Infinite Series Concerning Tails of Riemann Zeta Values

Abstract

Infinite series involving Riemann’s zeta and Dirichlet’s lambda tails, and weighted by three harmonic-like elementary symmetric functions are examined. By means of integral representations of zeta tails together with the telescopic approach, twelve general summation theorems are established that express these series as coefficients of the bivariate beta function $\mathrm{Beta}(u,v)$. By further expanding $\mathrm{Beta}(u,v)$ into Laurent series in u and v, several explicit summation formulae are shown as consequences.

1. Introduction and Outline

For $\alpha \in \mathbb{R}$ and $m,n\in {\mathbb{N}}_0$, define the parametric harmonic numbers by

$${\mathbf{H}}_n^{\langle m\rangle }\left(\alpha \right):=\sum _{k=0}^{n-1}\frac{1}{{(\alpha +k)}^m}.$$

When $\alpha =1$ and $\alpha =\frac{1}{2}$, they reduce to the usual harmonic numbers

$${\mathbf{H}}_n^{\langle m\rangle }:={\mathbf{H}}_n^{\langle m\rangle }\left(1\right)\mathrm{and}{\mathbf{O}}_n^{\langle m\rangle }:=2^{-m}{\mathbf{H}}_n^{\langle m\rangle }\left(\frac{1}{2}\right).$$

In case $m=1$, it will be suppressed from these notations.

For $n\in {\mathbb{N}}_0$ and an indeterminate x, the shifted factorial is defined by

$${\left(x\right)}_0=1\mathrm{and}{\left(x\right)}_n=x(x+1)\cdots (x+n-1)\mathrm{for}n\in \mathbb{N}.$$

It can also be expressed in terms of the $\Gamma$-function

$${\left(x\right)}_n=\frac{\Gamma (x+n)}{\Gamma \left(x\right)}\mathrm{with}\Gamma \left(x\right)={\int }_0^{\infty }{\tau }^{x-1}{e}^{-\tau }d\tau \mathrm{for}\Re \left(x\right)>0.$$

Associated with the $\Gamma$-function, there exists the beta function, that is defined, for $\Re \left(u\right)>0$ and $\Re \left(v\right)>0$, by

$$\mathrm{Beta}(u,v)=\mathrm{Beta}(v,u)={\int }_0^1{T}^{u-1}{(1-T)}^{v-1}dT=\frac{\Gamma \left(u\right)\Gamma \left(v\right)}{\Gamma (u+v)}.$$

Denote the Euler constant by ${\displaystyle \gamma =\underset{n\to \infty }{lim}\left({\mathbf{H}}_n-lnn\right)}$. Then, the logarithmic differentiation of the $\Gamma$-function results in the digamma function (cf. Rainville [1], §9)

$$\psi \left(z\right)=\frac{d}{dz}ln\Gamma \left(z\right)=\frac{{\Gamma }^{\prime }\left(z\right)}{\Gamma \left(z\right)}=-\gamma +\sum _{n=0}^{\infty }\frac{z-1}{(n+1)(n+z)}.$$

(1)

Let $\left[x^m\right]\phi \left(x\right)$ stand for the coefficient of $x^m$ in the formal power series $\phi \left(x\right)$. For a real number $\alpha \notin \mathbb{Z}\backslash \mathbb{N}$, we can extract the coefficients

$$\left[x\right]\frac{\Gamma (\alpha -x)}{\Gamma \left(\alpha \right)}=-\psi \left(\alpha \right) \mathrm{and} \left[x^2\right]\frac{\Gamma (\alpha -x)}{\Gamma \left(\alpha \right)}=\frac{{\psi }^2\left(\alpha \right)+{\psi }^{\prime }\left(\alpha \right)}{2}$$

from the exponential expression (cf. [2])

$$\frac{\Gamma (\alpha -x)}{\Gamma \left(\alpha \right)}=exp\left\{-x\psi \left(\alpha \right)+\sum _{k=2}^{\infty }\frac{x^k}{k}{\zeta }_k\left(\alpha \right)\right\}.$$

(2)

Henceforth, the Riemann and Hurwitz zeta functions are defined, respectively, by

$$\zeta \left(m\right)=\sum _{n=1}^{\infty }\frac{1}{n^m}\mathrm{and}{\zeta }_m\left(z\right)=\frac{{(-1)}^m}{(m-1)!}{\mathcal{D}}_z^{m-1}\psi \left(z\right)=\sum _{n=0}^{\infty }\frac{1}{{(n+z)}^m}.$$

In addition, we shall also make use of Dirichlet’s lambda function

$$\lambda \left(m\right)=\sum _{n=0}^{\infty }\frac{1}{{(2n+1)}^m}=(1-2^{-m})\zeta \left(m\right).$$

The Riemann zeta function and its variants play a pivotal role in number theory, and have important applications in mathematics and physics (cf. [3,4]). Recently, Furdui [5,6], followed by Boyadzhiev [7] and Nguyen [8], evaluated a few series involving the Riemann zeta tails, which are also related to harmonic numbers (cf. [9,10,11,12,13]) and multifold zeta values (cf. [14,15,16,17]). Motivated by these works, the objective of the present paper is to investigate infinite series containing Riemann’s zeta tails “$\zeta \left(n\right)-{\mathbf{H}}_k^{\langle n\rangle }$” and Dirichlet’s lambda tails “$\lambda \left(n\right)-{\mathbf{O}}_k^{\langle n\rangle }$”, and weighted by factors of three harmonic-like numbers $\left\{\sigma \right(m,k),\tau (m,k),\rho (m,k;\alpha \left)\right\}$. The remaining part of the paper is divided into three sections in accordance with these three-factor sequences. Several remarkable theorems will be shown that express the values of these series as coefficients of the bivariate beta function $\mathrm{Beta}(u,v)$, expanded into Laurent series of u and v. This will be fulfilled in two different approaches. One approach is combining the generating functions with two integral representation formulae $(m\in \mathrm{N},n\in {\mathbb{N}}_0)$

$$\begin{array}{cc}\hfill {\int }_0^1\frac{x^n{ln}^{m}x}{1-x}dx& ={(-1)}^{m}m!\left\{\zeta (m+1)-{\mathbf{H}}_n^{\langle m+1\rangle }\right\},\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill {\int }_0^1\frac{x^{2n}{ln}^{m}x}{1-x^2}dx& ={(-1)}^{m}m!\left\{\lambda (m+1)-{\mathbf{O}}_n^{\langle m+1\rangle }\right\};\hfill \end{array}$$

(4)

which can be proved, without difficulty, by integration by parts. The other approach is the series rearrangement together with telescoping and the following expression of the beta function:

$$\mathrm{Beta}(u,v)=\sum _{k=0}^{\infty }{(-1)}^k\left(\genfrac{}{}{0pt}{}{u-1}{k}\right){\int }_0^1{T}^{k+v-1}dT=\sum _{k=0}^{\infty }{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)}{k+v}.$$

(5)

In fact, by means of the binomial expansion, we can express

$$\begin{array}{cc}\hfill \mathrm{Beta}(u,v)& =\underset{\delta \to 1^{-}}{lim}{\int }_0^{\delta }{T}^{v-1}{(1-T)}^{u-1}dT\hfill \\ & =\underset{\delta \to 1^{-}}{lim}{\int }_0^{\delta }{T}^{v-1}\left\{\sum _{k=0}^{\infty }{(-T)}^k\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)\right\}dT\hfill \\ & =\underset{\delta \to 1^{-}}{lim}\sum _{k=0}^{\infty }{(-1)}^k\left(\genfrac{}{}{0pt}{}{u-1}{k}\right){\int }_0^{\delta }{T}^{k+v-1}dT\hfill \\ & =\underset{\delta \to 1^{-}}{lim}\sum _{k=0}^{\infty }{(-1)}^k\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)\frac{{\delta }^{k+v}}{k+v}=\sum _{k=0}^{\infty }{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)}{k+v}.\hfill \end{array}$$

This can be justified as follows. When $0<u<1$, it is trivial to check that $\left|\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)\right|<1$. For $0\le T\le \delta$, we have

$$T^{v-1}\sum _{k=0}^{\infty }{(-T)}^k\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)=T^{v-1}{(1-T)}^{u-1},$$

and the partial sum

$$\left|T^{v-1}\sum _{k=0}^n{(-T)}^k\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)\right|\le \sum _{k=0}^n{T}^{k+v-1}\left|\left(\genfrac{}{}{0pt}{}{u-1}{k}\right)\right|<\sum _{k=0}^{\infty }{T}^{k+v-1}$$

is dominated by the integrable function $\frac{T^{v-1}}{1-T}$ on $[0,\delta ]$. Then, exchanging orders of summation and integration is straightforwardly validated by applying Lebesgue’s dominated convergence theorem [18] (§11.32). Similar arguments can be utilized to treat exchanging orders of summation and integration for the rest of the paper, which will not be reproduced for brevity.

In order to ensure the accuracy, we have checked experimentally all the displayed equations throughout the paper using appropriately devised Mathematica (version 11) commands.

2. Four Series with Weight Factor $\mathbf{\sigma }(\mathit{m},\mathit{n})$

This section will be devoted to four classes of infinite series containing zeta and lambda tails with the weight factor $\sigma (m,n)$ introduced below. Four general theorems will be shown and several concrete summation formulae will be given as consequences.

For $m,n\in {\mathbb{N}}_0$, define the elementary symmetric function

$$\sigma (m,n)=\sum _{1\le k_1<k_2<\cdots <k_m\le n}\prod _{i=1}^m\frac{1}{k_i}=\left[x^m\right]\left(\genfrac{}{}{0pt}{}{x+n}{n}\right),$$

where the rightmost expression follows by

$$\sigma (m,n)=\left[x^m\right]\prod _{k=1}^n\left(1+\frac{x}{k}\right)=\left[x^m\right]\frac{{(x+1)}_n}{n!}=\left[x^m\right]\left(\genfrac{}{}{0pt}{}{x+n}{n}\right).$$

By applying the binomial equality

$$\left(\genfrac{}{}{0pt}{}{x+n+1}{n+1}\right)=\left(\genfrac{}{}{0pt}{}{x+n}{n}\right)+\frac{x}{n+1}\left(\genfrac{}{}{0pt}{}{x+n}{n}\right),$$

we can show the recurrence relation

$$\sigma (m+1,n+1)=\sigma (m+1,n)+\frac{\sigma (m,n)}{n+1}.$$

This sequence can be expressed as Bell polynomials (cf. [19,20], §3.3) of harmonic numbers

$$\begin{array}{cc}\hfill \sigma (0,n)& =1,\hfill \\ \hfill \sigma (1,n)& ={\mathbf{H}}_n,\hfill \\ \hfill \sigma (2,n)& =\frac{1}{2}\left\{{\mathbf{H}}_n^2-{\mathbf{H}}_n^{\langle 2\rangle }\right\},\hfill \\ \hfill \sigma (3,n)& =\frac{1}{6}\left\{{\mathbf{H}}_n^3-3{\mathbf{H}}_n{\mathbf{H}}_n^{\langle 2\rangle }+2{\mathbf{H}}_n^{\langle 3\rangle }\right\}.\hfill \end{array}$$

Furthermore, there exist logarithmic generating functions as in the lemma below.

Lemma 1

(Properties of $\sigma (m,n):\left|y\right|<1$).

$$\begin{array}{cc}\hfill \left(a\right)& \sum _{n=0}^{\infty }\sigma (m,n)y^n=\frac{{(-1)}^m}{m!}\frac{{ln}^m(1-y)}{1-y},\hfill \\ \hfill \left(b\right)& \sum _{n=0}^{\infty }\frac{\sigma (m,n)}{n+1}{y}^{n+1}=\frac{{(-1)}^{m+1}}{(m+1)!}{ln}^{m+1}(1-y).\hfill \end{array}$$

Proof. 

