Extended Hobson Series

Robert Reynolds

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Extended Hobson Series

Robert Reynolds

ABSTRACT. The mathematical technique known as the Hobson sum, which involves the sine and cosecant functions, has been expanded to include a finite sum using the Hurwitz-Lerch Zeta function. This expansion allows for the derivation of sums and products that involve both trigonometric and special functions. The underlying theorem used to obtain these sums and products takes the form of a finite sum over positive integers of the Hurwitz-Lerch Zeta function, with the parameters involved being general complex numbers. Furthermore, new recurrence identities for the Hurwitz-Lerch Zeta function, which relate consecutive terms, are derived. Additionally, specific examples of finite sum and product formulas are derived and evaluated, involving the cosine, tangent, and the product of cosine and tangent functions. Furthermore, extended product formulas are obtained, which involve the gamma function. These extended product formulas build upon previous work by Knar, Edreyli, Legendre, and Nielsen in this area.

# 1. Introduction

Numerous authors have dedicated significant time and effort to investigating the theory of sums and products that involve trigonometric and special functions throughout many years of research. Several investigations in this field were carried out by; Almodovar et. al [1], where the infinite products arising in paperfolding was published. In the work of krysicki [2] new properties of the beta distribution was studied. In the work by Kaluszka et. al [3] the decompositions of some random variables was published. In the work by Vignat et. al [4], finite generating functions for the sum-of-digits sequence was studied.

Work involving the sum of the Hurwitz-lerch zeta function is available in the work by Lerch [5], where original work by Lerch introduced the Lerch function. The study of finite sums involving the Hurwitz-Lerch zeta function was published in the work by Choi et. al [6]. In the book by Prudnikov et. al [7], a section on the Hurwitz-Lerch zeta function and its properties are tabled. in the work by Gorenflo et. al [8], the computational aspects of the hurwitz-Lerch zeta function and its derivative are discussed. The work by Kilbas et. al [9], includes a section on frac- tional sums and integrals involving the Hurwitz-Lerch zeta function.

2020 Mathematics Subject Classification. Primary 30E20, 33-01, 33-03, 33-04.

Key words and phrases. Finite and infinite sums and products, Cauchy integral, gamma function, Hurwitz-Lerch zeta function, log-gamma function.
Drawing from existing literature, it becomes evident that finite sums and products involving special functions and trigonometric functions serve a practical purpose. In this research, our objective is to establish a comprehensive theorem that encompasses the finite sum of Hurwitz-Lerch Zeta functions, expressed in relation to Hurwitz-Lerch Zeta functions themselves. This theorem entails a linear combination of Hurwitz-Lerch Zeta functions, offering potential advantages in the field of analysis.

In this current study, we aim to build upon previous research that focused on the finite sum of special functions. To achieve this, we employ the contour integral method [10], specifically applied to equation (5.74.4) mentioned in [11]. Consequently, we obtain the contour integral representation as a result given by;

$$ (1.1) \qquad \frac{1}{2\pi i} \int_C \sum_{p=1}^{n} 4 \, 3^{1-p} a^w w^{-k-1} \sin^3(3^{p-1}(m+w)) \sec(3^p(m+w)) dw \\ = \frac{1}{2\pi i} \int_C \frac{3}{2} a^w w^{-k-1} (3^{-n} \tan(3^n(m+w)) - \tan(m+w)) dw $$

where $a, m, k \in \mathbb{C}$, $Re(m+w) > 0$, $n \in \mathbb{Z}^+$. Using equation (1.1) the main Theorem to be derived and evaluated is given by

