{a+k+10q+4}{12q+6}\right)}{\Gamma\left(\frac{a+k-1}{12q+6}\right) \Gamma\left(\frac{a+k+4q+1}{12q+6}\right) \Gamma\left(\frac{a+k+8q+3}{12q+6}\right)} \right)^{(-1)^k}}{\prod_{k=1}^{2q} \left(2\pi\right)^{\frac{(-1)^k}{2}} q^{-\frac{(-1)^k (a+k-q-1)}{2q}} \Gamma\left(\frac{a+k-1}{2q}\right)^{(-1)^k}}
$$
PROOF. Use equation (5.3) and re-arrange and simplify. $\square$

EXAMPLE 5.5. Hurwitz-zeta function in terms of the finite product of the gamma function over even and odd intervals.

$$ (5.5) \quad \sqrt{6} \left( \zeta \left( -\frac{1}{2}, \frac{3}{2} \right) - \zeta \left( -\frac{1}{2}, n + \frac{3}{2} \right) \right) \\ = \sum_{q=1}^{n} \left( \frac{\prod_{k=1}^{2q+1} \left( \frac{\Gamma\left(\frac{a+k+2q}{12q+6}\right) \Gamma\left(\frac{a+k+6q+2}{12q+6}\right) \Gamma\left(\frac{a+k+10q+4}{12q+6}\right)}{\Gamma\left(\frac{a+k-1}{12q+6}\right) \Gamma\left(\frac{a+k+4q+1}{12q+6}\right) \Gamma\left(\frac{a+k+8q+3}{12q+6}\right)} \right)^{(-1)^k}}{\prod_{k=1}^{2q} (2\pi)^{\frac{(-1)^k}{2}} q^{-\frac{(-1)^k(a+k-q-1)}{2q}} \Gamma\left(\frac{a+k-1}{2q}\right)^{(-1)^k}} \right) $$

PROOF. Use equation (5.4) and take the finite sum of both sides over $q \in [0, n]$ and simplify using equation (25.11.1) in [5]. Similar forms with applications is given in [17]. $\square$

EXAMPLE 5.6. Finite series of the hypergeometric function.

$$ (5.6) \quad \sum_{k=1}^{q} \frac{z^k}{k} {}_2F_1 \left( 1, \frac{k}{q}; \frac{k+q}{q}; z^q \right) = -\log(z-1) $$

PROOF. Use equation (5.1) and set $s = -1$ and apply the method in section (8) in [10]. $\square$

EXAMPLE 5.7. Finite product of the hypergeometric function.

$$ (5.7) \quad \prod_{k=1}^{q} e^{\frac{z^k}{k} {}_2F_1\left(1, \frac{k}{q}; \frac{k+q}{q}; z^q\right)} = \frac{1}{1-z} $$

PROOF. Use equation (5.6) and take the exponential function of both sides and simplify. $\square$

EXAMPLE 5.8. Finite product of the hypergeometric function in terms of the gamma function.

$$ (5.8) \quad \prod_{k=1}^{q} \prod_{z=2}^{n} e^{\frac{z^k {}_2F_1\left(1, \frac{k}{q}, \frac{k+q}{q}; z^q\right)}{k}} = \frac{(-1)^n}{\Gamma(n)} $$

PROOF. Use equation (5.7) and take the finite product of both sides over $z \in [2, n]$ and simplify. $\square$

EXAMPLE 5.9. Finite series in terms of the polylogarithm function.

$$ (5.9) \quad \sum_{k=1}^{q} 3^{-s} z^{k-1} \left( 3z^{2q} \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) + \Phi \left( z^{3q}, s, \frac{k}{3q} \right) + z^q \left( \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{k}{q} + 1 \right) \right) - 2z^q \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) \right) \right) = \frac{q^s \mathrm{Li}_s(z)}{z} $$

PROOF. Use equation (5.1) and set $a = 1$ and simplify using equation (25.14.3) in [5]. $\square$
EXAMPLE 5.10. Finite series in terms of the Hurwitz zeta function.

