Alternating series in terms of Riemann zeta function and Dirichlet beta function

Zhiling Fan; Wenchang Chu; Zhiling Fan; Wenchang Chu

doi:10.3934/era.2024058

Electronic Research Archive

2024, Volume 32, Issue 2: 1227-1238. doi: 10.3934/era.2024058

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Alternating series in terms of Riemann zeta function and Dirichlet beta function

Zhiling Fan ¹,
Wenchang Chu ^{2
,
,}

1.
School of Mathematics and Statistics, Zhoukou Normal University, Zhoukou, China
2.
Via Dalmazio Birago 9/E, Lecce 73100, Italy

Received: 19 September 2023 Revised: 15 January 2024 Accepted: 25 January 2024 Published: 30 January 2024

By making use of the multisection series method, four classes of alternating infinite series are evaluated, in closed form, by the Riemann zeta function and the Dirichlet beta function.

Keywords:

Citation: Zhiling Fan, Wenchang Chu. Alternating series in terms of Riemann zeta function and Dirichlet beta function[J]. Electronic Research Archive, 2024, 32(2): 1227-1238. doi: 10.3934/era.2024058

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Abstract

By making use of the multisection series method, four classes of alternating infinite series are evaluated, in closed form, by the Riemann zeta function and the Dirichlet beta function.

1. Introduction and motivation

The Riemann zeta function and the Dirichlet beta function are important functions in number theory that are defined by (cf. [¹, Chapter 25])

$\zeta( \lambda) = \sum\limits_{k = 0}^\infty\frac{1}{(k+1)^ \lambda} \quad\text{and}\quad \beta( \lambda) = \sum\limits_{k = 0}^\infty\frac{(-1)^k}{(2k+1)^ \lambda},$

provided that $\Re(\lambda) > 1$ and $\Re(\lambda) > 0$ so that $\zeta(\lambda)$ and $\beta(\lambda)$ are convergent. There are numerous useful formulae for these functions in the literature (cf. ^[2,3,4,5]). For example, we shall frequently utilize the following relation:

$\sum\limits_{k = 0}^{\infty}\frac{(-1)^k}{(k+1)^ \lambda} = (1-2^{1- \lambda})\zeta( \lambda), \quad\Re( \lambda) > 0.$

In particular, there exist closed form expressions in $\pi$

$\zeta(2n) = \frac{(-1)^{n-1}(2\pi)^{2n}}{2(2n)!}B_{2n} \quad\text{and}\quad \beta(2n+1) = \frac{(-1)^n\pi^{2n+1}}{4^{n+1}(2n)!}E_{2n},$

where $B_n$ and $E_n$ are Bernoulli and Euler numbers (cf. [6, §6.5]), whose exponential generating functions are given by

$\begin{aligned}\frac{y}{e^y-1}& = \sum\limits_{n = 0}^{\infty}\frac{B_n}{n!}y^n, \quad |y| < 2\pi;\\ \frac{2e^y}{1+e^{2y}}& = \sum\limits_{n = 0}^{\infty}\frac{E_n}{n!}y^n, \quad |y| < \pi/2. \end{aligned}$

For a real number $x$ , denote by $\lfloor{{x}}\rfloor$ and $\lceil{{x}}\rceil$ the usual floor and ceiling functions. The aim this paper is to examine, for two natural numbers $m, n\in\mathbb{{N}}$ , the following four variants of the Riemann zeta function:

$\begin{align}& \mathrm{U}_m(n) = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k+m}{2m}}}\rfloor}}{(k+1)^{2n}}, &\qquad&\mathcal{{U}}_m(n) = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k+m}{2m}}}\rfloor}}{(2k+1)^{2n}};\\ & \mathrm{V}_m(n) = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{(k+1)^{2n-1}}, &&\mathcal{{V}}_m(n) = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{(2k+1)^{2n-1}}. \end{align}$

Several closed formulae will be established that express these alternating infinite series in terms of the Riemann zeta function and the Dirichlet beta function.