Suppose that both x and y are real variables with $\left|y\right|<1$. Recalling the equality

$$\sum _{m\ge 0}\sigma (m,n)x^m=\left(\genfrac{}{}{0pt}{}{x+n}{n}\right)$$

and applying the binomial series, we can determine the bivariate generating function:

$$\sum _{m,n\ge 0}\sigma (m,n)x^m{y}^n=\sum _{n\ge 0}\left(\genfrac{}{}{0pt}{}{x+n}{n}\right)y^n={(1-y)}^{-1-x}=\frac{exp(-xln(1-y\left)\right)}{1-y}.$$

Then, the first formula (a) can be derived by extracting the coefficient $x^m$ across the above equality. The second generating function (b) results from integrating the first one (a).    □

2.1. Series ${\mathrm{U}}_1(m,n)$ with Zeta Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with zeta tails

$$\begin{array}{cc}\hfill {\mathrm{U}}_1(m,n)& :=\sum _{k=0}^{\infty }\sigma (m,k)\left\{\zeta (n+3)-{\mathbf{H}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^n}{(n+2)!}{\int }_0^1\frac{{ln}^{n+2}(1-x)}{x}dx\sum _{k=0}^{\infty }\sigma (m,k){(1-x)}^k\hfill \\ & =\frac{{(-1)}^{m+n}}{m!(n+2)!}{\mathcal{U}}_1(m,n),\hfill \end{array}$$

where according to Lemma 1(a), ${\mathcal{U}}_1(m,n)$ is the definite integral

$${\mathcal{U}}_1(m,n)={\int }_0^1\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{x^2}dx.$$

The convergence of this and all the subsequent improper integrals in this paper can exclusively be verified by making use of the following almost trivial limiting relation. For $m\in {\mathbb{N}}_0$ and a sufficiently small real number $\epsilon >0$, we have

$$\underset{x\to 0^{+}}{lim}{x}^{\epsilon }{ln}^{m}x=0,$$

which is equivalent to that there is another small real number $\delta >0$ such that

$$\left|{ln}^{m}x\right|<\frac{\epsilon }{x^{\epsilon }}\mathrm{for}0<x<\delta .$$

Now, split the improper integral for ${\mathcal{U}}_1(m,n)$ into three ${\mathcal{U}}_1(m,n)=I_1+I_2+I_3$, where

$$\begin{array}{cc}\hfill I_1& :={\int }_{\delta }^{1-\delta }\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{x^2}dx,\hfill \\ \hfill I_2& :={\int }_0^{\delta }\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{x^2}dx,\hfill \\ \hfill I_3& :={\int }_{1-\delta }^1\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{x^2}dx.\hfill \end{array}$$

The first integral $I_1$ is well-defined. For the second integral $I_2$, we can estimate its integrand by

$$\left|\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{x^2}\right|<\left|\frac{\epsilon {ln}^{n+2}(1-x)}{x^{2+\epsilon }}\right|<\frac{\epsilon A}{x^{\epsilon }}\mathrm{for}0<x<\delta ,$$

where keeping in mind that ${lim}_{x\to 0}\frac{ln(1-x)}{x}=-1$, the last step is justified by the fact that $\frac{{ln}^{n+2}(1-x)}{x^2}$ is bounded for $x\in (0,\delta ]$, i.e., there exists a constant $A>0$ such that

$$\left|\frac{{ln}^{n+2}(1-x)}{x^2}\right|=\left|\frac{{ln}^2(1-x)}{x^2}{ln}^n(1-x)\right|<A\mathrm{for}0<x<\delta .$$

Then, $I_2$ converges since $\frac{1}{x^{\epsilon }}$ is integrable over $[0,\delta ]$. Finally, by making the change of variables $x\to 1-y$, we can express

$$I_3={\int }_0^{\delta }\frac{{ln}^m(1-y){ln}^{n+2}\left(y\right)}{{(1-y)}^2}dy.$$

Analogously, we can estimate the above integrand by

$$\left|\frac{{ln}^m(1-y){ln}^{n+2}\left(y\right)}{{(1-y)}^2}\right|<\left|\frac{\epsilon {ln}^m(1-y)}{y^{\epsilon }{(1-y)}^2}\right|<\frac{\epsilon B}{y^{\epsilon }}\mathrm{for}0<y<\delta ,$$

because $\frac{{ln}^m(1-y)}{{(1-y)}^2}$ is also bounded for $y\in [0,\delta ]$ which is equivalent to the existence of a constant $B>0$ such that

$$\left|\frac{{ln}^m(1-y)}{{(1-y)}^2}\right|<B\mathrm{for}0<y<\delta .$$

We affirm consequently that $I_3$ is convergent since $\frac{1}{y^{\epsilon }}$ is integrable over $[0,\delta ]$. Summing up, we have shown that the improper integral for ${\mathcal{U}}_1(m,n)$ is always convergent for $m,n\in {\mathbb{N}}_0$.

In order to evaluate the above integral, we examine the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{U}}_1(m,n)={\int }_0^1{x}^{u-2}{(1-x)}^v{ln}^2(1-x)dx\hfill \\ & =\frac{{\partial }^2}{\partial v^2}{\int }_0^1{x}^{u-2}{(1-x)}^{v}dx=\frac{{\partial }^2}{\partial v^2}\mathrm{Beta}(u-1,v+1),\hfill \end{array}$$

where exchanging orders of summation and integration is justified by Lebesgue’s dominated convergence theorem [18] (§11.32) as described at the end of the introduction. Then, by extracting the coefficient of $u^m{v}^n$, we have

$$\begin{array}{cc}\hfill {\mathcal{U}}_1(m,n)& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial v^2}\mathrm{Beta}(u-1,v+1)\hfill \\ & =m!(n+2)!\left[u^m{v}^{n+2}\right]\mathrm{Beta}(u-1,v+1),\hfill \end{array}$$

which leads us to the following remarkable expression.

Theorem 1.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{U}}_1(m,n)& =\sum _{k=0}^{\infty }\sigma (m,k)\left\{\zeta (n+3)-{\mathbf{H}}_k^{\langle n+3\rangle }\right\}\hfill \\ & ={(-1)}^{m+n}\left[u^m{v}^{n+2}\right]\mathrm{Beta}(u-1,v+1).\hfill \end{array}$$

Observing the expression of the $\Gamma$-function quotient

$$\mathrm{Beta}\left(u-1,v+1\right)=\frac{(u+v)\Gamma (1+u)\Gamma (1+v)}{u(u-1)\Gamma (1+u+v)}$$

and then applying (2), we can expand the above beta function into the Laurent series in u and v. By extracting coefficients of monomials “$u^m{v}^{n+2}$”, we can derive numerous infinite series identities for ${\mathrm{U}}_1(m,n)$ by running the following Mathematica commands:

u1x[m_,n_]:=(-1)^(m+n)*SeriesCoefficient[Beta[u-1,1+v],{u,0,m},{v,0,n+2}]

u1y[m_,n_]:=FullSimplify[u1x[m,n]/.PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]]

For instance, three values ${\mathrm{U}}_1(0,0)$, ${\mathrm{U}}_1(0,1)$ and ${\mathrm{U}}_1(1,0)$ are produced as follows:

In[9]:= Partition[Flatten[Table[{m,n,u1y[m,n]},{m,0,1},{n,0,1-m}]],3]

Out[9]= {{0,0,\[Pi]^2/6},{0,1,Zeta[3]},{1,0,-(\[Pi]^2/6)+2 Zeta[3]}}

Some particular formulae related to Theorem 1 are recorded in the corollary below.

Corollary 1

(Particular values of ${\mathrm{U}}_1(m,n)$).

$$\begin{array}{cc}& \begin{array}{cc}{\mathrm{U}}_1(0,0)=\frac{{\pi }^2}{6},\hfill & {\mathrm{U}}_1(1,0)=2\zeta \left(3\right)-\frac{{\pi }^2}{6},\hfill \\ {\mathrm{U}}_1(0,1)=\zeta \left(3\right),\hfill & {\mathrm{U}}_1(1,1)=\frac{{\pi }^4}{72}-\zeta \left(3\right),\hfill \\ {\mathrm{U}}_1(0,2)=\frac{{\pi }^4}{90},\hfill & {\mathrm{U}}_1(1,2)=3\zeta \left(5\right)-\frac{{\pi }^4}{90}-\frac{{\pi }^2}{6}\zeta \left(3\right),\hfill \\ {\mathrm{U}}_1(0,3)=\zeta \left(5\right),\hfill & {\mathrm{U}}_1(1,3)=\frac{{\pi }^6}{540}-\frac{1}{2}{\zeta }^2\left(3\right)-\zeta \left(5\right),\hfill \end{array}\hfill \\ & {\mathrm{U}}_1(2,0)=\frac{{\pi }^2}{6}+\frac{{\pi }^4}{72}-2\zeta \left(3\right),\hfill \\ & {\mathrm{U}}_1(2,1)=\zeta \left(3\right)+4\zeta \left(5\right)-\frac{{\pi }^4}{72}-\frac{{\pi }^2\zeta \left(3\right)}{3},\hfill \\ & {\mathrm{U}}_1(2,2)=\frac{{\pi }^4}{90}+\frac{{\pi }^6}{432}+\frac{{\pi }^2\zeta \left(3\right)}{6}-\frac{3{\zeta }^2\left(3\right)}{2}-3\zeta \left(5\right),\hfill \\ & {\mathrm{U}}_1(2,3)=\frac{{\zeta }^2\left(3\right)}{2}-\frac{{\pi }^6}{540}-\frac{{\pi }^4\zeta \left(3\right)}{40}-\frac{{\pi }^2\zeta \left(5\right)}{2}+\zeta \left(5\right)+8\zeta \left(7\right).\hfill \end{array}$$

2.2. Series ${\mathrm{U}}_2(m,n)$ with Lambda Tails

Alternatively, for $m,n\in {\mathbb{N}}_0$, consider the next series containing Dirichlet’s lambda tails

$$\begin{array}{cc}\hfill {\mathrm{U}}_2(m,n)& :=\sum _{k=0}^{\infty }\sigma (m,k)\left\{\lambda (n+3)-{\mathbf{O}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^n}{(n+2)!}{\int }_0^1\frac{{ln}^{n+2}x}{1-x^2}dx\sum _{k=0}^{\infty }\sigma (m,k)x^{2k}\hfill \\ & =\frac{{(-1)}^{m+n}}{m!(n+2)!}{\mathcal{U}}_2(m,n),\hfill \end{array}$$

where in view of Lemma 1(a), ${\mathcal{U}}_2(m,n)$ is the definite integral

$${\mathcal{U}}_2(m,n)={\int }_0^1\frac{{ln}^m(1-x^2){ln}^{n+2}x}{{(1-x^2)}^2}dx.$$

By manipulating the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{U}}_2(m,n)={\int }_0^1{(1-x^2)}^{u-2}{x}^v{ln}^{2}xdx\hfill \\ & =\frac{{\partial }^2}{\partial v^2}{\int }_0^1{(1-x^2)}^{u-2}{x}^{v}dx=\frac{{\partial }^2}{\partial v^2}\frac{\mathrm{Beta}(u-1,\frac{v+1}{2})}{2}\hfill \end{array}$$

and then, by extracting the coefficient of $u^m{v}^n$, we have

$$\begin{array}{cc}\hfill {\mathcal{U}}_2(m,n)& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial v^2}\frac{\mathrm{Beta}(u-1,\frac{v+1}{2})}{2}\hfill \\ & =m!(n+2)!\left[u^m{v}^{n+2}\right]\frac{\mathrm{Beta}(u-1,\frac{v+1}{2})}{2},\hfill \end{array}$$

which yields the following compact formula.