$$ (1.2) \qquad \sum_{p=1}^{n} \sum_{p=1}^{n} -i3^{1-p} \left( \log^k(a) + 2^k (i3^p)^k e^{2im3^{p-1}} \left( -3\Phi \left( -e^{2i3^p m}, -k, \frac{1}{3} - \frac{1}{2}i3^{-p} \log(a) \right) \right. \right. \\ \left. \left. +3e^{2im3^{p-1}} \Phi \left( -e^{2i3^p m}, -k, \frac{2}{3} - \frac{1}{2}i3^{-p} \log(a) \right) - 2e^{4im3^{p-1}} \Phi \left( -e^{2i3^p m}, -k, 1 - \frac{1}{2}i3^{-p} \log(a) \right) \right) \right) \\ = -\frac{1}{2}i3^{1-n} \left( (3^n - 1) \log^k(a) + 2^{k+1} \left( (i3^n)^k e^{2im3^n} \Phi \left( -e^{2i3^n m}, -k, 1 - \frac{1}{2}i3^{-n} \log(a) \right) \right. \right. \\ \left. \left. -i^k e^{2im3^n} \Phi \left( -e^{2im}, -k, 1 - \frac{1}{2}i \log(a) \right) \right) \right) $$

The expression provided involves the variables $k, a, m$, which can be any complex numbers, and $n$, which represents any positive integer. This expression is subsequently utilized to obtain specific instances using trigonometric functions. The derivations employed in this process are based on the approach presented in our previous work referenced as [10]. This approach incorporates a variant of the generalized Cauchy's integral formula, as given by;

$$ (1.3) \qquad \frac{y^k}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C \frac{e^{wy}}{w^{k+1}} dw, $$

In this context, we consider variables $y$ and $w$ belonging to the set of complex numbers $(\mathbb{C})$, while $C$ represents a general open contour in the complex plane. The bilinear concomitant, as described in the reference [10], vanishes at the endpoints of the contour. This approach involves utilizing a specific form of equation (1.3), multiplying both sides by a function, and subsequently summing the resulting finite terms on both sides. As a result, a finite sum expressed in terms of a contour integral is obtained. Additionally, by multiplying both sides of equation (1.3) by another function and summing infinitely on both sides, the contour integrals in both equations become identical.
## 2. The Hurwitz-Lerch Zeta Function

In our analysis, we employ equation (1.11.3) from the reference [12], where $\Phi(z, s, v)$ denotes the Hurwitz-Lerch zeta function. This function serves as a generalization of both the Hurwitz zeta function $\zeta(s, v)$ and the Polylogarithm function $\mathrm{Li}_n(z)$. The Lerch function can be expressed using a series representation, which is given by.

$$ (2.1) \qquad \Phi(z, s, v) = \sum_{n=0}^{\infty} (v+n)^{-s} z^n $$

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

$$ (2.2) \qquad \Phi(z, s, v) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-vt}}{1 - ze^{-t}} dt = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-(v-1)t}}{e^t - z} dt $$

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

## 3. Contour Integral Representation for the Finite Sum of the Hurwitz-Lerch Zeta Functions

In this section we derive the contour integral representations of the left-hand side and right-hand side of equation (1.1) in terms of the Hurwitz-Lerch zeta and trigonometric functions.

### 3.1. Derivation of the left-hand side first contour integral.
We use the method in [10]. The cut and contour are in the first quadrant of the complex w-plane with $0 < \mathrm{Re}(w + m)$. The cut approaches the origin from the interior of the first quadrant and goes to infinity vertically and the contour goes round the origin with zero radius and is on opposite sides of the cut. Using a generalization of Cauchy's integral formula (1.3) we first replace $y$ by $\log(a) + ix + y$ then multiply both sides by $e^{itx}$ then form a second equation by replacing $x$ by $-x$ and subtracting both equations to get

$$ (3.1) \qquad \frac{e^{-imx} ((\log(a) - ix + y)^k - e^{2imx} (\log(a) + ix + y)^k)}{k!} \\ = -\frac{1}{2\pi i} \int_C 2ia^w w^{-k-1} e^{wy} \sin(x(m+w)) dw $$