Throughout the paper, we shall make use of the following polynomial expression for higher derivatives of cotangent function (cf. ^[7,8])

$\frac{d^n\cot x}{dx^n} = P_n(\cot x):\; P_n(x) = (2 \mathbf{i})^n(x- \mathbf{i})\sum\limits_{k = 0}^n\frac{k!}{2^k}\bigg\{\genfrac{}{}{0mm}{0}{{n}}{{k}}\bigg\}( \mathbf{i}x-1)^k,$

where $\bigg\{\genfrac{}{}{0mm}{0}{{n}}{{k}}\bigg\}$ are the Stirling numbers of the second kind (cf. [6, §6.1]). The first five polynomials read explicitly as follows:

$\begin{aligned}P_1(x)& = -x^2-1, \\ P_2(x)& = 2 x^3+2 x, \\ P_3(x)& = -6 x^4-8 x^2-2, \\ P_4(x)& = 24 x^5+40 x^3+16 x, \\ P_5(x)& = -120 x^6-240 x^4-136x^2-16.\end{aligned}$

By putting these in conjunction with the partial fraction decomposition (see, for example, Stromberg [⁹, Theorem 5.43] and Higgins ^[10])

$\cot(\pi x) = \lim\limits_{n\to\infty}\frac1{\pi}\sum\limits_{k = -n}^{n}\frac1{k+x} = \frac{1}{\pi x}+\frac{2x}{\pi}\sum\limits_{k = 1}^{\infty}\frac1{x^2-k^2},$

we deduce for $m\in\mathbb{{N}}$ , by computing $m$ th derivative with respect to $x$ , the following compact expression for the bilateral series

$\begin{equation} { \sum\limits_{k = -\infty}^{\infty}\frac1{(k+x)^{m+1}} = \frac{(-1)^m\pi^{m+1}}{m!}P_m\big(\cot(\pi x)\big).} \end{equation}$

(1)

2. Evaluation of $\mathrm{U}_m(n)$ and $\mathcal{{U}}_m(n)$

In this section, we shall examine the series $\mathrm{U}_m(n)$ and $\mathcal{{U}}_m(n)$ by dividing them into multisection series. Two general summation theorems will be proved and several explicit formulae will be presented as consequences.

2.1. $\mathrm{U}_m(n)$

Since the series $\mathrm{U}_m(n)$ is absolutely convergent, in order to evaluate it explicitly, we can regroup its terms. For example $\mathrm{U}_1(n)$ and $\mathrm{U}_2(n)$ can be expressed as follows:

$\begin{align} \mathrm{U}_1(n)& = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k+1}2}}\rfloor}}{(k+1)^{2n}} = 1-\frac1{2^{2n}}-\frac1{3^{2n}}+\frac1{4^{2n}}+\frac1{5^{2n}} -\frac1{6^{2n}}-\frac1{7^{2n}}+\frac1{8^{2n}}+\cdots\\ & = \bigg\{1-\frac1{3^{2n}}+\frac1{5^{2n}}-\frac1{7^{2n}}+\cdots\bigg\} -\bigg\{\frac1{2^{2n}}-\frac1{4^{2n}}+\frac1{6^{2n}}-\frac1{8^{2n}}+\cdots\bigg\}, \\ \mathrm{U}_2(n)& = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k+2}4}}\rfloor}}{(k+1)^{2n}} = 1+\frac1{2^{2n}}-\frac1{3^{2n}}-\frac1{4^{2n}}-\frac1{5^{2n}} -\frac1{6^{2n}}+\frac1{7^{2n}}+\frac1{8^{2n}}+\frac1{9^{2n}}\\ & \qquad \qquad \qquad \qquad +\frac1{10^{2n}}-\frac1{11^{2n}}-\frac1{12^{2n}}-\frac1{13^{2n}} -\frac1{14^{2n}}+\frac1{15^{2n}}+\frac1{16^{2n}}+\cdots\\ & = \bigg\{1-\frac1{5^{2n}}+\frac1{9^{2n}}-\frac1{13^{2n}}+\cdots\bigg\} +\bigg\{\frac1{2^{2n}}-\frac1{6^{2n}}+\frac1{10^{2n}}-\frac1{14^{2n}}+\cdots\bigg\}\\ &-\bigg\{\frac1{3^{2n}}-\frac1{7^{2n}}+\frac1{11^{2n}}-\frac1{15^{2n}}+\cdots\bigg\} -\bigg\{\frac1{4^{2n}}-\frac1{8^{2n}}+\frac1{12^{2n}}-\frac1{16^{2n}}+\cdots\bigg\}. \end{align}$

In general, by dividing the summation index $k$ according to the residue classes modulo $4m$ , i.e., making the replacements $k\to 4km+j-1$ with $1\le j\le4m$ , we can express $\mathrm{U}_m(n)$ in terms of multisection series