$$
(5.10) \qquad
\begin{aligned}
& \sum_{k=1}^{q} 3^{-s} z^{k-1} \left( \Phi \left( z^{3q}, s, \frac{k}{3q} \right) + z^q \left( \Phi \left( z^{3q}, s, \frac{k+q}{3q} \right) + z^q \Phi \left( z^{3q}, s, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) \right) \right) \\
& = \frac{\left( \frac{i}{2\pi} \right)^{1-s} q^s \Gamma(1-s) \left( \zeta \left( 1-s, \frac{\pi-i\log(-z)}{2\pi} \right) - i^{2s} \zeta \left( 1-s, \frac{i\log(-z)+\pi}{2\pi} \right) \right)}{z}
\end{aligned}
$$

PROOF. Use equation (5.9) and simplify the Polylogarithm function in terms of the Hurwitz zeta function using equation (6) in [19]. $\square$

EXAMPLE 5.11. Finite product of the Hurwitz-Lerch zeta function.

$$
(5.11) \qquad \prod_{k=1}^{q} \exp \left( \frac{z^k \left( \Phi \left( z^{3q}, 1, \frac{k}{3q} \right) + z^q \left( \Phi \left( z^{3q}, 1, \frac{k+q}{3q} \right) + z^q \Phi \left( z^{3q}, 1, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) \right) \right)}{3q} \right) \\
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad = \frac{1}{1-z}
$$

PROOF. Use equation (5.10) and take the limit of both sides as $s \to -1$ and simplify in terms of the log-gamma function using equation (64:10:2) in [4]. Next take the exponential function of both sides and simplify. $\square$

EXAMPLE 5.12. Finite double product of the Hurwitz-Lerch zeta function in terms of the Pochhammer symbol.

$$
(5.12) \quad \prod_{k=1}^{q} \prod_{p=0}^{n} \exp \left( \frac{(p+z)^{k+2q} \Phi \left( (p+z)^{3q}, 1, \frac{1}{3} \left(2+\frac{k}{q}\right) \right)}{3q} \right) \\
\exp \left( \frac{(p+z)^k \Phi \left( (p+z)^{3q}, 1, \frac{k}{3q} \right)}{3q} \right) \exp \left( \frac{(p+z)^{k+q} \Phi \left( (p+z)^{3q}, 1, \frac{k+q}{3q} \right)}{3q} \right) \\
= - \frac{(-1)^n}{(z-1)(z)_n}
$$

PROOF. Use equation (5.11) and replace $z$ by $z+p$ and take the finite sum over $p \in [0, n]$ and simplify. Similar forms are given in [18]. $\square$

EXAMPLE 5.13. Finite series in terms of the Riemann zeta function.

$$
(5.13) \quad \sum_{k=1}^{q} (-1)^k 3^{-s} \left( (-1)^{2q} \Phi \left( (-1)^{3q}, s, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) + \Phi \left( (-1)^{3q}, s, \frac{k}{3q} \right) \right. \\
\left. \qquad + (-1)^q \Phi \left( (-1)^{3q}, s, \frac{k+q}{3q} \right) \right) = -2^{-s} (2^s - 2) q^s \zeta(s)
$$

PROOF. Use equation (5.9) and set $s = -1$ and simplify using equation (25.11.2) in [5]. $\square$

EXAMPLE 5.14. Special case of the finite product in terms of the Riemann zeta function.
$$ (5.14) \quad \sum_{k=1}^{q} (-1)^k \left( \begin{aligned} & (-1)^{2q} \Phi' \left( (-1)^{3q}, 0, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) + \Phi' \left( (-1)^{3q}, 0, \frac{k}{3q} \right) \\ & + (-1)^q \Phi' \left( (-1)^{3q}, 0, \frac{k+q}{3q} \right) + \frac{\log(3)}{(-1)^q - 1} \end{aligned} \right) = -\frac{1}{2} \log \left( \frac{\pi q}{2} \right) $$

PROOF. Use equation (5.13) and take the first partial derivative with respect to $s$ and set $s = 0$ and simplify using equation (25.6.11) in [5]. □

## 6. Special cases of the Hurwitz-Lerch zeta function in terms of fundamental constants

In this section we will evaluate equation (5.1) and its first partial derivative using special cases of the parameters involved and simplifying in terms of fundamental constants.