Theorem 2.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{U}}_2(m,n)& =\sum _{k=0}^{\infty }\sigma (m,k)\left\{\lambda (n+3)-{\mathbf{O}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{m+n}}{2}\left[u^m{v}^{n+2}\right]\mathrm{Beta}\left(u-1,\frac{v+1}{2}\right).\hfill \end{array}$$

By executing the following Mathematica commands:

u2x[m_,n_]:=(-1)^(m+n)/2*SeriesCoefficient[Beta[u-1,(1+v)/2],{u,0,m},{v,0,n+2}]

u2y[m_,n_]:=FullSimplify[u2x[m,n]/.PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]]

we display some summation formulae as examples.

Corollary 2

(Particular values of ${\mathrm{U}}_2(m,n)$).

$$\begin{array}{cc}& \begin{array}{cc}{\mathrm{U}}_2(0,0)=\frac{{\pi }^2}{16}+\frac{7\zeta \left(3\right)}{16},\hfill & {\mathrm{U}}_2(0,1)=\frac{{\pi }^4}{192}+\frac{7\zeta \left(3\right)}{16},\hfill \\ {\mathrm{U}}_2(0,2)=\frac{{\pi }^4}{192}+\frac{31\zeta \left(5\right)}{64},\hfill & {\mathrm{U}}_2(0,3)=\frac{{\pi }^6}{1920}+\frac{31\zeta \left(5\right)}{64},\hfill \end{array}\hfill \\ & {\mathrm{U}}_2(1,0)=\frac{{\pi }^4}{128}-\frac{{\pi }^2}{16}-\frac{{\pi }^2}{8}ln2+\frac{21\zeta \left(3\right)}{16}-\frac{7ln2}{8}\zeta \left(3\right),\hfill \\ & {\mathrm{U}}_2(1,1)=\frac{5{\pi }^4}{384}-\frac{{\pi }^4}{96}ln2-\frac{7\zeta \left(3\right)}{16}-\frac{7{\pi }^2}{64}\zeta \left(3\right)-\frac{7ln2}{8}\zeta \left(3\right)+\frac{31\zeta \left(5\right)}{16},\hfill \\ & {\mathrm{U}}_2(1,2)=\frac{{\pi }^6}{768}-\frac{{\pi }^4}{192}-\frac{{\pi }^4}{96}ln2-\frac{7{\pi }^2}{64}\zeta \left(3\right)-\frac{49{\zeta }^2\left(3\right)}{128}+\frac{155\zeta \left(5\right)}{64}-\frac{31ln2}{32}\zeta \left(5\right),\hfill \\ & {\mathrm{U}}_2(1,3)=\frac{7{\pi }^6}{3840}-\frac{{\pi }^{6}ln2}{960}-\frac{7{\pi }^4}{768}\zeta \left(3\right)-\frac{49{\zeta }^2\left(3\right)}{128}-\frac{31\zeta \left(5\right)}{64}-\frac{31{\pi }^2}{256}\zeta \left(5\right)-\frac{31ln2}{32}\zeta \left(5\right)+\frac{381\zeta \left(7\right)}{128},\hfill \\ & {\mathrm{U}}_2(2,2)=\frac{19{\pi }^6}{4608}-\frac{{\pi }^{6}ln2}{384}+\frac{{\pi }^4}{192}+\frac{{\pi }^4}{96}ln2+\frac{{\pi }^4}{96}{ln}^{2}2-\frac{49{\pi }^4}{1536}\zeta \left(3\right)+\frac{7{\pi }^2}{64}\zeta \left(3\right)+\frac{7{\pi }^{2}ln2}{32}\zeta \left(3\right)\hfill \\ & +\frac{49ln2}{64}{\zeta }^2\left(3\right)-\frac{245{\zeta }^2\left(3\right)}{128}-\frac{155\zeta \left(5\right)}{64}-\frac{217{\pi }^2}{384}\zeta \left(5\right)-\frac{155ln2}{32}\zeta \left(5\right)+\frac{31{ln}^{2}2}{32}\zeta \left(5\right)+\frac{635\zeta \left(7\right)}{64}.\hfill \end{array}$$

2.3. Series ${\mathrm{U}}_3(m,n)$ with Zeta Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with zeta tails

$$\begin{array}{cc}\hfill {\mathrm{U}}_3(m,n)& :=\sum _{k=0}^{\infty }\frac{\sigma (m,k)}{k+1}\left\{\zeta (n+2)-{\mathbf{H}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{n+1}}{(n+1)!}{\int }_0^1\frac{{ln}^{n+1}(1-x)}{x}dx\sum _{k=0}^{\infty }\frac{\sigma (m,k)}{k+1}{(1-x)}^k\hfill \\ & =\frac{{(-1)}^{m+n}}{(m+1)!(n+1)!}{\mathcal{U}}_3(m,n),\hfill \end{array}$$

where in view of Lemma 1(b), ${\mathcal{U}}_3(m,n)$ is the definite integral given by

$${\mathcal{U}}_3(m,n)={\int }_0^1\frac{{ln}^{m+1}x{ln}^{n+1}(1-x)}{x(1-x)}dx.$$

In order to evaluate the above integral, we examine the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{U}}_3(m,n)={\int }_0^1{x}^{u-1}{(1-x)}^{v-1}lnxln(1-x)dx\hfill \\ & =\frac{{\partial }^2}{\partial u\partial v}{\int }_0^1{x}^{u-1}{(1-x)}^{v-1}dx=\frac{{\partial }^2}{\partial u\partial v}\mathrm{Beta}(u,v).\hfill \end{array}$$

By extracting the coefficient of $u^m{v}^n$

$$\begin{array}{cc}\hfill {\mathcal{U}}_3(m,n)& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial u\partial v}\mathrm{Beta}(u,v)\hfill \\ & =(m+1)!(n+1)!\left[u^{m+1}{v}^{n+1}\right]\mathrm{Beta}(u,v),\hfill \end{array}$$

we obtain the following simplified formula.

Theorem 3.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{U}}_3(m,n)& =\sum _{k=0}^{\infty }\frac{\sigma (m,k)}{k+1}\left\{\zeta (n+2)-{\mathbf{H}}_k^{\langle n+2\rangle }\right\}\hfill \\ & ={(-1)}^{m+n}\left[u^{m+1}{v}^{n+1}\right]\mathrm{Beta}(u,v).\hfill \end{array}$$

According to this theorem, there is an interesting symmetry: ${\mathrm{U}}_3(m,n)={\mathrm{U}}_3(n,m)$. By carrying out the following Mathematica commands:

u3x[m_,n_]:=(-1)^(m+n)*SeriesCoefficient[Beta[u,v],{u,0,m+1},{v,0,n+1}]

u3y[m_,n_]:=FullSimplify[u3x[m,n]/.PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]]

we find the initial identities in the following corollary.

Corollary 3

(Particular values of ${\mathrm{U}}_3(m,n)$).

$$\begin{array}{cc}& {\mathrm{U}}_3(0,0)=2\zeta \left(3\right),\hfill \\ & {\mathrm{U}}_3(0,1)={\mathrm{U}}_3(1,0)=\frac{{\pi }^4}{72},\hfill \\ & {\mathrm{U}}_3(0,2)={\mathrm{U}}_3(2,0)=3\zeta \left(5\right)-\frac{{\pi }^2}{6}\zeta \left(3\right),\hfill \\ & {\mathrm{U}}_3(0,3)={\mathrm{U}}_3(3,0)=\frac{{\pi }^6}{540}-\frac{{\zeta }^2\left(3\right)}{2},\hfill \\ & {\mathrm{U}}_3(1,1)=4\zeta \left(5\right)-\frac{{\pi }^2}{3}\zeta \left(3\right),\hfill \\ & {\mathrm{U}}_3(1,2)={\mathrm{U}}_3(2,1)=\frac{{\pi }^6}{432}-\frac{3}{2}{\zeta }^2\left(3\right),\hfill \\ & {\mathrm{U}}_3(1,3)={\mathrm{U}}_3(3,1)=8\zeta \left(7\right)-\frac{{\pi }^4\zeta \left(3\right)}{40}-\frac{{\pi }^2\zeta \left(5\right)}{2},\hfill \\ & {\mathrm{U}}_3(2,2)=10\zeta \left(7\right)-\frac{2{\pi }^2}{3}\zeta \left(5\right)-\frac{{\pi }^4}{36}\zeta \left(3\right),\hfill \\ & {\mathrm{U}}_3(2,3)={\mathrm{U}}_3(3,2)=\frac{{\pi }^2{\zeta }^2\left(3\right)}{4}-7\zeta \left(3\right)\zeta \left(5\right)+\frac{47{\pi }^8}{86400},\hfill \\ & {\mathrm{U}}_3(3,3)=28\zeta \left(9\right)+{\zeta }^3\left(3\right)-\frac{{\pi }^6\zeta \left(3\right)}{216}-\frac{7{\pi }^4\zeta \left(5\right)}{90}-\frac{5{\pi }^2\zeta \left(7\right)}{3}.\hfill \end{array}$$

2.4. Series ${\mathrm{U}}_4(m,n)$ with Lambda Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with lambda tails

$$\begin{array}{cc}\hfill {\mathrm{U}}_4(m,n)& :=\sum _{k=0}^{\infty }\frac{\sigma (m,k)}{k+1}\left\{\lambda (n+2)-{\mathbf{O}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{n+1}}{(n+1)!}{\int }_0^1\frac{{ln}^{n+1}x}{1-x^2}dx\sum _{k=0}^{\infty }\frac{\sigma (m,k)}{k+1}{x}^{2k}\hfill \\ & =\frac{{(-1)}^{m+n}}{(m+1)!(n+1)!}{\mathcal{U}}_4(m,n),\hfill \end{array}$$

where, by Lemma 1(b), ${\mathcal{U}}_4(m,n)$ is explicitly given by the definite integral

$${\mathcal{U}}_4(m,n)={\int }_0^1\frac{{ln}^{m+1}(1-x^2){ln}^{n+1}x}{x^2(1-x^2)}dx.$$

By reformulating the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{U}}_4(m,n)={\int }_0^1{(1-x^2)}^{u-1}{x}^{v-2}lnxln(1-x^2)dx\hfill \\ & =\frac{{\partial }^2}{\partial u\partial v}{\int }_0^1{(1-x^2)}^{u-1}{x}^{v-2}dx=\frac{{\partial }^2}{\partial u\partial v}\frac{\mathrm{Beta}(u,\frac{v-1}{2})}{2},\hfill \end{array}$$

and then, by extracting the coefficient of $u^m{v}^n$

$$\begin{array}{cc}\hfill {\mathcal{U}}_4(m,n)& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial u\partial v}\frac{\mathrm{Beta}(u,\frac{v-1}{2})}{2}\hfill \\ & =(m+1)!(n+1)!\left[u^{m+1}{v}^{n+1}\right]\frac{\mathrm{Beta}(u,\frac{v-1}{2})}{2},\hfill \end{array}$$

we establish the following compact formula.

Theorem 4.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{U}}_4(m,n)& =\sum _{k=0}^{\infty }\frac{\sigma (m,k)}{k+1}\left\{\lambda (n+2)-{\mathbf{O}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{m+n}}{2}\left[u^{m+1}{v}^{n+1}\right]\mathrm{Beta}\left(u,\frac{v-1}{2}\right).\hfill \end{array}$$

By fulfilling the following Mathematica commands:

u4x[m_,n_]:=(-1)^(m+n)/2*SeriesCoefficient[Beta[u,(v-1)/2],{u,0,m+1},{v,0,n+1}]

u4y[m_,n_]:=FullSimplify[u4x[m,n]/.PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]]

some explicit identities are shown as consequences in the corollary below.