$\begin{align} \mathrm{U}_m(n)& = \sum\limits_{k = 0}^\infty \nonumber \frac{(-1)^{\lfloor{{\frac{k+m}{2m}}}\rfloor}}{(k+1)^{2n}} = \sum\limits_{j = 1}^{4m}\sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{4km+m+j-1}{2m}}}\rfloor}}{(4km+j)^{2n}}\\ & = \sum\limits_{j = 1}^{m}\sum\limits_{k = 0}^{\infty} \Bigg\{\frac{(-1)^{\lfloor{{\frac{m+j-1}{2m}}}\rfloor}}{(4km+j)^{2n}} +\frac{(-1)^{\lfloor{{\frac{2m+j-1}{2m}}}\rfloor}}{(4km+m+j)^{2n}}\Bigg\}\\ &+\sum\limits_{j = 1}^{m}\sum\limits_{k = 0}^{\infty} \Bigg\{\frac{(-1)^{\lfloor{{\frac{3m+j-1}{2m}}}\rfloor}}{(4km+2m+j)^{2n}} +\frac{(-1)^{\lfloor{{\frac{4m+j-1}{2m}}}\rfloor}}{(4km+3m+j)^{2n}}\Bigg\}.\end{align}$

For $1\le j\le m$ , observing that

$\begin{aligned}(-1)^{\lfloor{{\frac{m+j-1}{2m}}}\rfloor} = &(-1)^{\lfloor{{\frac{4m+j-1}{2m}}}\rfloor} = 1, \\ (-1)^{\lfloor{{\frac{2m+j-1}{2m}}}\rfloor} = &(-1)^{\lfloor{{\frac{3m+j-1}{2m}}}\rfloor} = -1;\end{aligned}$

and then putting the summation terms corresponding to $j = m$ aside, we can reformulate further

$\begin{align} \mathrm{U}_m(n)& = \sum\limits_{k = 0}^\infty \Bigg\{\frac{(-1)^k}{(2km+m)^{2n}}-\frac{(-1)^k}{(2km+2m)^{2n}}\Bigg\}\\ &+\sum\limits_{j = 1}^{m-1}\sum\limits_{k = 0}^{\infty} \Bigg\{\frac{1}{(4km+j)^{2n}}-\frac{1}{(4km+m+j)^{2n}}\Bigg\}\\ &+\sum\limits_{j = 1}^{m-1}\sum\limits_{k = 0}^{\infty} \Bigg\{\frac{1}{(4km+3m+j)^{2n}}-\frac{1}{(4km+2m+j)^{2n}}\Bigg\}.\end{align}$

Writing the first line directly in $\beta(2n)$ and $\zeta(2n)$ , then inverting simultaneously the summation indices by $j\to m-j$ and $k\to-k-1$ in the last line, we derive the following bilateral series expression

$\begin{align} \mathrm{U}_m(n)& = \frac{\beta(2n)}{m^{2n}} -\frac{\zeta(2n)}{(2m)^{2n}}\bigg\{1-\frac{2}{4^n}\bigg\}\\ &+\sum\limits_{j = 1}^{m-1}\sum\limits_{k = -\infty}^{\infty} \Bigg\{\frac{1}{(4km+j)^{2n}}-\frac{1}{(4km+m+j)^{2n}}\Bigg\}.\end{align}$

By applying Eq (1), we find the simplified formula as in the following theorem.

Theorem 1 ( $m, n\in\mathbb{{N}}$ ).

$\begin{aligned} \mathrm{U}_m(n)& = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k+m}{2m}}}\rfloor}}{(k+1)^{2n}} = \frac{\beta(2n)}{m^{2n}} +\frac{\zeta(2n)}{(2m)^{2n}}\bigg\{\frac{2}{4^n}-1\bigg\}\\ &+(-1)^{n}\frac{4n\; \zeta(2n)}{4^{3n}m^{2n}B_{2n}}\sum\limits_{j = 1}^{m-1} \bigg\{P_{2n-1}\Big(\cot\big(\tfrac{j\pi}{4m}\big)\Big) -P_{2n-1}\Big(\cot\big(\tfrac{m+j}{4m}\pi\big)\Big)\bigg\}.\end{aligned}$

For small $m, n\in\mathbb{{N}}$ , this theorem can be utilized to compute exact values of $\mathrm{U}_m(n)$ . The following interesting formulae are exhibited as examples.

Example 1.

$\begin{align} \mathrm{U}_1(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(k+1)^{2}} = \beta(2)-\frac{\zeta(2)}{8}, \\ \mathrm{U}_1(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(k+1)^{4}} = \beta(4)-\frac{7\zeta(4)}{128}, \\ \mathrm{U}_1(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(k+1)^{6}} = \beta(6)-\frac{31\zeta(6)}{2048}, \\ \mathrm{U}_1(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(k+1)^{8}} = \beta(8)-\frac{127\zeta(8)}{32768}, \\ \mathrm{U}_1(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(k+1)^{10}} = \beta(10)-\frac{511\zeta(10)}{524288}.\end{align}$

Example 2.