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

$$ (3.2) \qquad \sum_{p=1}^{n} \frac{i2^k 3^{2-p} (i3^p)^k e^{im(3^p-3^{p-1})}}{k!} \left( \Phi\left(-e^{2i3^p m}, -k, \frac{1}{2}3^{-p}(-i\log(a)-3^{p-1}+3^p)\right) \right. \\ \left. -e^{2im3^{p-1}} \Phi\left(-e^{2i3^p m}, -k, \frac{1}{2}3^{-p}(-i\log(a)+3^{p-1}+3^p)\right) \right) $$
$$
\begin{align*}
&= -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \sum_{p=1}^{n} \int_C 2i(-1)^y a^w w^{-k-1} e^{ib(2y+1)(m+w)} \sin(x(m+w)) dw \\
&= -\frac{1}{2\pi i} \sum_{p=0}^{n-1} \int_C \sum_{y=0}^{\infty} 2i(-1)^y a^w w^{-k-1} e^{ib(2y+1)(m+w)} \sin(x(m+w)) dw \\
&= -\frac{1}{2\pi i} \int_C \sum_{p=1}^{n} 2i(-1)^y a^w w^{-k-1} e^{ib(2y+1)(m+w)} \sin(x(m+w)) dw \\
&= \frac{1}{2\pi i} \int_C 3^{2-p} a^w w^{-k-1} \sin(3^{p-1}(m+w)) \sec(3^p(m+w)) dw
\end{align*}
$$

from equation (1.232.1) in [13] where we replaced $x = 3^{p-1}$, $b = 3^p$ and $Im(n(m+w)) > 0$ and $Re(n(m+w)) > 0$. We apply Tonelli's theorem for multiple sums, see page 177 in [14] as the summands are of bounded measure over the space $C \times [0, n]$.

**3.2. Derivation of the generalized Hurwitz-Lerch zeta contour integral.** Using a generalization of Cauchy's integral formula (1.3), 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 to get

$$
\begin{align*}
& (3.3) - \frac{i 2^{k+1} (ib)^k e^{2ibm} \Phi\left(-e^{2ibm}, -k, 1 - \frac{i \log(a)}{2b}\right)}{k!} \\
&= \frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C (-1)^y a^w w^{-k-1} e^{2ib(y+1)(m+w)} dw \\
&= -\frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} (-1)^y a^w w^{-k-1} e^{2ib(y+1)(m+w)} dw \\
&= \frac{1}{2\pi i} \int_C \left( a^w w^{-k-1} \tan(b(m+w)) - ia^w w^{-k-1} \right) dw
\end{align*}
$$

from equation (1.232.1) in [13] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177 in [14] as the summands are of bounded measure over the space $C \times [0, \infty)$.

**3.2.1. Derivation of the additional contour.** Using a generalization of Cauchy's integral formula (1.3), first replace $y \to \log(a)$ then multiply both sides by $-i$ and simplify to get

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

**3.3. Derivation of the Left-hand side second contour integral.** Here we use equation (3.3) substitute $b \to 3^p$ and take the finite sum of both sides over $p \in [1, n]$ to get;

$$
\begin{align*}
& (3.5) \quad \sum_{p=1}^{n} \frac{i 2^{k+1} 3^{1-p} (i 3^p)^k e^{2im3^p} \Phi\left(-e^{2i3^p m}, -k, 1 - \frac{1}{2}i3^{-p} \log(a)\right)}{k!} \\
&= -\frac{1}{2\pi i} \int_C \sum_{p=1}^{n} 3^{1-p} a^w w^{-k-1} \tan(3^p(m+w)) + i 3^{1-p} a^w w^{-k-1} dw
\end{align*}
$$
where the additional contour integral is given by;

$$ (3.6) \qquad \sum_{p=1}^{n} \frac{i3^{1-p} \log^k(a)}{k!} = \frac{1}{2\pi i} \int_C \sum_{p=1}^{n} i3^{1-p} a^w w^{-k-1} $$