Corollary 4

(Particular values of ${\mathrm{U}}_4(m,n)$).

$$\begin{array}{cc}\hfill {\mathrm{U}}_4(0,0)& =\frac{{\pi }^2}{4}-2ln2-\frac{{\pi }^2}{4}ln2+\frac{7}{4}\zeta \left(3\right),\hfill \\ \hfill {\mathrm{U}}_4(0,1)& =\frac{{\pi }^4}{64}-\frac{{\pi }^2}{4}+2ln2+\frac{7}{4}\zeta \left(3\right)-\frac{7}{4}\zeta \left(3\right)ln2,\hfill \\ \hfill {\mathrm{U}}_4(0,2)& =\frac{{\pi }^2}{4}+\frac{{\pi }^4}{48}-2ln2-\frac{{\pi }^4}{48}ln2-\frac{7}{4}\zeta \left(3\right)-\frac{7{\pi }^2}{32}\zeta \left(3\right)+\frac{31}{8}\zeta \left(5\right),\hfill \\ \hfill {\mathrm{U}}_4(1,0)& =\frac{{\pi }^4}{48}-\frac{{\pi }^2}{6}-\frac{{\pi }^2}{2}ln2+2{ln}^{2}2+\frac{{\pi }^2}{4}{ln}^{2}2+\frac{7}{2}\zeta \left(3\right)-\frac{7ln2}{2}\zeta \left(3\right),\hfill \\ \hfill {\mathrm{U}}_4(1,1)& =\frac{{\pi }^2}{6}+\frac{{\pi }^4}{32}+\frac{{\pi }^2}{2}ln2-\frac{{\pi }^4}{32}ln2-2{ln}^{2}2-\frac{7}{2}\zeta \left(3\right)-\frac{7{\pi }^2}{12}\zeta \left(3\right)-\frac{7ln2}{2}\zeta \left(3\right)+\frac{7}{4}\zeta \left(3\right){ln}^{2}2+\frac{31}{4}\zeta \left(5\right),\hfill \\ \hfill {\mathrm{U}}_4(1,2)& =\frac{13{\pi }^6}{2304}-\frac{{\pi }^4}{32}-\frac{{\pi }^2}{6}-\frac{{\pi }^2}{2}ln2-\frac{{\pi }^4}{24}ln2+2{ln}^{2}2+\frac{{\pi }^4}{48}{ln}^{2}2+\frac{7}{2}\zeta \left(3\right)\hfill \\ & -\frac{7{\pi }^2}{16}\zeta \left(3\right)+\frac{7ln2}{2}\zeta \left(3\right)+\frac{7{\pi }^2}{16}\zeta \left(3\right)ln2-\frac{49}{16}\zeta {\left(3\right)}^2+\frac{31}{4}\zeta \left(5\right)-\frac{31ln2}{4}\zeta \left(5\right),\hfill \\ \hfill {\mathrm{U}}_4(2,2)& =\frac{{\pi }^4}{24}+\frac{{\pi }^4}{16}ln2+\frac{{\pi }^4}{24}{ln}^{2}2-\frac{{\pi }^4}{72}{ln}^{3}2-\frac{43{\pi }^4}{384}\zeta \left(3\right)+\frac{13{\pi }^6}{1152}-\frac{13{\pi }^6}{1152}ln2+\frac{7{\pi }^{2}ln2}{8}\zeta \left(3\right)\hfill \\ & +\frac{{\pi }^2}{3}ln2+\frac{{\pi }^2}{2}{ln}^{2}2-\frac{7{\pi }^2{ln}^{2}2}{16}\zeta \left(3\right)+\frac{7{\pi }^2\zeta \left(3\right)}{6}-\frac{31{\pi }^2}{12}\zeta \left(5\right)-\frac{7{ln}^{2}2}{2}\zeta \left(3\right)-\frac{31ln2}{2}\zeta \left(5\right)\hfill \\ & -\frac{4{ln}^{3}2}{3}+\frac{31{ln}^{2}2}{4}\zeta \left(5\right)-\frac{49{\zeta }^2\left(3\right)}{8}+\frac{49ln2}{8}{\zeta }^2\left(3\right)-\frac{31\zeta \left(5\right)}{2}+\frac{635\zeta \left(7\right)}{16}-2\zeta \left(3\right)-7ln2\zeta \left(3\right).\hfill \end{array}$$

3. Four Series with Weight Factor $\mathbf{\tau }(\mathit{m},\mathit{n})$

For $m,n\in {\mathbb{N}}_0$, define the elementary symmetric function

$$\tau (m,n)=\sum _{1\le k_1<k_2<\cdots <k_m\le n}\prod _{i=1}^m\frac{1}{2k_i-1}=\left[x^m\right]\frac{{\left(\frac{1+x}{2}\right)}_n}{{\left(\frac{1}{2}\right)}_n},$$

where the rightmost expression is justified by

$$\tau (m,n)=\left[x^m\right]\prod _{k=1}^n\left(1+\frac{x}{2k-1}\right)=\left[x^m\right]\prod _{k=1}^n\frac{x+2k-1}{2k-1}=\left[x^m\right]\frac{{\left(\frac{1+x}{2}\right)}_n}{{\left(\frac{1}{2}\right)}_n}.$$

By extracting the coefficient of $x^{m+1}$ across the equality

$$\frac{{\left(\frac{1+x}{2}\right)}_{n+1}}{{\left(\frac{1}{2}\right)}_{n+1}}=\frac{{\left(\frac{1+x}{2}\right)}_n}{{\left(\frac{1}{2}\right)}_n}+\frac{x}{2n+1}\frac{{\left(\frac{1+x}{2}\right)}_n}{{\left(\frac{1}{2}\right)}_n},$$

we deduce the following recurrence relation

$$\tau (m+1,n+1)=\tau (m+1,n)+\frac{\tau (m,n)}{2n+1}.$$

This sequence can explicitly be expressed in terms of skew harmonic numbers:

$$\begin{array}{cc}\hfill \tau (0,n)& =1,\hfill \\ \hfill \tau (1,n)& ={\mathbf{O}}_n,\hfill \\ \hfill \tau (2,n)& =\frac{1}{2}\left\{{\mathbf{O}}_n^2-{\mathbf{O}}_n^{\langle 2\rangle }\right\},\hfill \\ \hfill \tau (3,n)& =\frac{1}{6}\left\{{\mathbf{O}}_n^3-3{\mathbf{O}}_n{\mathbf{O}}_n^{\langle 2\rangle }+2{\mathbf{O}}_n^{\langle 3\rangle }\right\}.\hfill \end{array}$$

Two further useful properties are highlighted in the lemma below.

Lemma 2

(Properties of $\tau (m,n)$: $\left|x\right|<1$ and $\left|y\right|<1$).

$$\begin{array}{cccc}& \mathrm{(a)}Generatingfunction:\hfill & & \sum _{n=0}^{\infty }{\left(\frac{y}{4}\right)}^n\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\tau (m,n)=\frac{{ln}^m(1-y)}{m!\sqrt{1-y}}{\left(\frac{-1}{2}\right)}^m.\hfill \\ & \mathrm{(b)}Finite\mathrm{sum}umidentity:\hfill & & \sum _{k=0}^j\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}=2-2\left[x^m\right]\frac{\left(\genfrac{}{}{0pt}{}{\frac{1+x}{2}+j}{1+j}\right)}{1-x}.\hfill \end{array}$$

These relations will be utilized in this section to examine four classes of infinite series containing zeta and lambda tails weighted by the factor $\tau (m,n)$. Four general theorems will be proven and several concrete summation formulae will be deduced as consequences.

Proof. 

According to the equality

$$\sum _{m\ge 0}\tau (m,n)x^m=\frac{{\left(\frac{1+x}{2}\right)}_n}{{\left(\frac{1}{2}\right)}_n},$$

we consider the bivariate generating function

$$\sum _{m,n\ge 0}{\left(\frac{y}{4}\right)}^n\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\tau (m,n)x^m=\sum _{n\ge 0}\left(\genfrac{}{}{0pt}{}{\frac{1+x}{-2}}{n}\right){(-y)}^n={(1-y)}^{-\frac{1+x}{2}}=\frac{exp(-\frac{x}{2}ln(1-y))}{\sqrt{1-y}}.$$

By extracting the coefficient $x^m$ across, we deduce the first formula (a).

The identity (b) can be validated by telescoping as follows:

$$\begin{array}{cc}& \sum _{k=0}^j\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}=\left[x^m\right]\sum _{k=0}^j\frac{{\left(\frac{1+x}{2}\right)}_k}{(k+1)!}\hfill \\ \hfill =& \left[x^m\right]\sum _{k=0}^j\frac{2}{1-x}\left\{\frac{{\left(\frac{1+x}{2}\right)}_k}{k!}-\frac{{\left(\frac{1+x}{2}\right)}_{k+1}}{(k+1)!}\right\}\hfill \\ \hfill =& \left[x^m\right]\frac{2}{1-x}\left\{1-\frac{{\left(\frac{1+x}{2}\right)}_{j+1}}{(j+1)!}\right\}=2-2\left[x^m\right]\frac{\left(\genfrac{}{}{0pt}{}{\frac{1+x}{2}+j}{1+j}\right)}{1-x}.\hfill \end{array}$$

This completes the proof of Lemma 2.    □

3.1. Series ${\mathrm{V}}_1(m,n)$ with Zeta Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with zeta tails

$$\begin{array}{cc}\hfill {\mathrm{V}}_1(m,n)& :=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k}\left\{\zeta (n+3)-{\mathbf{H}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^n}{(n+2)!}{\int }_0^1\frac{{ln}^{n+2}(1-x)}{x}dx\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k}{(1-x)}^k\hfill \\ & =\frac{{(-1)}^{m+n}}{m!(n+2)!2^m}{\mathcal{V}}_1(m,n),\hfill \end{array}$$

where in view of Lemma 2(a), ${\mathcal{V}}_1(m,n)$ is the definite integral given by

$${\mathcal{V}}_1(m,n)={\int }_0^1\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{\sqrt{x^3}}dx.$$

In order to evaluate ${\mathcal{V}}_1(m,n)$, we examine the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{V}}_1(m,n)={\int }_0^1{x}^{u-\frac{3}{2}}{(1-x)}^v{ln}^2(1-x)dx\hfill \\ & =\frac{{\partial }^2}{\partial v^2}{\int }_0^1{x}^{u-\frac{3}{2}}{(1-x)}^{v}dx=\frac{{\partial }^2}{\partial v^2}\mathrm{Beta}(u-\frac{1}{2},v+1).\hfill \end{array}$$

By extracting the coefficient of $u^m{v}^n$, we have

$$\begin{array}{cc}\hfill {\mathcal{V}}_1(m,n)& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial v^2}\mathrm{Beta}(u-\frac{1}{2},v+1)\hfill \\ & =m!(n+2)!\left[u^m{v}^{n+2}\right]\mathrm{Beta}(u-\frac{1}{2},v+1),\hfill \end{array}$$

which results in the following general summation formula.