$\begin{align} \mathrm{U}_2(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(k+1)^{2}} = \frac{\beta(2)}{4}+\frac{3 \zeta(2)}{4\sqrt{2}}-\frac{\zeta(2)}{32}, \\ \mathrm{U}_2(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(k+1)^{4}} = \frac{\beta(4)}{16}+\frac{165 \zeta(4)}{128 \sqrt{2}}-\frac{7\zeta(4)}{2048}, \\ \mathrm{U}_2(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(k+1)^{6}} = \frac{\beta(6)}{64}+\frac{22743 \zeta(6)}{16384 \sqrt{2}}-\frac{31\zeta(6)}{131072}, \\ \mathrm{U}_2(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(k+1)^{6}} = \frac{\beta(8)}{256}+\frac{369165 \zeta(8)}{262144 \sqrt{2}}-\frac{127\zeta(8)}{8388608}, \\ \mathrm{U}_2(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(k+1)^{10}} = \frac{\beta(10)}{1024}+\frac{94810353 \zeta(10)}{67108864 \sqrt{2}}-\frac{511\zeta(10)}{536870912}.\end{align}$

Example 3.

$\begin{align} \mathrm{U}_3(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(k+1)^{2}} = \frac{\beta(2)}{9}+\frac{\zeta(2)}{\sqrt{3}}+\frac{7\zeta(2)}{72}, \\ \mathrm{U}_3(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(k+1)^{4}} = \frac{\beta(4)}{81}+\frac{115 \zeta(4)}{72 \sqrt{3}}+\frac{553 \zeta(4)}{10368}, \\ \mathrm{U}_3(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(k+1)^{6}} = \frac{\beta(6)}{729}+\frac{11767 \zeta(6)}{6912 \sqrt{3}}+\frac{22537 \zeta(6)}{1492992}, \\ \mathrm{U}_3(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(k+1)^{8}} = \frac{\beta(8)}{6561}+\frac{1287715 \zeta(8)}{746496 \sqrt{3}}+\frac{832993\zeta(8)}{214990848}, \\ \mathrm{U}_3(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(k+1)^{10}} = \frac{\beta(10)}{59049}+\frac{744010091 \zeta(10)}{429981696 \sqrt{3}}+\frac{30173017 \zeta(10)}{30958682112}.\end{align}$

2.2. $\mathcal{{U}}_m(n)$

Performing the replacement $k\to 4km+j-1$ with $1\le j\le4m$ , we can express $\mathcal{{U}}_m(n)$ as the following multisection series

$\begin{align}\mathcal{{U}}_m(n)& = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k+m}{2m}}}\rfloor}}{(2k+1)^{2n}} = \sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{4m} \frac{(-1)^{\lfloor{{\frac{m+j-1}{2m}}}\rfloor}}{(8km+2j-1)^{2n}}\\ & = \sum\limits_{k = 0}^\infty \Bigg\{\sum\limits_{j = 1}^{m} \frac{(-1)^{\lfloor{{\frac{m+j-1}{2m}}}\rfloor}}{(8km+2j-1)^{2n}} +\sum\limits_{j = 1}^{m} \frac{(-1)^{\lfloor{{\frac{2m+j-1}{2m}}}\rfloor}}{(8km+2m+2j-1)^{2n}}\\ &+\sum\limits_{j = 1}^{m} \frac{(-1)^{\lfloor{{\frac{3m+j-1}{2m}}}\rfloor}}{(8km+4m+2j-1)^{2n}} +\sum\limits_{j = 1}^{m} \frac{(-1)^{\lfloor{{\frac{4m+j-1}{2m}}}\rfloor}}{(8km+6m+2j-1)^{2n}} \Bigg\}.\end{align}$

The above expression can be further reduced as follows:

$\begin{align}\mathcal{{U}}_m(n)& = \sum\limits_{k = 0}^\infty \sum\limits_{j = 1}^{m} \Bigg\{\frac{1}{(8km+2j-1)^{2n}} -\frac{1}{(8km+2m+2j-1)^{2n}}\Bigg\}\\ &+\sum\limits_{k = 0}^\infty \sum\limits_{j = 1}^{m}\Bigg\{ \frac{1}{(8km+6m+2j-1)^{2n}} -\frac{1}{(8km+4m+2j-1)^{2n}}\Bigg\}.\end{align}$

For the two sums on the ultimate line, making the changes on summation indices by $j\to1+m-j$ and $k\to-k-1$