**3.4. Derivation of Right-hand side first contour integral.** Here we use equation (3.3) substitute $b=1$ to get;

$$ (3.7) \qquad \frac{3i(2i)^k e^{2im\Phi} (-e^{2im}, -k, 1 - \frac{1}{2}i \log(a))}{k!} = -\frac{1}{2\pi i} \int_C \frac{3}{2} a^w w^{-k-1} \tan(m+w) + \frac{3}{2} ia^w w^{-k-1} dw $$

where the additional contour integral is given by;

$$ (3.8) \qquad \frac{3i \log^k(a)}{2k!} = \frac{1}{2\pi i} \int_C \frac{3}{2} ia^w w^{-k-1} dw $$

**4. Derivation of Right-hand side second contour integral**

Here we use equation (3.3) substitute $b=3^n$ to get;

$$ (4.1) \qquad - \frac{i2^k 3^{1-n} (i3^n)^k e^{2im3^n} \Phi(-e^{2i3^pm}, -k, 1 - \frac{1}{2}i3^{-n} \log(a))}{k!} = \frac{1}{2\pi i} \int_C \left( \frac{1}{2} 3^{1-n} a^w w^{-k-1} \tan(3^n(m+w)) - \frac{1}{2} i3^{1-n} a^w w^{-k-1} \right) dw $$

where the additional contour integral is given by;

$$ (4.2) \qquad -\frac{i3^{1-n} \log^k(a)}{2k!} = -\frac{1}{2\pi i} \int_C \frac{1}{2} i3^{1-n} a^w w^{-k-1} dw $$

**5. The sum and product involving Special functions.**

In the proceeding section we will evaluate and derive sum and product formulae involving special functions. Similar forms were derived by authors and publication details are provided for each derivation.

**THEOREM 5.1.** For all $k, a, m \in \mathbb{C}$ then,

$$ (5.1) \qquad \sum_{p=1}^{n} -i3^{1-p} \left( \log^k(a) + 2^k (i3^p)^k e^{2im3^{p-1}} \left( -3\Phi\left(-e^{2i3^pm}, -k, \frac{1}{3} - \frac{1}{2}i3^{-p} \log(a)\right) \right. \right. \\ \left. \left. + 3e^{2im3^{p-1}} \Phi\left(-e^{2i3^pm}, -k, \frac{2}{3} - \frac{1}{2}i3^{-p} \log(a)\right) - 2e^{4im3^{p-1}} \Phi\left(-e^{2i3^pm}, -k, 1 - \frac{1}{2}i3^{-p} \log(a)\right) \right) \right) \\ = -\frac{1}{2}i3^{1-n} \left( (3^n - 1) \log^k(a) + 2^{k+1} \left( (i3^n)^k e^{2im3^n} \Phi\left(-e^{2i3^n m}, -k, 1 - \frac{1}{2}i3^{-n} \log(a)\right) \right. \right. \\ \left. \left. -i^k e^{2im3^n} \Phi\left(-e^{2im}, -k, 1 - \frac{1}{2}i \log(a)\right) \right) \right) $$
PROOF. Observe that the addition of the right-hand sides of equations (3.2) and (3.5), (3.6), is equal to the addition of the right-hand sides of equations (3.7), (3.8), (4.1) and (4.2) so we may equate the left-hand sides and simplify using equation (5.74.4) on page 90 in [11] and the Gamma function to yield the stated result. □

EXAMPLE 5.2. The Degenerate Case.

$$ (5.2) \quad \sum_{p=1}^{n} 3^{-p} \sin^3 (m3^{p-1}) \sec (m3^p) = \frac{1}{8} (3^{-n} \tan (m3^n) - \tan(m)) $$

PROOF. Use equation (5.1) and set $k = 0$ and simplify using entry (2) in Table below (64:12:7) in [15]. □

EXAMPLE 5.3. Extended Nielsen-Legendre Form.