Theorem 5.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{V}}_1(m,n)& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k}\left\{\zeta (n+3)-{\mathbf{H}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{m+n}}{2^m}\left[u^m{v}^{n+2}\right]\mathrm{Beta}(u-\frac{1}{2},v+1).\hfill \end{array}$$

By expressing the above beta function in terms of $\Gamma$-function quotient

$$\mathrm{Beta}\left(u-\frac{1}{2},v+1\right)=\frac{\Gamma (\frac{1}{2}+u)\Gamma (1+v)}{(u-\frac{1}{2})\Gamma (\frac{1}{2}+u+v)}$$

and then applying (2), we can expand the above beta function into power series in u and v. By extracting coefficients of monomials “$u^m{v}^{n+2}$”, we can derive numerous infinite series identities by running the following Mathematica commands:

v1x[m_,n_]:=(-1)^(m+n)/2^m*SeriesCoefficient[Beta[u-1/2,1+v],{u,0,m},{v,0,n+2}]

v1y[m_,n_]:=FullSimplify[v1x[m,n]/.PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]]

Some initial ones are displayed in the following corollary.

Corollary 5

(Particular values of ${\mathrm{V}}_1(m,n)$).

$$\begin{array}{cc}\hfill {\mathrm{V}}_1(0,0)& =\frac{{\pi }^2}{3}-4{ln}^{2}2,\hfill \\ \hfill {\mathrm{V}}_1(0,1)& =\frac{8}{3}{ln}^{3}2-\frac{2{\pi }^2}{3}ln2+4\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_1(0,2)& =\frac{{\pi }^4}{20}+\frac{2{\pi }^2}{3}{ln}^{2}2-\frac{4}{3}{ln}^{4}2-8\zeta \left(3\right)ln2,\hfill \\ \hfill {\mathrm{V}}_1(1,0)& =4{ln}^{2}2-\frac{{\pi }^2}{3}-{\pi }^{2}ln2+7\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_1(1,1)& =\frac{{\pi }^4}{12}+\frac{2{\pi }^2}{3}ln2+{\pi }^2{ln}^{2}2-\frac{8}{3}{ln}^{3}2-4\zeta \left(3\right)-14\zeta \left(3\right)ln2,\hfill \\ \hfill {\mathrm{V}}_1(1,2)& =\frac{4}{3}{ln}^{4}2-\frac{{\pi }^4}{20}-\frac{{\pi }^4}{6}ln2-\frac{2{\pi }^2}{3}{ln}^{2}2-\frac{2{\pi }^2}{3}{ln}^{3}2\hfill \\ & -\frac{13{\pi }^2}{6}\zeta \left(3\right)+8\zeta \left(3\right)ln2+14\zeta \left(3\right){ln}^{2}2+31\zeta \left(5\right),\hfill \\ \hfill {\mathrm{V}}_1(2,0)& =\frac{{\pi }^2}{3}+\frac{{\pi }^4}{16}+{\pi }^{2}ln2-4{ln}^{2}2-7\zeta \left(3\right)-7ln2\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_1(2,1)& =\frac{8{ln}^{3}2}{3}-\frac{{\pi }^4}{12}-\frac{{\pi }^4}{8}ln2-\frac{2{\pi }^2}{3}ln2-\frac{7{\pi }^2}{3}\zeta \left(3\right)-{\pi }^2{ln}^{2}2\hfill \\ & +4\zeta \left(3\right)+14ln2\zeta \left(3\right)+7{ln}^{2}2\zeta \left(3\right)+31\zeta \left(5\right),\hfill \\ \hfill {\mathrm{V}}_1(2,2)& =\frac{{\pi }^6}{32}+\frac{{\pi }^4}{20}+\frac{{\pi }^4}{6}ln2+\frac{{\pi }^4}{8}{ln}^{2}2+\frac{2{\pi }^2}{3}{ln}^{2}2+\frac{2{\pi }^2}{3}{ln}^{3}2+\frac{13{\pi }^2}{6}\zeta \left(3\right)-\frac{14{ln}^{3}2}{3}\zeta \left(3\right)\hfill \\ & +\frac{14{\pi }^{2}ln2}{3}\zeta \left(3\right)-\frac{77}{4}{\zeta }^2\left(3\right)-\frac{4{ln}^{4}2}{3}-8ln2\zeta \left(3\right)-14{ln}^{2}2\zeta \left(3\right)-62ln2\zeta \left(5\right)-31\zeta \left(5\right).\hfill \end{array}$$

3.2. Series ${\mathrm{V}}_2(m,n)$ with Lambda Tails

For $m,n\in {\mathbb{N}}_0$, we consider the next infinite series that contains Dirichlet’s lambda tails in the summand

$$\begin{array}{cc}\hfill {\mathrm{V}}_2(m,n)& :=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k}\left\{\lambda (n+3)-{\mathbf{O}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^n}{(n+2)!}{\int }_0^1\frac{{ln}^{n+2}x}{1-x^2}dx\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k}{x}^{2k}\hfill \\ & =\frac{{(-1)}^{m+n}}{m!(n+2)!2^m}{\mathcal{V}}_2(m,n),\hfill \end{array}$$

where according to Lemma 2(a), ${\mathcal{V}}_2(m,n)$ is the definite integral given by

$${\mathcal{V}}_2(m,n)={\int }_0^1\frac{{ln}^m(1-x^2){ln}^{n+2}x}{{(1-x^2)}^{\frac{3}{2}}}dx.$$

By examining the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{V}}_2(m,n)={\int }_0^1{(1-x^2)}^{u-\frac{3}{2}}{x}^v{ln}^{2}xdx\hfill \\ & =\frac{{\partial }^2}{\partial v^2}{\int }_0^1{(1-x^2)}^{u-\frac{3}{2}}{x}^{v}dx=\frac{{\partial }^2}{\partial v^2}\frac{\mathrm{Beta}(u-\frac{1}{2},\frac{v+1}{2})}{2},\hfill \end{array}$$

and then, by extracting the coefficient of $u^m{v}^n$

$$\begin{array}{cc}\hfill {\mathcal{V}}_2(m,n)& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial v^2}\frac{\mathrm{Beta}(u-\frac{1}{2},\frac{v+1}{2})}{2}\hfill \\ & =m!(n+2)!\left[u^m{v}^{n+2}\right]\frac{\mathrm{Beta}(u-\frac{1}{2},\frac{v+1}{2})}{2},\hfill \end{array}$$

we establish the compact formula as in the following theorem.

Theorem 6.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{V}}_2(m,n)& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k}\left\{\lambda (n+3)-{\mathbf{O}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{m+n}}{2^{m+1}}\left[u^m{v}^{n+2}\right]\mathrm{Beta}\left(u-\frac{1}{2},\frac{v+1}{2}\right).\hfill \end{array}$$

By executing the following Mathematica commands:

v2x[m_,n_]:=(-1)^(m+n)/2^(m+1)*SeriesCoefficient[Beta[u-1/2,(1+v)/2],{u,0,m},{v,0,n+2}]

v2y[m_,n_]:=FullSimplify[v2x[m,n]/.PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]]

we establish the identities below as examples.

Corollary 6

(Particular values of ${\mathrm{V}}_2(m,n)$).

$$\begin{array}{cc}\hfill {\mathrm{V}}_2(0,0)& =\frac{\pi }{2}ln2,\hfill \\ \hfill {\mathrm{V}}_2(0,1)& =\frac{{\pi }^3}{48}+\frac{\pi }{4}{ln}^{2}2,\hfill \\ \hfill {\mathrm{V}}_2(0,2)& =\frac{{\pi }^3}{48}ln2+\frac{\pi }{12}{ln}^{3}2+\frac{\pi }{8}\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_2(1,0)& =\frac{3\pi }{4}{ln}^{2}2-\frac{\pi }{2}ln2,\hfill \\ \hfill {\mathrm{V}}_2(1,1)& =\frac{{\pi }^3}{48}ln2-\frac{{\pi }^3}{48}-\frac{\pi }{4}{ln}^{2}2+\frac{\pi }{3}{ln}^{3}2+\frac{\pi }{16}\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_2(1,2)& =\frac{{\pi }^5}{2304}-\frac{{\pi }^3}{48}ln2+\frac{{\pi }^3}{48}{ln}^{2}2-\frac{\pi }{12}{ln}^{3}2+\frac{5\pi }{48}{ln}^{4}2-\frac{\pi }{8}\zeta \left(3\right)+\frac{3\pi }{16}\zeta \left(3\right)ln2,\hfill \\ \hfill {\mathrm{V}}_2(2,0)& =\frac{\pi }{2}ln2-\frac{3\pi }{4}{ln}^{2}2+\frac{\pi }{2}{ln}^{3}2-\frac{\pi }{8}\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_2(2,1)& =\frac{{\pi }^3}{48}-\frac{{\pi }^5}{2304}-\frac{{\pi }^3}{48}ln2+\frac{\pi }{4}{ln}^{2}2+\frac{{\pi }^3}{96}{ln}^{2}2-\frac{\pi }{3}{ln}^{3}2+\frac{5\pi }{24}{ln}^{4}2-\frac{\pi \zeta \left(3\right)}{16}-\frac{\pi ln2}{16}\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_2(2,2)& =\frac{{\pi }^3}{48}ln2-\frac{{\pi }^3}{48}{ln}^{2}2+\frac{{\pi }^3}{96}{ln}^{3}2-\frac{{\pi }^5}{2304}+\frac{\pi }{12}{ln}^{3}2-\frac{5\pi }{48}{ln}^{4}2+\frac{\pi }{16}{ln}^{5}2\hfill \\ & +\frac{\pi }{8}\zeta \left(3\right)-\frac{{\pi }^3}{384}\zeta \left(3\right)-\frac{3\pi ln2}{16}\zeta \left(3\right)+\frac{\pi {ln}^{2}2}{16}\zeta \left(3\right)-\frac{3\pi }{64}\zeta \left(5\right).\hfill \end{array}$$

3.3. Series ${\mathrm{V}}_3(m,n)$ with Zeta Tails

For $m,n\in {\mathbb{N}}_0$, define the infinite series with zeta tails

$${\mathrm{V}}_3(m,n):=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}\left\{\zeta (n+2)-{\mathbf{H}}_k^{\langle n+2\rangle }\right\}.$$

Since the integration method fails, we have to deal with this series by the series rearrangement:

$$\begin{array}{cc}\hfill {\mathrm{V}}_3(m,n)& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}\sum _{j\ge k}\frac{1}{{(j+1)}^{n+2}}\hfill \\ & =\sum _{j=0}^{\infty }\frac{1}{{(j+1)}^{n+2}}\sum _{k=0}^j\frac{{\left(\frac{1}{2}\right)}_k}{(k+1)!}\tau (m,k).\hfill \end{array}$$

Evaluating the rightmost sum by Lemma 2(b), we find the expression below:

$${\mathrm{V}}_3(m,n)=2\zeta (n+2)-2\sum _{j=0}^{\infty }\frac{\left[x^m\right]}{{(j+1)}^{n+2}}\frac{\left(\genfrac{}{}{0pt}{}{\frac{1+x}{2}+j}{1+j}\right)}{1-x}.$$

Then, by making use of (5), we can determine also the generating function

$$\begin{array}{cc}\hfill \sum _{n=-1}^{\infty }{y}^{n+2}\sum _{j=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{\frac{1+x}{2}+j}{1+j}\right)}{{(j+1)}^{n+2}}& =\sum _{j=0}^{\infty }{(-1)}^{j+1}\frac{\left(\genfrac{}{}{0pt}{}{-\frac{1+x}{2}}{1+j}\right)y}{1+j-y}\begin{array}{|c|}\hline j+1\to k\\ \hline\end{array}\hfill \\ & =1+\sum _{k=0}^{\infty }{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{-\frac{1+x}{2}}{k}\right)y}{k-y}\hfill \\ & =1+y\mathrm{Beta}\left(\frac{1-x}{2},-y\right).\hfill \end{array}$$

Consequently, we find the formula as in the following theorem.