$\sum\limits_{k = -\infty}^{-1} \sum\limits_{j = 1}^{m}\Bigg\{ \frac{1}{(8km+2j-1)^{2n}} -\frac{1}{(8km+2m+2j-1)^{2n}}\Bigg\}$

and then unifying this resulting expression to the first line, we find the bilateral series expression

$\begin{aligned}\mathcal{{U}}_m(n) & = \sum\limits_{j = 1}^{m}\sum\limits_{k = -\infty}^\infty \Bigg\{\frac{1}{(8km+2j-1)^{2n}} -\frac{1}{(8km+2m+2j-1)^{2n}}\Bigg\}\\ & = \sum\limits_{j = 1}^{m}\frac{1}{(8m)^{2n}} \sum\limits_{k = -\infty}^\infty \Bigg\{\frac{1}{(k+\frac{2j-1}{8m})^{2n}} -\frac{1}{(k+\frac{2m+2j-1}{8m})^{2n}}\Bigg\}.\end{aligned}$

By invoking Eq (1), we establish the following summation formula.

Theorem 2 ( $m, n\in\mathbb{{N}}$ ).

$\begin{aligned}\mathcal{{U}}_m(n)& = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k+m}{2m}}}\rfloor}}{(2k+1)^{2n}} = (-1)^{n}\frac{4n\; \zeta(2n)}{4^{4n}m^{2n}B_{2n}}\\ &\times\sum\limits_{j = 1}^{m} \bigg\{P_{2n-1}\Big(\cot\big(\tfrac{2j-1}{8m}\pi\big)\Big) -P_{2n-1}\Big(\cot\big(\tfrac{2m+2j-1}{8m}\pi\big)\Big)\bigg\}.\end{aligned}$

For $m = 1, 2, 3$ , we record the following closed formulae as applications.

Example 4 ( $m = 1$ ).

$\begin{align}\mathcal{{U}}_1(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(2k+1)^{2}} = \frac{3\zeta(2)}{4\sqrt2}, \\ \mathcal{{U}}_1(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(2k+1)^{4}} = \frac{165\zeta(4)}{128 \sqrt{2}}, \\ \mathcal{{U}}_1(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(2k+1)^{6}} = \frac{22743\zeta(6)}{16384\sqrt{2}}, \\ \mathcal{{U}}_1(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(2k+1)^{8}} = \frac{369165\zeta(8)}{262144 \sqrt{2}}, \\ \mathcal{{U}}_1(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+1}{2} \rfloor}}{(2k+1)^{10}} = \frac{94810353\zeta(10)}{67108864\sqrt{2}}.\end{align}$

Example 5 ( $m = 2$ ).

$\begin{align}\mathcal{{U}}_2(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(2k+1)^{2}} = \frac{3\zeta(2)}{16} \sqrt{10+\sqrt{2}}, \\ \mathcal{{U}}_2(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(2k+1)^{4}} = \frac{15\zeta(4)}{2048} \sqrt{15898+241 \sqrt{2}}, \\ \mathcal{{U}}_2(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(2k+1)^{6}} = \frac{63 \zeta(6)}{1048576} \sqrt{267655210+494881 \sqrt{2}}, \\ \mathcal{{U}}_2(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(2k+1)^{8}} = \frac{15 \zeta(8)}{67108864} \sqrt{19853759178298+4207474321 \sqrt{2}}, \\ \mathcal{{U}}_2(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+2}{4} \rfloor}}{(2k+1)^{10}} = \frac{33 \zeta(10) }{68719476736} \sqrt{4327811705145000010+103023433654081 \sqrt{2}} .\end{align}$

Example 6 ( $m = 3$ ).

$\begin{align}\mathcal{{U}}_3(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(2k+1)^{2}} = \frac{\zeta(2)}{24}\sqrt{218+24\sqrt3}, \\ \mathcal{{U}}_3(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(2k+1)^{4}} = \frac{\zeta(4)}{6912}\sqrt{1635146+22968\sqrt3}, \\ \mathcal{{U}}_3(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(2k+1)^{6}} = \frac{\zeta(6)}{2654208}\sqrt{138927209978+219727704\sqrt3}, \\ \mathcal{{U}}_3(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(2k+1)^{8}} = \frac{\zeta(8)}{1146617856} \sqrt{52163307935186666+9179017200888 \sqrt3} , \\ \mathcal{{U}}_3(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor\frac{k+3}{6} \rfloor}}{(2k+1)^{10}} = \frac{\zeta(10)}{2641807540224} \sqrt{57564351813037136458778+1125649024023711384 \sqrt3} .\end{align}$

3. Evaluation of $\mathrm{V}_m(n)$ and $\mathcal{{V}}_m(n)$

This section will be devoted to the two remaining series $\mathrm{V}_m(n)$ and $\mathcal{{V}}_m(n)$ . We shall establish two general summation theorems. Several explicit formulae in terms of the Riemann zeta function and the Dirichlet beta function will be deduced as particular cases.