EXAMPLE 6.1. Finite series of the Hurwitz-Lerch zeta function in terms of Catalan's constant $C$.

$$ (6.1) \quad \sum_{k=1}^{q} (-1)^k \left( \Phi \left( (-1)^{3q}, 2, \frac{2k-1}{6q} \right) + (-1)^q \left( \Phi \left( (-1)^{3q}, 2, \frac{2k+2q-1}{6q} \right) \right. \right. \\ \left. \left. + (-1)^q \Phi \left( (-1)^{3q}, 2, \frac{2k+4q-1}{6q} \right) \right) \right) = -36Cq^2 $$

PROOF. Use equation (5.1) and set $z = -1, s = 2, a = 1/2$ and simplify using equation (21) in [20]. □

EXAMPLE 6.2. Finite series of the Hurwitz-Lerch zeta function in terms of the Glaisher-Kinkelin constant $A$.

$$ (6.2) \quad \sum_{k=1}^{q} (-1)^k \left( \Phi' \left( (-1)^{3q}, -1, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) + \Phi' \left( (-1)^{3q}, -1, \frac{k}{3q} \right) \right. \\ \left. + (-1)^q \Phi' \left( (-1)^{3q}, -1, \frac{k+q}{3q} \right) \right) = -\frac{1}{3q} \log \left( \frac{\sqrt[4]{\frac{3}{e}} A^3 \sqrt[4]{q}}{\sqrt[3]{2}} \right) $$

PROOF. Use equation (5.1) and take the first derivative with respect to $s$ and set $z = -1, s = -1, a = 1$ and simplify using equation (18) in [20]. □

EXAMPLE 6.3. Finite series of the Hurwitz-Lerch zeta function in terms of the is Apéry's constant $\zeta(3)$.

$$ (6.3) \quad \sum_{k=1}^{q} (-1)^k \left( \Phi' \left( (-1)^{3q}, -2, \frac{1}{3} \left( \frac{k}{q} + 2 \right) \right) + \Phi' \left( (-1)^{3q}, -2, \frac{k}{3q} \right) \right. \\ \left. + (-1)^q \Phi' \left( (-1)^{3q}, -2, \frac{k+q}{3q} \right) \right) = -\frac{7\zeta(3)}{36\pi^2 q^2} $$

PROOF. Use equation (5.1) and take the first derivative with respect to $s$ and set $z = -1, s = -2, a = 1$ and simplify using equation (19) in [20]. □
EXAMPLE 6.4. Finite series of the Hurwitz-Lerch zeta function in terms of the is Catalan's constant C.

$$ (6.4) \qquad \sum_{k=1}^{q} (-1)^k \left( \Phi' \left( (-1)^{3q}, -1, \frac{2k-1}{6q} \right) + (-1)^q \left( \Phi' \left( (-1)^{3q}, -1, \frac{2k+2q-1}{6q} \right) + (-1)^q \Phi' \left( (-1)^{3q}, -1, \frac{2k+4q-1}{6q} \right) \right) \right) = -\frac{C}{3\pi q} $$

PROOF. Use equation (5.1) and take the first derivative with respect to s and set $z = -1, s = -1, a = 1/2$ and simplify using equation (20) in [20]. □

EXAMPLE 6.5. The first partial derivative of the Hurwitz-Lerch zeta function in terms of the log-gamma function.

$$ (6.5) \qquad \Phi'\left(\sqrt[3]{-1}, 0, a\right) = \log \left( \frac{6^{-\frac{1}{2}i(\sqrt{3}-i)} \Gamma\left(\frac{a}{6}\right) \left(\frac{\Gamma\left(\frac{a+1}{6}\right)}{\Gamma\left(\frac{a+4}{6}\right)}\right)^{\frac{3}{2}-1} \left(\frac{\Gamma\left(\frac{a+2}{6}\right)}{\Gamma\left(\frac{a+5}{6}\right)}\right)^{-(1)^{2/3}}}{\Gamma\left(\frac{a+3}{6}\right)} \right) $$

PROOF. Use equation (4.12) and set $z = (-1)^{1/3}$, then take the first derivative with respect to s and set $s = 0$ and simplify using equation (25.11.18) in [5]. □

## 7. Exercise and examples

Product involving trigonometric functions with arguments of the form $2^{\pm kx}$. Use example (17) on page (416) in [12] along with the contour integral method [1] to derive the following finite product.