Theorem 7.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$${\mathrm{V}}_3(m,n)=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}\left\{\zeta (n+2)-{\mathbf{H}}_k^{\langle n+2\rangle }\right\}=2\zeta (n+2)-2\left[x^m{y}^{n+1}\right]\frac{\mathrm{Beta}(\frac{1-x}{2},-y)}{1-x}.$$

By carrying out the following Mathematica commands:

v3x[m_,n_]:=SeriesCoefficient[2/(1-x) Beta[(1-x)/2,-y],{x,0,m},{y,0,n+1}]

v3y[m_,n_]:=FullSimplify[2Zeta[n+2]-v3x[m,n]]

we record the initial formulae in the corollary below.

Corollary 7

(Particular values of ${\mathrm{V}}_3(m,n)$).

$$\begin{array}{cc}\hfill {\mathrm{V}}_3(0,0)& =4{ln}^{2}2\hfill \\ \hfill {\mathrm{V}}_3(0,1)& =\frac{2{\pi }^2}{3}ln2-\frac{8{ln}^{3}2}{3}-2\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_3(0,2)& =8\zeta \left(3\right)ln2+\frac{4{ln}^{4}2}{3}-\frac{2{\pi }^2}{3}{ln}^{2}2-\frac{{\pi }^4}{36},\hfill \\ \hfill {\mathrm{V}}_3(0,3)& =\frac{2{\pi }^2}{3}\zeta \left(3\right)+\frac{4{\pi }^2}{9}{ln}^{3}2+\frac{{\pi }^4}{10}ln2-\frac{8{ln}^{5}2}{15}-8\zeta \left(3\right){ln}^{2}2-10\zeta \left(5\right),\hfill \\ \hfill {\mathrm{V}}_3(1,0)& ={\pi }^{2}ln2+4{ln}^{2}2-7\zeta \left(3\right),\hfill \\ \hfill {\mathrm{V}}_3(1,1)& =14ln2\zeta \left(3\right)-2\zeta \left(3\right)-{\pi }^2{ln}^{2}2-\frac{{\pi }^4}{12}+\frac{2{\pi }^2}{3}ln2-\frac{8{ln}^{3}2}{3},\hfill \\ \hfill {\mathrm{V}}_3(1,2)& =8ln2\zeta \left(3\right)-14{ln}^{2}2\zeta \left(3\right)-31\zeta \left(5\right)-\frac{{\pi }^4}{36}+\frac{{\pi }^4}{6}ln2\hfill \\ & +\frac{2{\pi }^2}{3}{ln}^{3}2-\frac{2{\pi }^2}{3}{ln}^{2}2+\frac{13{\pi }^2}{6}\zeta \left(3\right)+\frac{4{ln}^{4}2}{3},\hfill \\ \hfill {\mathrm{V}}_3(2,0)& =7ln2\zeta \left(3\right)-7\zeta \left(3\right)+4{ln}^{2}2+{\pi }^{2}ln2-\frac{{\pi }^4}{16},\hfill \\ \hfill {\mathrm{V}}_3(2,1)& =14\zeta \left(3\right)ln2-2\zeta \left(3\right)-7{ln}^{2}2\zeta \left(3\right)-31\zeta \left(5\right)-{\pi }^2{ln}^{2}2\hfill \\ & +\frac{7{\pi }^2}{3}\zeta \left(3\right)-\frac{{\pi }^4}{12}+\frac{{\pi }^4}{8}ln2+\frac{2{\pi }^2}{3}ln2-\frac{8{ln}^{3}2}{3},\hfill \\ \hfill {\mathrm{V}}_3(3,0)& =\frac{7{\pi }^2\zeta \left(3\right)}{8}-\frac{31\zeta \left(5\right)}{2}-\frac{{\pi }^4}{16}+\frac{{\pi }^4}{12}ln2+7\zeta \left(3\right)ln2-7\zeta \left(3\right)+4{ln}^{2}2+{\pi }^{2}ln2,\hfill \\ \hfill {\mathrm{V}}_3(2,2)& =\frac{13{\pi }^2}{6}\zeta \left(3\right)+\frac{77}{4}{\zeta }^2\left(3\right)+\frac{14{ln}^{3}2}{3}\zeta \left(3\right)-\frac{14{\pi }^{2}ln2}{3}\zeta \left(3\right)-14{ln}^{2}2\zeta \left(3\right)+8ln2\zeta \left(3\right)\hfill \\ & -31\zeta \left(5\right)+62ln2\zeta \left(5\right)-\frac{{\pi }^6}{32}-\frac{{\pi }^4}{36}-\frac{{\pi }^4}{8}{ln}^{2}2+\frac{{\pi }^4}{6}ln2-\frac{2{\pi }^2}{3}{ln}^{2}2+\frac{2{\pi }^2}{3}{ln}^{3}2+\frac{4{ln}^{4}2}{3}.\hfill \end{array}$$

3.4. Series ${\mathrm{V}}_4(m,n)$ with Lambda Tails

For $m,n\in {\mathbb{N}}_0$, define the infinite series with lambda tails

$${\mathrm{V}}_4(m,n):=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}\left\{\lambda (n+2)-{\mathbf{O}}_k^{\langle n+2\rangle }\right\}.$$

By means of the series rearrangement, we have

$$\begin{array}{cc}\hfill {\mathrm{V}}_4(m,n)& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}\sum _{j\ge k}\frac{1}{{(2j+1)}^{n+2}}\hfill \\ & =\sum _{j=0}^{\infty }\frac{1}{{(2j+1)}^{n+2}}\sum _{k=0}^j\frac{{\left(\frac{1}{2}\right)}_k}{(k+1)!}\tau (m,k)\hfill \\ & =2\lambda (n+2)-2\sum _{j=0}^{\infty }\frac{\left[x^m\right]}{{(2j+1)}^{n+2}}\frac{\left(\genfrac{}{}{0pt}{}{\frac{1+x}{2}+j}{j+1}\right)}{1-x},\hfill \end{array}$$

where the last line is justified by Lemma 2(b). Keeping in mind (5), we can analogously determine the generating function

$$\begin{array}{cc}\hfill \sum _{n=-1}^{\infty }{y}^{n+2}\sum _{j=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{\frac{1+x}{2}+j}{1+j}\right)}{{(2j+1)}^{n+2}}& =\sum _{j=0}^{\infty }{(-1)}^{j+1}\frac{\left(\genfrac{}{}{0pt}{}{-\frac{1+x}{2}}{1+j}\right)y}{1+2j-y}\begin{array}{|c|}\hline j+1\to k\\ \hline\end{array}\hfill \\ & =\frac{y}{1+y}+\sum _{k=0}^{\infty }{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{-\frac{1+x}{2}}{k}\right)y}{2k-(1+y)}\hfill \\ & =\frac{y}{1+y}+\frac{y}{2}\mathrm{Beta}\left(\frac{1-x}{2},-\frac{1+y}{2}\right).\hfill \end{array}$$

Consequently, we established the formula as in the following theorem.

Theorem 8.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{V}}_4(m,n)& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\frac{\tau (m,k)}{4^k(k+1)}\left\{\lambda (n+2)-{\mathbf{O}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =2\lambda (n+2)+2{(-1)}^n-\left[x^m{y}^{n+1}\right]\frac{\mathrm{Beta}\left(\frac{1-x}{2},-\frac{1+y}{2}\right)}{1-x}.\hfill \end{array}$$

By fulfilling the following Mathematica commands:

v4x[m_,n_]:=SeriesCoefficient[1/(1-x) Beta[(1-x)/2,-((1+y)/2)],{x,0,m},{y,0,n+1}]

v4y[m_,n_]:=FullSimplify[2(-1)^n+2DirichletLambda[n+2]-v4x[m,n]]

we highlight a few summation formulae below as consequences.

Corollary 8

(Particular values of ${\mathrm{V}}_4(m,n)$).

$$\begin{array}{cc}\hfill {\mathrm{V}}_4(0,0)& =2-\pi +\frac{{\pi }^2}{4},\hfill \\ \hfill {\mathrm{V}}_4(0,1)& =\pi -2-\pi ln2+\frac{7\zeta \left(3\right)}{4},\hfill \\ \hfill {\mathrm{V}}_4(0,2)& =2-\pi +\pi ln2-\frac{{\pi }^3}{24}+\frac{{\pi }^4}{48}-\frac{\pi }{2}{ln}^{2}2,\hfill \\ \hfill {\mathrm{V}}_4(0,3)& =\pi -2-\pi ln2+\frac{{\pi }^3}{24}-\frac{{\pi }^3}{24}ln2+\frac{\pi }{2}{ln}^{2}2-\frac{\pi }{6}{ln}^{3}2-\frac{\pi \zeta \left(3\right)}{4}+\frac{31\zeta \left(5\right)}{16},\hfill \\ \hfill {\mathrm{V}}_4(1,0)& =2-2\pi ln2+\frac{{\pi }^2}{4},\hfill \\ \hfill {\mathrm{V}}_4(1,1)& =\pi ln2-2-\frac{3\pi }{2}{ln}^{2}2+\frac{7\zeta \left(3\right)}{4},\hfill \\ \hfill {\mathrm{V}}_4(1,2)& =2-\pi ln2+\pi {ln}^{2}2-\frac{{\pi }^3}{24}+\frac{{\pi }^4}{48}-\frac{{\pi }^3}{24}ln2-\frac{2\pi }{3}{ln}^{3}2-\frac{\pi \zeta \left(3\right)}{8},\hfill \\ \hfill {\mathrm{V}}_4(2,0)& =2+\frac{{\pi }^2}{4}-\frac{3}{2}\pi {ln}^{2}2-\pi ln2\hfill \\ \hfill {\mathrm{V}}_4(2,1)& =\frac{7\zeta \left(3\right)}{4}+\frac{\pi \zeta \left(3\right)}{4}-2-\pi {ln}^{3}2\hfill \\ \hfill {\mathrm{V}}_4(3,0)& =2-\pi ln2-\pi {ln}^{2}2-\frac{\pi }{8}\zeta \left(3\right)-\frac{2\pi }{3}{ln}^{3}2+\frac{{\pi }^2}{4}+\frac{{\pi }^3}{24}-\frac{{\pi }^3}{24}ln2\hfill \\ \hfill {\mathrm{V}}_4(2,2)& =2+\frac{\pi ln2}{8}\zeta \left(3\right)-\frac{3\pi }{8}\zeta \left(3\right)-\frac{{\pi }^3}{24}+\frac{{\pi }^4}{48}+\frac{{\pi }^5}{1152}\hfill \\ & -\frac{5\pi }{12}{ln}^{4}2+\frac{\pi }{3}{ln}^{3}2-\frac{\pi }{2}{ln}^{2}2-\frac{{\pi }^3}{48}{ln}^{2}2-\frac{{\pi }^3}{24}ln2.\hfill \end{array}$$

4. Four Series with Weight Factor $\mathit{\rho }(\mathit{m},\mathit{n};\mathit{\alpha })$

In this section, we are going to investigate four classes of infinite series of zeta and lambda tails weighted by the function $\rho (m,n;\alpha )$ as below. For each class of the series, a very general summation theorem will be proved and four specific identities will be collected (due to complexities).