3.1. $\mathrm{V}_m(n)$

Making the replacement $k\to km+j-1$ with $1\le j\le m$ , we can rewrite $\mathrm{V}_m(n)$ as

$\begin{align} \mathrm{V}_m(n) & = \sum\limits_{k = 0}^\infty \nonumber \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{(k+1)^{2n-1}} = \sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^m \frac{(-1)^{\lfloor{{\frac{km+j-1}{m}}}\rfloor}}{(km+j)^{2n-1}}\\ & = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{km+m-1}{m}}}\rfloor}}{(km+m)^{2n-1}} +\sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{m-1} \frac{(-1)^{\lfloor{{\frac{km+j-1}{m}}}\rfloor}}{(km+j)^{2n-1}}.\end{align}$

We remark that the above equalities are valid for $n\geq1$ . When $n = 1$ , the corresponding formula in Theorem 3 can be shown analogously by examining the truncated sum below

$\mathrm{V}_m(1) = \lim\limits_{n\to\infty}\sum\limits_{k = 0}^n \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{k+1}.$

The first series in the last line can be evaluated directly as follows:

$\begin{align} \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{km+m-1}{m}}}\rfloor}}{(km+m)^{2n-1}} = \frac1{m^{2n-1}}\sum\limits_{k = 0}^\infty\frac{(-1)^k}{(k+1)^{2n-1}} = \begin{cases} \dfrac{\ln2}{m}, &n = 1;\\ \dfrac{2^{2n-1}-2}{(2m)^{2n-1}}\zeta(2n-1), &n > 1. \end{cases}\end{align}$

According to the parity of $k$ , the double series can be split into two

$\begin{align}\sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{m-1} \frac{(-1)^{\lfloor{{\frac{km+j-1}{m}}}\rfloor}}{(km+j)^{2n-1}} & = \sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{m-1} \frac{(-1)^k}{(km+j)^{2n-1}}\\ & = \sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{m-1} \frac1{(2km+j)^{2n-1}} -\sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{m-1} \frac1{(2km+m+j)^{2n-1}}.\end{align}$

Now making the replacements on summation indices by $j\to m-j$ and $k\to-k-1$ in the latter double sum and then unifying it to the former, we can simplify the resulting expression into the following bilateral series:

$\begin{align}\sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{m-1} \frac{(-1)^{\lfloor{{\frac{km+j-1}{m}}}\rfloor}}{(km+j)^{2n-1}} = &\sum\limits_{j = 1}^{m-1}\frac1{(2m)^{2n-1}}\sum\limits_{k = -\infty}^\infty \frac1{(k+\frac j{2m})^{2n-1}}.\end{align}$

Finally, by making use of (1), we can express the rightmost bilateral series in terms of Dirichlet beta function:

$\begin{align}\sum\limits_{k = -\infty}^\infty \frac1{(k+\frac j{2m})^{2n-1}} & = \frac{\pi^{2n-1}}{(2n-2)!} P_{2n-2}\Big(\cot\Big(\frac{j\pi}{2m}\Big)\Big)\\ & = (-1)^{n-1}\frac{4^n\beta(2n-1)}{E_{2n-2}} P_{2n-2}\Big(\cot\Big(\frac{j\pi}{2m}\Big)\Big).\end{align}$

Summing up, we have proved the following theorem.

Theorem 3 ( $m, n\in\mathbb{{N}}$ ).

$\begin{aligned}\sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{(k+1)^{2n-1}} & = (-1)^{n-1}\frac{2\beta(2n-1)}{m^{2n-1}E_{2n-2}}\sum\limits_{j = 1}^{m-1} P_{2n-2}\Big(\cot\Big(\frac{j\pi}{2m}\Big)\Big)\\ &+\begin{cases} \dfrac{\ln2}{m}, &n = 1;\\ \dfrac{2^{2n-1}-2}{(2m)^{2n-1}}\zeta(2n-1), &n\ge2. \end{cases}\end{aligned}$

When $m = 1$ , it is trivial to check that

$\mathrm{V}_1(n) = \sum\limits_{k = 0}^\infty\frac{(-1)^k}{(k+1)^{2n-1}} = \begin{cases} \ln2, &n = 1;\\ \big(1-4^{1-n}\big)\zeta(2n-1), &n\ge2. \end{cases}$

For $m = 2, 3$ and $1\le n\le 5$ , the closed formulae are highlighted as examples.