EXAMPLE 7.1.

$$ (7.1) \qquad \prod_{p=0}^{n-1} (1 - 2 \cos (2m3^p)) = (-1)^n \sec(m) \cos(m3^n) $$

EXAMPLE 7.2.

$$ (7.2) \qquad \prod_{p=0}^{n-1} (1 - 2 \cos (2m3^{-p})) = (-1)^n \cos(3m) \sec(m3^{-n}) $$

EXAMPLE 7.3. Finite product involving trigonometric functions with arguments of the form $2^{\pm kx}$.

$$ (7.3) \qquad \prod_{p=0}^{n-1} \frac{1 - 2 \cos (2m3^p)}{1 + 2 \cos (2m3^p)} = (-1)^n \tan(m) \cot(m3^n) $$

PROOF. Use equations (4.5) and (7.1) and multiply and simplify. □

EXAMPLE 7.4. Finite product involving trigonometric functions with arguments of the form $2^{\pm kx}$.

$$ (7.4) \qquad \prod_{p=0}^{n-1} \frac{1 - 2 \cos (2m3^{-p})}{1 + 2 \cos (2m3^{-p})} = (-1)^n \cot(3m) \sin(m3^{1-n}) \sec(m3^{-n}) $$

PROOF. Use equations (4.6) and (7.2) and multiply and simplify. □
8. Table of series and products

$$
\sum_{p=0}^{n-1} \frac{3^p \sin(2m3^p)}{2 \cos(2m3^p) + 1} = \frac{1}{4} (\cot(m) - 3^n \cot(m3^n))
$$

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

$$
\prod_{p=0}^{n-1} \frac{1}{1 + 2 \cos(2 \times 3^p m)} = \frac{\sin(m)}{\sin(3^m m)}
$$

$$
\prod_{p=0}^{n-1} \frac{1}{2 \cos(2m3^{-p-1}) + 1} = \csc(m) \sin(m3^{-n})
$$

$$
\prod_{p=0}^{n-1} \frac{1+2 \cos(2 \times 3^p r)}{1+2 \cos(2 \times 3^p m)} = \frac{\sin(3^n r) \sin(m)}{\sin(3^n m) \sin(r)}
$$

$$
\prod_{p=0}^{n-1} \frac{1+2 \cos(2 \times 3^{-1+p} x)}{1+2 \cos(2 \times 3^p x)} = \frac{\sin(x) \sin(3^{-1+n} x)}{\sin(3^n x) \sin\left(\frac{x}{3}\right)}
$$

$$
\sqrt{6q+3} = \frac{\prod_{k=1}^{2q+1} \left( \frac{\Gamma\left(\frac{a+k+2q}{12q+6}\right) \Gamma\left(\frac{a+k+6q+2}{12q+6}\right) \Gamma\left(\frac{a+k+10q+4}{12q+6}\right)}{\Gamma\left(\frac{a+k-1}{12q+6}\right) \Gamma\left(\frac{a+k+4q+1}{12q+6}\right) \Gamma\left(\frac{a+k+8q+3}{12q+6}\right)} \right)^{(-1)^k}}{\prod_{k=1}^{2q} (2\pi)^{\frac{(-1)^k}{2}} q^{\frac{(-1)^k}{2q} \frac{(a+k-q-1)}{2q}} \Gamma\left(\frac{a+k-1}{2q}\right)^{(-1)^k}}
$$

$$
\sqrt{6} \left( \zeta \left(-\frac{1}{2}, \frac{3}{2}\right) - \zeta \left(-\frac{1}{2}, n + \frac{3}{2}\right) \right) = \sum_{q=1}^{n} \left( \frac{\prod_{k=1}^{2q+1} \left( \frac{\Gamma\left(\frac{a+k+2q}{12q+6}\right) \Gamma\left(\frac{a+k+6q+2}{12q+6}\right) \Gamma\left(\frac{a+k+10q+4}{12q+6}\right)}{\Gamma\left(\frac{a+k-1}{12q+6}\right) \Gamma\left(\frac{a+k+4q+1}{12q+6}\right) \Gamma\left(\frac{a+k+8q+3}{12q+6}\right)} \right)^{(-1)^k}}{\prod_{k=1}^{2q} (2\pi)^{\frac{(-1)^k}{2}} q^{\frac{(-1)^k}{2q} \frac{(a+k-q-1)}{2q}} \Gamma\left(\frac{a+k-1}{2q}\right)^{(-1)^k}} \right)
$$