For a real number $\alpha$ subject to $0<\alpha \le 1$, define the elementary symmetric function

$$\rho (m,n;\alpha )=\sum _{0\le k_1<k_2<\cdots <k_m<n}\prod _{i=1}^m\frac{1}{\alpha +k_i}=\left[x^m\right]\frac{{(\alpha +x)}_n}{{\left(\alpha \right)}_n},$$

where the rightmost expression is shown as follows:

$$\rho (m,n;\alpha )=\left[x^m\right]\prod _{k=0}^{n-1}\left(1+\frac{x}{\alpha +k}\right)=\left[x^m\right]\prod _{k=0}^{n-1}\frac{x+\alpha +k}{\alpha +k}=\left[x^m\right]\frac{{(x+\alpha )}_n}{{\left(\alpha \right)}_n}.$$

By means of the equality

$$\frac{{(x+\alpha )}_{n+1}}{{\left(\alpha \right)}_{n+1}}=\frac{{(x+\alpha )}_n}{{\left(\alpha \right)}_n}+\frac{x}{\alpha +n}\frac{{(x+\alpha )}_n}{{\left(\alpha \right)}_n},$$

we can check the following recurrence relation

$$\rho (m+1,n+1;\alpha )=\rho (m+1,n;\alpha )+\frac{\rho (m,n;\alpha )}{\alpha +n}.$$

This sequence can be expressed in terms of generalized harmonic numbers:

$$\begin{array}{cc}\hfill \rho (0,n;\alpha )& =1,\hfill \\ \hfill \rho (1,n;\alpha )& ={\mathbf{H}}_n\left(\alpha \right),\hfill \\ \hfill \rho (2,n;\alpha )& =\frac{1}{2}\left\{{\mathbf{H}}_n^2\left(\alpha \right)-{\mathbf{H}}_n^{\langle 2\rangle }\left(\alpha \right)\right\},\hfill \\ \hfill \rho (3,n;\alpha )& =\frac{1}{6}\left\{{\mathbf{H}}_n^3\left(\alpha \right)-3{\mathbf{H}}_n\left(\alpha \right){\mathbf{H}}_n^{\langle 2\rangle }\left(\alpha \right)+2{\mathbf{H}}_n^{\langle 3\rangle }\left(\alpha \right)\right\}.\hfill \end{array}$$

The following lemma will play a crucial role subsequently in this section.

Lemma 3

(Properties of $\rho (m,n;\alpha )$: $\left|x\right|<1$ and $\left|y\right|<1$).

$$\begin{array}{cccc}& (a)Generatingfunction:\hfill & & \sum _{n=0}^{\infty }{(-y)}^n\left(\genfrac{}{}{0pt}{}{-\alpha }{n}\right)\rho (m,n;\alpha )={(-1)}^m\frac{{ln}^m(1-y)}{m!{(1-y)}^{\alpha }}.\hfill \\ & (b)Finitesumidentity:\hfill & & \sum _{k=0}^j\frac{{\left(\alpha \right)}_k}{(k+1)!}\rho (m,k;\alpha )=\frac{\left[x^m\right]}{1-x-\alpha }\left\{1-\frac{{(\alpha +x)}_{j+1}}{(j+1)!}\right\}.\hfill \end{array}$$

Proof. 

Analogously to the proofs of Lemmas 1 and 2, the first formula (a) follows by extracting the coefficient $x^m$ across the bivariate generating function

$$\sum _{m,n\ge 0}{(-y)}^n\left(\genfrac{}{}{0pt}{}{-\alpha }{n}\right)\rho (m,n;\alpha )x^m=\sum _{n\ge 0}\left(\genfrac{}{}{0pt}{}{-\alpha -x}{n}\right){(-y)}^n={(1-y)}^{-\alpha -x}.$$

By telescoping, we can manipulate the sum in (b) as follows:

$$\begin{array}{cc}& \sum _{k=0}^j\frac{{\left(\alpha \right)}_k}{(k+1)!}\rho (m,k;\alpha )=\sum _{k=0}^j\frac{{\left(\alpha \right)}_k}{(k+1)!}\left[x^m\right]\frac{{(\alpha +x)}_k}{{\left(\alpha \right)}_k}\hfill \\ \hfill =& \left[x^m\right]\sum _{k=0}^j\frac{1}{1-\alpha -x}\left\{\frac{{(\alpha +x)}_k}{k!}-\frac{{(\alpha +x)}_{k+1}}{(k+1)!}\right\}\hfill \\ \hfill =& \left[x^m\right]\frac{1}{1-\alpha -x}\left\{1-\frac{{(\alpha +x)}_{j+1}}{(j+1)!}\right\},\hfill \end{array}$$

which confirms the second formula (b) displayed in Lemma 3.    □

4.1. Series ${\mathrm{W}}_1(m,n;\alpha )$ with Zeta Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with zeta tails

$$\begin{array}{cc}\hfill {\mathrm{W}}_1(m,n;\alpha )& :=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\rho (m,k;\alpha )\left\{\zeta (n+3)-{\mathbf{H}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^n}{(n+2)!}{\int }_0^1\frac{{ln}^{n+2}(1-x)}{x}dx\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\rho (m,k;\alpha ){(1-x)}^k\hfill \\ & =\frac{{(-1)}^{m+n}}{m!(n+2)!}{\mathcal{W}}_1(m,n;\alpha ),\hfill \end{array}$$

where in view of Lemma 3(a), ${\mathcal{W}}_1(m,n;\alpha )$ is the definite integral given by

$${\mathcal{W}}_1(m,n;\alpha )={\int }_0^1\frac{{ln}^{m}x{ln}^{n+2}(1-x)}{x^{1+\alpha }}dx.$$

In order to evaluate the above integral, we examine the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{W}}_1(m,n;\alpha )={\int }_0^1{x}^{u-1-\alpha }{(1-x)}^v{ln}^2(1-x)dx\hfill \\ & =\frac{{\partial }^2}{\partial v^2}{\int }_0^1{x}^{u-1-\alpha }{(1-x)}^{v}dx=\frac{{\partial }^2}{\partial v^2}\mathrm{Beta}(u-\alpha ,v+1).\hfill \end{array}$$

By extracting the coefficient of $u^m{v}^n$, we have

$$\begin{array}{cc}\hfill {\mathcal{W}}_1(m,n;\alpha )& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial v^2}\mathrm{Beta}(u-\alpha ,v+1)\hfill \\ & =m!(n+2)!\left[u^m{v}^{n+2}\right]\mathrm{Beta}(u-\alpha ,v+1),\hfill \end{array}$$

which yields the compact formula as in the following theorem.

Theorem 9.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{W}}_1(m,n;\alpha )& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\rho (m,k;\alpha )\left\{\zeta (n+3)-{\mathbf{H}}_k^{\langle n+3\rangle }\right\}\hfill \\ & ={(-1)}^{m+n}\left[u^m{v}^{n+2}\right]\mathrm{Beta}(u-\alpha ,v+1).\hfill \end{array}$$

By running the following Mathematica commands:

w1x[a_,m_,n_]:=(-1)^(m+n)*SeriesCoefficient[Beta[x-a,1+y],{x,0,m},{y,0,n+2}]

w1y[a_,m_,n_]:=FunctionExpand[w1x[a,m,n]]/.{PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]}

we record four summation formulae as examples in the following corollary.

Corollary 9

(Particular values of ${\mathrm{W}}_1(m,n;\alpha )$).

$$\begin{array}{cc}\hfill {\mathrm{W}}_1\left(1,0;\frac{1}{3}\right)& =\frac{1}{2}{\zeta }_2\left(\frac{2}{3}\right)\left(\pi \sqrt{3}-9-9ln3\right)\hfill \\ & +\frac{1}{24}\left\{936\zeta \left(3\right)+27{\pi }^2-16\sqrt{3}{\pi }^3+243{ln}^{2}3-54\sqrt{3}\pi ln3\right\},\hfill \\ \hfill {\mathrm{W}}_1\left(1,0;\frac{2}{3}\right)& =-\frac{1}{8}{\zeta }_2\left(\frac{1}{3}\right)\left(9+2\sqrt{3}\pi +18ln3\right)\hfill \\ & +\frac{1}{96}\left\{1872\zeta \left(3\right)+27{\pi }^2+32\sqrt{3}{\pi }^3+243{ln}^{2}3+54\sqrt{3}\pi ln3\right\},\hfill \\ \hfill {\mathrm{W}}_1\left(0,1;\frac{1}{4}\right)& =36\zeta \left(3\right)-8G(\pi -6ln2)-\frac{1}{12}\left\{7{\pi }^3+42{\pi }^{2}ln2+108\pi {ln}^{2}2-216{ln}^{3}2\right\},\hfill \\ \hfill {\mathrm{W}}_1\left(0,1;\frac{3}{4}\right)& =12\zeta \left(3\right)-\frac{8G}{3}(\pi +6ln2)+\frac{1}{36}\left\{7{\pi }^3-42{\pi }^{2}ln2+108\pi {ln}^{2}2+216{ln}^{3}2\right\}.\hfill \end{array}$$

4.2. Series ${\mathrm{W}}_2(m,n;\alpha )$ with Lambda Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with lambda tails

$$\begin{array}{cc}\hfill {\mathrm{W}}_2(m,n;\alpha )& :=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\rho (m,k;\alpha )\left\{\lambda (n+3)-{\mathbf{O}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^n}{(n+2)!}{\int }_0^1\frac{{ln}^{n+2}x}{1-x^2}dx\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\rho (m,k;\alpha )x^{2k}\hfill \\ & =\frac{{(-1)}^{m+n}}{m!(n+2)!}{\mathcal{W}}_2(m,n;\alpha ),\hfill \end{array}$$

where according to Lemma 3(a), ${\mathcal{W}}_2(m,n;\alpha )$ reads as

$${\mathcal{W}}_2(m,n;\alpha )={\int }_0^1\frac{{ln}^m(1-x^2){ln}^{n+2}x}{{(1-x^2)}^{1+\alpha }}dx.$$

By examining the exponential generating function

$$\begin{array}{cc}& \sum _{m,n=0}^{\infty }\frac{u^m{v}^n}{m!n!}{\mathcal{W}}_2(m,n;\alpha )={\int }_0^1{x}^v{(1-x^2)}^{u-1-\alpha }{ln}^{2}xdx\hfill \\ & =\frac{{\partial }^2}{\partial v^2}{\int }_0^1{x}^v{(1-x^2)}^{u-1-\alpha }dx=\frac{{\partial }^2}{\partial v^2}\frac{\mathrm{Beta}(u-\alpha ,\frac{v+1}{2})}{2}\hfill \end{array}$$

and then, by extracting the coefficient of $u^m{v}^n$

$$\begin{array}{cc}\hfill {\mathcal{W}}_2(m,n;\alpha )& =m!n!\left[u^m{v}^n\right]\frac{{\partial }^2}{\partial v^2}\frac{\mathrm{Beta}(u-\alpha ,\frac{v+1}{2})}{2}\hfill \\ & =\frac{m!(n+2)!}{2}\left[u^m{v}^{n+2}\right]\mathrm{Beta}\left(u-\alpha ,\frac{v+1}{2}\right),\hfill \end{array}$$

we arrive at the following closed-form expression.

Theorem 10.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{W}}_2(m,n;\alpha )& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\rho (m,k;\alpha )\left\{\lambda (n+3)-{\mathbf{O}}_k^{\langle n+3\rangle }\right\}\hfill \\ & =\frac{{(-1)}^{m+n}}{2}\left[u^m{v}^{n+2}\right]\mathrm{Beta}\left(u-\alpha ,\frac{v+1}{2}\right).\hfill \end{array}$$

By executing the following Mathematica commands:

w2x[a_,m_,n_]:=(-1)^(m+n)/2*SeriesCoefficient[Beta[x-a,(1+y)/2],{x,0,m},{y,0,n+2}]

w2y[a_,m_,n_]:=FunctionExpand[w2x[a,m,n]]/.{PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]}

we obtain four concrete infinite series identities as applications.