Example 7 ( $m = 2$ ).

$\begin{align} \mathrm{V}_2(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{2}}}\rfloor}}{k+1} = \beta(1)+\frac{\ln2}2, \\ \mathrm{V}_2(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{2}}}\rfloor}}{(k+1)^{3}} = \beta(3)+\frac{3}{32}\zeta(3), \\ \mathrm{V}_2(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{2}}}\rfloor}}{(k+1)^{5}} = \beta(5)+\frac{15}{512}\zeta(5), \\ \mathrm{V}_2(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{2}}}\rfloor}}{(k+1)^{7}} = \beta(7)+\frac{63}{8192}\zeta(7), \\ \mathrm{V}_2(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{2}}}\rfloor}}{(k+1)^{9}} = \beta(9)+\frac{255}{131072}\zeta(7).\end{align}$

Example 8 ( $m = 3$ ).

$\begin{align} \mathrm{V}_3(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{3}}}\rfloor}}{k+1} = \frac{8\beta(1)}{3 \sqrt{3}}+\frac{\ln2}{3}, \\ \mathrm{V}_3(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{3}}}\rfloor}}{(k+1)^{3}} = \frac{160 \beta(3)}{81 \sqrt{3}}+\frac{\zeta(3)}{36}, \\ \mathrm{V}_3(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{3}}}\rfloor}}{(k+1)^{5}} = \frac{2176 \beta(5)}{1215 \sqrt{3}}+\frac{5\zeta(5)}{1296}, \\ \mathrm{V}_3(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{3}}}\rfloor}}{(k+1)^{7}} = \frac{232960 \beta(7)}{133407 \sqrt{3}}+\frac{7\zeta(7)}{15552}, \\ \mathrm{V}_3(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor{{\frac{k}{3}}}\rfloor}}{(k+1)^{9}} = \frac{425805824 \beta(9)}{245348595 \sqrt{3}}+\frac{85\zeta(9)}{1679616}.\end{align}$

3.2. $\mathcal{{V}}_m(n)$

Under the replacements $k\to 2km+j-1$ with $1\leq j\leq2m$ , for $n > 1$ , we can rewrite $\mathcal{{V}}_m(n)$ as the following multisection series

$\begin{align}\mathcal{{V}}_m(n)& = \sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{(2k+1)^{2n-1}} = \sum\limits_{k = 0}^\infty\sum\limits_{j = 1}^{2m} \frac{(-1)^{\lfloor{{\frac{j-1}{m}}}\rfloor}}{(4km+2j-1)^{2n-1}}\\ & = \sum\limits_{k = 0}^\infty\Bigg\{ \sum\limits_{j = 1}^{m} \frac{(-1)^{\lfloor{{\frac{j-1}{m}}}\rfloor}}{(4km+2j-1)^{2n-1}} +\sum\limits_{j = 1}^{m} \frac{(-1)^{\lfloor{{\frac{m+j-1}{m}}}\rfloor}}{(4km+2m+2j-1)^{2n-1}}\Bigg\}\\ & = \sum\limits_{j = 1}^{m}\sum\limits_{k = 0}^\infty \frac{1}{(4km+2j-1)^{2n-1}} -\sum\limits_{j = 1}^{m}\sum\limits_{k = 0}^\infty \frac{1}{(4km+2m+2j-1)^{2n-1}}.\end{align}$

When $n = 1$ , the corresponding formula in Theorem 4 can be confirmed by examining the following truncated sum

$\mathcal{{V}}_m(1) = \lim\limits_{\ell\to\infty}\sum\limits_{k = 0}^\ell \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{2k+1}.$

Now, making the replacements $j\to m+1-j$ and $k\to-k-1$ simultaneously for the latter double sums and then unifying it to the former one, we obtain the bilateral series expression below

$\begin{align} \mathcal{{V}}_m(n)& = \sum\limits_{k = 0}^\infty \sum\limits_{j = 1}^{m} \frac{1}{(4km+2j-1)^{2n-1}} +\sum\limits_{k = -\infty}^{-1}\sum\limits_{j = 1}^{m} \frac{1}{(4km+2j-1)^{2n-1}}\\ & = \sum\limits_{j = 1}^{m}\sum\limits_{k = -\infty}^\infty \frac{1}{(4km+2j-1)^{2n-1}} = \sum\limits_{j = 1}^{m}\frac1{(4m)^{2n-1}}\sum\limits_{k = -\infty}^\infty \frac{1}{(k+\frac{2j-1}{4m})^{2n-1}}.\end{align}$