$$
\sum_{k=1}^{q} \frac{z^k}{k} {}_2F_1 \left( 1, \frac{k}{q}; \frac{k+q}{q}; z^q \right) = -\log(z-1)
$$

$$
\prod_{k=1}^{q} e^{\frac{z^k}{k} {}_2F_1(1, \frac{k}{q}; \frac{k+q}{q}; z^q)} = \frac{1}{1-z}
$$

$$
\prod_{k=1}^{q} \prod_{z=2}^{n} e^{\frac{z^k 2 F_1 \left(1, \frac{k}{q}, \frac{k+q}{q}; z^q\right)}{\Gamma(n)}} = - \frac{(-1)^n}{\Gamma(n)}
$$

$$
\prod_{k=1}^{q} \exp \left( \frac{z^k \left( \Phi(z^{3q}, 1, \frac{k}{3q}) + z^q \left( \Phi(z^{3q}, 1, \frac{k+q}{3q}) + z^q \Phi(z^{3q}, 1, \frac{1}{3}(\frac{k}{q}+2)) \right) \right)}{3q} \right) = \frac{1}{1-z}
$$

$$
\Phi'\left(\sqrt[3]{-1}, 0, a\right) = \log\left(\frac{6^{-\frac{1}{2}i(\sqrt{3}-i)}\Gamma\left(\frac{a}{6}\right)\left(\frac{\Gamma\left(\frac{a+1}{6}\right)}{\Gamma\left(\frac{a+4}{6}\right)}\right)^{\sqrt[3]{-1}}\left(\frac{\Gamma\left(\frac{a-2}{6}\right)}{\Gamma\left(\frac{a+5}{6}\right)}\right)^{(-1)^{2/3}}}{\Gamma\left(\frac{a+3}{6}\right)}\right)
$$

$$
\prod_{p=0}^{n-1} (1 - 2 \cos (2 \times 3^p m)) = (-1)^n \sec(m) \cos(m3^n)
$$

$$
\prod_{p=0}^{n-1} (1 - 2 \cos(2m3^{-p})) = (-1)^n \cos(3m) \sec(m3^{-n})
$$

$$
\prod_{p=0}^{n-1} \frac{1 - 2 \cos(2m3^p)}{2 \cos(2m3^p) + 1} = (-1)^n \tan(m) \cot(m3^n)
$$

$$
\prod_{p=0}^{n-1} \frac{1 - 2 \cos(2m3^{-p})}{1 + 2 \cos(2m3^{-p})} = (-1)^n \cot(3m) \sin(m3^{1-n}) \sec(m3^{-n})
$$

$$
\prod_{p=0}^{n-1} \frac{2 \cos(3^p x) + 1}{2 \cos(3^{p-1} x) + 1} = \frac{2 \cos(3^{n-1} x) + 1}{2 \cos\left(\frac{x}{3}\right) + 1}
$$

$$
\prod_{p=0}^{n-1} \frac{2 \cos(2m3^p) + 1}{2 \cos(2r3^p) + 1} = \csc(m) \sin(r) \sin(m3^n) \csc(3^n r)
$$

$$
\prod_{p=0}^{n-1} (2 \cos(m3^p) + 1) = \csc\left(\frac{m}{2}\right) \sin\left(\frac{m3^n}{2}\right)
$$
9. Conclusion

In this paper, we have presented derivations and evaluations of the Hurwitz-
Lerch zeta function in terms of special functions and fundamental constants, using
contour integration. We started with a well known finite series tabled by Loney
in 1893 and extended this series which involved special functions. Special cases
using algebraic techniques were employed to achieve new results which could prove
useful to the academic community. The results presented were numerically verified
for both real and imaginary and complex values of the parameters in the integrals
using Mathematica by Wolfram.

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(Robert Reynolds) DEPARTMENT OF MATHEMATICS AND STATISTICS, YORK UNIVERSITY, TORONTO, ON, CANADA, M3J1P3

Email address, Corresponding author: milver73@gmail.com