$$ (5.3) \quad \prod_{p=1}^{n} (3^{-p}z - 1)^{-3^{-p}} e^{-3^{-p}\log\Gamma(3^{-p}z-1)} \left( \frac{\Gamma(3^{1-p}z)}{\Gamma(3^{-p}z + \frac{1}{6}) \Gamma(3^{-p}z + \frac{5}{6})} \right)^{3^{1-p}} \\ \qquad \left( \left(1 - \frac{3^p}{2z}\right) \Gamma\left(\frac{1}{2}(23^{-p}z - 1)\right) \right)^{-23^{-p}} \\ \qquad = \frac{1}{(2z-1)\Gamma(z-\frac{1}{2})} 2^{\frac{1}{2}(3^{1-n}-1)} \pi^{\frac{1}{2}(3^{1-n}-3)} z^{3-\frac{3}{8}(3(9^{-n}-1)z-4)} 3^{-n+4} \left(1 - \frac{3^n}{2z}\right)^{3^{-n}} \\ \qquad \qquad \Gamma(z)\Gamma(3^{-n}z)^{-3^{-n}} \Gamma\left(3^{-n}z - \frac{1}{2}\right)^{3^{-n}} $$

PROOF. Use equation (5.1) and set $m = 0$ and simplify to yield the Hurwitz zeta function using entry (4) in Table below (64:12:7) in [15]. Next we 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 [15]. Next we take the exponential function of both sides and simplify in terms of the gamma function. The $\log \Gamma(z)$ is analytic throughout the complex $z$ plane, except for a single branch cut discontinuity along the negative real axis. The $\log(\Gamma(z))$ has a more complex branch cut structure. Similar finite product forms featuring powers of 2 are given in equation (1.6.7) in [16] and equation (1.3.5.1) in [12]. □

EXAMPLE 5.4. Extended Nielsen-Erdeyli-Knar form.

$$ (5.4) \quad 3^{-9z/8} \prod_{p=1}^{\infty} \sqrt[3p]{\frac{\Gamma(3^{1-p}z)^3 B(\frac{1}{6}, 3^{-p}z)^3 B(\frac{1}{2}, 3^{-p}z)^2 B(\frac{5}{6}, 3^{-p}z)^3}{\pi \Gamma(3^{-p}z)^9}} = \frac{\Gamma(z)}{\Gamma(z + \frac{1}{2})} $$

PROOF. Use equation (5.3) and take the limit as $n \to \infty$ and simplify the right-hand side. Similar forms are given in equations (1.6.8-9) in [16], equation (1.3.6) in [12] and equation (7) in [17]. □
<figure><img src="image_2.png" /><figcaption>FIGURE 1. Plot of $\frac{\Gamma(z)}{\Gamma(z+\frac{1}{2})}$, $z \in \mathbb{R}$.</figcaption></figure>

<figure><img src="image_3.png" /><figcaption>FIGURE 2. Plot of $Re \left( \frac{\Gamma(z)}{\Gamma(z+\frac{1}{2})} \right)$, $z \in \mathbb{C}$.</figcaption></figure>
<figure><img src="image_4.png" /><figcaption>FIGURE 3. Plot of Im ($\frac{\Gamma(z)}{\Gamma(z+\frac{1}{2})}$), $z \in \mathbb{C}$.</figcaption></figure>

<figure><img src="image_5.png" /><figcaption>FIGURE 4. Plot of Abs ($\frac{\Gamma(z)}{\Gamma(z+\frac{1}{2})}$), $z \in \mathbb{C}$.</figcaption></figure>

EXAMPLE 5.5. Finite product of the exponential function to a power.

$$ (5.5) \quad \prod_{p=1}^{n} (e^{-2im3^{p-1}})^{3^{1-2p}} (e^{4im3^{p-1}})^{\frac{1}{2}3^{1-2p}} (e^{-2im3^p})^{9-p} = e^{-im} (e^{2im3^n})^{\frac{9-n}{2}} $$

PROOF. Use equation (5.1) and form a second equation by replacing $m \to -m$. Next take the difference of the two equations. Next set $k = -1, a = 1$ and simplify in terms of the logarithm function using entry (1) in Table below (64:12:7) in [15]. Next take the exponential function of both sides and simplify. Other forms of this finite product is given by equation (9.34a) in [18]. □
EXAMPLE 5.6. A functional equation for the Hurwitz-Lerch zeta function.
(5.6)