Keeping in mind (1) and writing the rightmost sum in terms of the Dirichlet beta function

$\begin{align}\sum\limits_{k = -\infty}^\infty \frac{1}{(k+\frac{2j-1}{4m})^{2n-1}} & = \frac{\pi^{2n-1}}{(2n-2)!} P_{2n-2}\Big(\cot\Big(\frac{2j-1}{4m}\pi\Big)\Big)\\ & = \frac{(-1)^{n-1}4^n \beta(2n-2)}{E_{2n-2}} P_{2n-2}\Big(\cot\Big(\frac{2j-1}{4m}\pi\Big)\Big), \end{align}$

we arrive at the compact expression as in the following theorem.

Theorem 4 ( $m, n\in\mathbb{{N}}$ ).

$\begin{aligned}\sum\limits_{k = 0}^\infty \frac{(-1)^{\lfloor{{\frac{k}{m}}}\rfloor}}{(2k+1)^{2n-1}} & = \frac{\beta(2n-1)}{m^{2n-1}E_{2n-2}} \Big(\frac{-1}{4}\Big)^{n-1}\sum\limits_{j = 1}^{m} P_{2n-2}\Big(\cot\Big(\frac{2j-1}{4m}\pi\Big)\Big).\end{aligned}$

When $m = 1$ , it is obvious to see that

$\mathcal{{V}}_1(n) = \sum\limits_{k = 0}^\infty\frac{(-1)^k}{(2k+1)^{2n-1}} = \beta(2n-1).$

For $m = 2, 3$ , the corresponding closed formulae are displayed as follows.

Example 9 ( $m = 2$ ).

$\begin{align} \mathcal{{V}}_2(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{2} \rfloor }}{2k+1} = \sqrt2\beta(1), \\ \mathcal{{V}}_2(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{2} \rfloor }}{(2k+1)^{3}} = \frac{3 \beta(3)}{2 \sqrt{2}}, \\ \mathcal{{V}}_2(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{2} \rfloor }}{(2k+1)^{5}} = \frac{57\beta(5)}{40 \sqrt{2}}, \\ \mathcal{{V}}_2(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{2} \rfloor }}{(2k+1)^{7}} = \frac{2763\beta(7)}{1952 \sqrt{2}}, \\ \mathcal{{V}}_2(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{2} \rfloor }}{(2k+1)^{9}} = \frac{250737\beta(9)}{177280 \sqrt{2}}.\end{align}$

Example 10 ( $m = 3$ ).

$\begin{align} \mathcal{{V}}_3(1) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{3} \rfloor }}{2k+1} = \frac{5\beta(1)}{3}, \\ \mathcal{{V}}_3(2) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{3} \rfloor }}{(2k+1)^{3}} = \frac{29 \beta(3)}{27}, \\ \mathcal{{V}}_3(3) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{3} \rfloor }}{(2k+1)^{5}} = \frac{245 \beta(5)}{243}, \\ \mathcal{{V}}_3(4) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{3} \rfloor }}{(2k+1)^{7}} = \frac{2189\beta(7)}{2187}, \\ \mathcal{{V}}_3(5) & = \sum\limits_{k = 0}^\infty\frac{(-1)^{\lfloor \frac{k}{3} \rfloor }}{(2k+1)^{9}} = \frac{19685 \beta(9)}{19683}.\end{align}$

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References

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Alternating series in terms of Riemann zeta function and Dirichlet beta function

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1. Introduction and motivation

2. Evaluation of $\mathrm{U}_m(n)$ and $\mathcal{{U}}_m(n)$

2.1. $\mathrm{U}_m(n)$

2.2. $\mathcal{{U}}_m(n)$

3. Evaluation of $\mathrm{V}_m(n)$ and $\mathcal{{V}}_m(n)$

3.1. $\mathrm{V}_m(n)$

3.2. $\mathcal{{V}}_m(n)$

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Alternating series in terms of Riemann zeta function and Dirichlet beta function

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2.1. Um(n) \mathrm{U}_m(n)

2.2. Um(n) \mathcal{{U}}_m(n)

3. Evaluation of Vm(n) \mathrm{V}_m(n) and Vm(n) \mathcal{{V}}_m(n)

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2. Evaluation of $\mathrm{U}_m(n)$ and $\mathcal{{U}}_m(n)$

2.1. $\mathrm{U}_m(n)$

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3. Evaluation of $\mathrm{V}_m(n)$ and $\mathcal{{V}}_m(n)$

3.1. $\mathrm{V}_m(n)$

3.2. $\mathcal{{V}}_m(n)$