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

PROOF. Use equation (5.1) and set $n = 1$ and simplify. Next replace $m \to \log(z)/i, k \to -s, a \to e^{ai}$ and simplify. Next replace $z \to iz^{1/2}$ and simplify. $\square$

EXAMPLE 5.7. Special case of the first partial derivative of the Hurwitz-Lerch zeta function.

$$ (5.7) \qquad \Phi'(\sqrt[3]{-1}, 0, a) = \log\left(\frac{\Gamma\left(\frac{a}{6}\right)}{\Gamma\left(\frac{a+3}{6}\right)}\right) + \sqrt[3]{-1} \log\left(\frac{\Gamma\left(\frac{a+1}{6}\right)}{\Gamma\left(\frac{a+4}{6}\right)}\right) + (-1)^{2/3} \log\left(\frac{\Gamma\left(\frac{a+2}{6}\right)}{\Gamma\left(\frac{a+5}{6}\right)}\right) \\ \qquad \qquad \qquad \qquad \qquad - \frac{1}{2}i(\sqrt{3} - i)\log(6) $$

PROOF. Use equation (5.6) and set $z = i^{2/3}$ and simplify using entry (4) in Table below (64:12:7) in [15]. Next take the first partial derivative with respect to $s$ and set $s = 0$ and simplify in terms of the log-gamma function using equation (64:10:2) in [15]. $\square$

EXAMPLE 5.8. Finite product involving the cosine function raised to a complex power.

$$ (5.8) \quad \prod_{p=1}^{n} \cos^8 9^{-p} (3^{p-1}x) (2 \cos (2 \sqrt{3}^p x) - 1)^{-9-p} \cos^{8b9^{-p}} \left( \frac{3^{p-1}x}{b^2} \right) \\ \qquad \left( 2 \cos \left( \frac{2 \sqrt{3}^{p-1}x}{b^2} \right) - 1 \right)^{b(-9^{-p})} \cos^{-8(b+1)9^{-p}} \left( \frac{3^{p-1}x}{b} \right) \left( 2 \cos \left( \frac{2 \sqrt{3}^{p-1}x}{b} \right) - 1 \right)^{(b+1)9^{-p}} \\ \qquad = \cos(x) \cos^b \left( \frac{x}{b^2} \right) \cos^{-b-1} \left( \frac{x}{b} \right) \cos^{-9-n} (3^n x) \cos^b(-9^{-n}) \left( \frac{3^n x}{b^2} \right) \cos^{(b+1)9^{-n}} \left( \frac{3^n x}{b} \right) $$

PROOF. Use equation (5.1) and set $k = 1, a = 1, m = x$ and simplify using the method in section (8.1) in [19]. $\square$

## 6. Conclusion

In this paper, we have presented a method for deriving sums, products and transformation formulae involving trigonometric and special functions along with some interesting definite integrals similar to those published by Nielsen, Knar and Erdeyli, using contour integration. 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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[15] Oldham, K.B.; Myland, J.C.; Spanier, J. *An Atlas of Functions: With Equator, the Atlas Function Calculator*, 2nd ed.; Springer: New York, NY, USA, 2009.

[16] N. Nielsen (1906a) *Handbuch der Theorie der Gammafunktion*, B. G. Teubner, Leipzig (German).

[17] John Pearson (2018) *Viète's Formula, Knar's Formula, and the Geometry of the Gamma Function*, The American Mathematical Monthly, 125:8, 704-714, DOI: 10.1080/00029890.2018.1496704

[18] Braga da Costa Campos, Luis Manuel. *Transcendental Representations with Applications to Solids and Fluids*, United States: CRC Press, 2012.

[19] Reynolds, R., & Stauffer, A. (2022)., *A Note on the Infinite Sum of the Lerch function*. European Journal of Pure and Applied Mathematics, 15(1), 158-168. https://doi.org/10.29020/nybg.ejpam.v15i1.4137

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

Email address, Corresponding author: milver73@gmail.com