Corollary 10

(Particular values of ${\mathrm{W}}_2(m,n;\alpha )$).

$$\begin{array}{cc}\hfill {\mathrm{W}}_2\left(1,0;\frac{1}{4}\right)& =\frac{{\Gamma }^2\left(\frac{3}{4}\right)}{16\sqrt{2\pi }}\times \left\{\begin{array}{c}224\zeta \left(3\right)-4{\pi }^2+3{\pi }^3-64G(2+\pi +ln2)\\ +16{ln}^{2}2+4\pi {ln}^{2}2+16\pi ln2-4{\pi }^{2}ln2\end{array}\right\},\hfill \\ \hfill {\mathrm{W}}_2\left(1,0;\frac{3}{4}\right)& =\frac{{\Gamma }^2\left(\frac{1}{4}\right)}{576\sqrt{2\pi }}\times \left\{\begin{array}{c}64G(16+3\pi -3ln2)-672\zeta \left(3\right)-64\pi -56{\pi }^2+9{\pi }^3\\ +32{ln}^{2}2+12\pi {ln}^{2}2+128ln2-128\pi ln2+12{\pi }^{2}ln2\end{array}\right\},\hfill \\ \hfill {\mathrm{W}}_2\left(0,0;\frac{1}{6}\right)& =\frac{\sqrt{\pi }\Gamma \left(\frac{5}{6}\right)}{32\Gamma \left(\frac{1}{3}\right)}\left\{12{\zeta }_2\left(\frac{1}{3}\right)-7{\pi }^2-\left[2\pi \sqrt{3}+3ln\left(\frac{27}{16}\right)\right]ln\left(\frac{27}{16}\right)\right\},\hfill \\ \hfill {\mathrm{W}}_2\left(0,0;\frac{5}{6}\right)& =\frac{\sqrt{\pi }\Gamma \left(\frac{1}{6}\right)}{480\Gamma \left(\frac{2}{3}\right)}\left\{12{\zeta }_2\left(\frac{1}{3}\right)-9{\pi }^2+12\pi \sqrt{3}-\left[36+2\pi \sqrt{3}-3ln\left(\frac{27}{16}\right)\right]ln\left(\frac{27}{16}\right)\right\}.\hfill \end{array}$$

4.3. Series ${\mathrm{W}}_3(m,n;\alpha )$ with Zeta Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with zeta tails

$$\begin{array}{cc}\hfill {\mathrm{W}}_3(m,n;\alpha )& :=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\frac{\rho (m,k;\alpha )}{k+1}\left\{\zeta (n+2)-{\mathbf{H}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\sum _{j=0}^{\infty }\frac{1}{{(j+1)}^{n+2}}\sum _{k=0}^j\frac{{\left(\alpha \right)}_k}{(k+1)!}\rho (m,k;\alpha ).\hfill \end{array}$$

Evaluating the rightmost sum by Lemma 3(b) leads us to the expression below

$${\mathrm{W}}_3(m,n;\alpha )=\frac{\zeta (n+2)}{{(1-\alpha )}^{m+1}}-\sum _{j=0}^{\infty }\frac{\left[x^m\right]}{{(j+1)}^{n+2}}\frac{\left(\genfrac{}{}{0pt}{}{\alpha +x+j}{j+1}\right)}{1-\alpha -x}.$$

Then, taking into account (5), we can determine the generating function

$$\begin{array}{cc}\hfill \sum _{n=-1}^{\infty }{y}^{n+2}\sum _{j=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{\alpha +x+j}{j+1}\right)}{{(j+1)}^{n+3}}& =\sum _{j=0}^{\infty }{(-1)}^{j+1}\frac{\left(\genfrac{}{}{0pt}{}{-\alpha -x}{j+1}\right)y}{1+j-y}\begin{array}{|c|}\hline j+1\to k\\ \hline\end{array}\hfill \\ & =1+\sum _{k=0}^{\infty }{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{-\alpha -x}{k}\right)y}{k-y}\hfill \\ & =1+y\mathrm{Beta}(1-\alpha -x,-y).\hfill \end{array}$$

By substitution, we derive the following compact formula.

Theorem 11.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{W}}_3(m,n;\alpha )& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\frac{\rho (m,k;\alpha )}{k+1}\left\{\zeta (n+2)-{\mathbf{H}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\frac{\zeta (n+2)}{{(1-\alpha )}^{m+1}}-\left[x^m{y}^{n+1}\right]\frac{\mathrm{Beta}(1-\alpha -x,-y)}{1-\alpha -x}.\hfill \end{array}$$

By carrying out the following Mathematica commands:

w3x[a_,m_,n_]:=Zeta[n+2]/(1-a)^(m+1)-SeriesCoefficient[Beta[1-a-x,-y]/(1-a-x),{x,0,m},{y,0,n+1}]

w3y[a_,m_,n_]:=FunctionExpand[w3x[a,m,n]]/.{PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]}

four summation formulae are produced in the following corollary.

Corollary 11

(Particular values of ${\mathrm{W}}_3(m,n;\alpha )$).

$$\begin{array}{cc}\hfill {\mathrm{W}}_3\left(0,1;\frac{1}{3}\right)& =\frac{9ln3-\sqrt{3}\pi }{8}{\zeta }_2\left(\frac{2}{3}\right)-\frac{1}{96}\left\{432\zeta \left(3\right)-13\sqrt{3}{\pi }^3+81{ln}^{3}3-27\sqrt{3}\pi {ln}^{2}3+27{\pi }^{2}ln3\right\},\hfill \\ \hfill {\mathrm{W}}_3\left(0,1;\frac{2}{3}\right)& =\frac{9ln3+\sqrt{3}\pi }{4}{\zeta }_2\left(\frac{1}{3}\right)-\frac{1}{48}\left\{432\zeta \left(3\right)+13\sqrt{3}{\pi }^3+81{ln}^{3}3+27\sqrt{3}\pi {ln}^{2}3+27{\pi }^{2}ln3\right\},\hfill \\ \hfill {\mathrm{W}}_3\left(1,0;\frac{1}{4}\right)& =\frac{16G}{9}(4+3\pi -18ln2)-\frac{112\zeta \left(3\right)}{3}+\frac{2}{9}\left\{3{\pi }^3-{\pi }^2-12\pi ln2+18{\pi }^{2}ln2+36{ln}^{2}2\right\},\hfill \\ \hfill {\mathrm{W}}_3\left(1,0;\frac{3}{4}\right)& =16G(\pi -4+6ln2)-112\zeta \left(3\right)-2\left\{{\pi }^2+{\pi }^3-12\pi ln2-6{\pi }^{2}ln2-36{ln}^{2}2\right\}.\hfill \end{array}$$

4.4. Series ${\mathrm{W}}_4(m,n;\alpha )$ with Lambda Tails

For $m,n\in {\mathbb{N}}_0$, consider the series with lambda tails

$$\begin{array}{cc}\hfill {\mathrm{W}}_4(m,n;\alpha )& :=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\frac{\rho (m,k;\alpha )}{k+1}\left\{\lambda (n+2)-{\mathbf{O}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\sum _{j=0}^{\infty }\frac{1}{{(2j+1)}^{n+2}}\sum _{k=0}^j\frac{{\left(\alpha \right)}_k}{(k+1)!}\rho (m,k;\alpha ).\hfill \end{array}$$

By appealing to Lemma 3(b), we can derive the expression below

$${\mathrm{W}}_4(m,n;\alpha )=\frac{\lambda (n+2)}{{(1-\alpha )}^{m+1}}-\sum _{j=0}^{\infty }\frac{\left[x^m\right]}{{(2j+1)}^{n+2}}\frac{\left(\genfrac{}{}{0pt}{}{\alpha +x+j}{j+1}\right)}{1-\alpha -x}.$$

Then, we can determine, by making use of (5), the generating function

$$\begin{array}{cc}\hfill \sum _{n=-1}^{\infty }{y}^{n+2}\sum _{j=0}^{\infty }\frac{\left(\genfrac{}{}{0pt}{}{\alpha +x+j}{j+1}\right)}{{(2j+1)}^{n+2}}& =\sum _{j=0}^{\infty }{(-1)}^{j+1}\frac{\left(\genfrac{}{}{0pt}{}{-\alpha -x}{j+1}\right)y}{1+2j-y}\begin{array}{|c|}\hline j+1\to k\\ \hline\end{array}\hfill \\ & =\frac{y}{1+y}+\sum _{k=0}^{\infty }{(-1)}^k\frac{\left(\genfrac{}{}{0pt}{}{-\alpha -x}{k}\right)y}{2k-1-y}\hfill \\ & =\frac{y}{1+y}+\frac{y}{2}\mathrm{Beta}\left(1-\alpha -x,-\frac{1+y}{2}\right).\hfill \end{array}$$

By substitution, we establish the following formula.

Theorem 12.

For any $m,n\in {\mathbb{N}}_0$, the following identity holds:

$$\begin{array}{cc}\hfill {\mathrm{W}}_4(m,n;\alpha )& =\sum _{k=0}^{\infty }\left(\genfrac{}{}{0pt}{}{\alpha +k-1}{k}\right)\frac{\rho (m,k;\alpha )}{k+1}\left\{\lambda (n+2)-{\mathbf{O}}_k^{\langle n+2\rangle }\right\}\hfill \\ & =\frac{\lambda (n+2)+{(-1)}^n}{{(1-\alpha )}^{m+1}}-\left[x^m{y}^{n+1}\right]\frac{\mathrm{Beta}(1-\alpha -x,-\frac{1+y}{2})}{2(1-\alpha -x)}.\hfill \end{array}$$

By fulfilling the following Mathematica commands:

w4x[a_,m_,n_]:=((-1)^n+DirichletLambda[n+2])/(1-a)^(m+1)-SeriesCoefficient[Beta[1-a-x,-((1+y)/2)]/(2(1-a-x)),{x,0,m},{y,0,n+1}]

w4y[a_,m_,n_]:=FunctionExpand[w4x[a,m,n]]/.{PolyGamma[k_,x_]->(-1)^(k+1)*k!*Zeta[k+1,x]}

four infinite series identities are exhibited in the corollary below.

Corollary 12

(Particular values of ${\mathrm{W}}_4(m,n;\alpha )$).

$$\begin{array}{cc}\hfill {\mathrm{W}}_4\left(0,1;\frac{1}{4}\right)& =\frac{7\zeta \left(3\right)}{6}-\frac{4}{3}+\frac{{\Gamma }^2\left(\frac{3}{4}\right)}{24\sqrt{2\pi }}\left\{32-32G+8\pi -{\pi }^2+16ln2+4\pi ln2+4{ln}^{2}2\right\},\hfill \\ \hfill {\mathrm{W}}_4\left(0,1;\frac{3}{4}\right)& =\frac{7\zeta \left(3\right)}{2}-4+\frac{{\Gamma }^2\left(\frac{1}{4}\right)}{32\sqrt{2\pi }}\left\{32-32G-8\pi +{\pi }^2+16ln2+4\pi ln2-4{ln}^{2}2\right\},\hfill \\ \hfill {\mathrm{W}}_4\left(0,0;\frac{1}{6}\right)& =\frac{6}{5}+\frac{3{\pi }^2}{20}-\frac{\sqrt{\pi }\Gamma \left(\frac{5}{6}\right)}{10\Gamma \left(\frac{1}{3}\right)}\left\{12+\sqrt{3}\pi -12ln2+9ln3\right\},\hfill \\ \hfill {\mathrm{W}}_4\left(0,0;\frac{5}{6}\right)& =6+\frac{3{\pi }^2}{4}-\frac{\sqrt{\pi }\Gamma \left(\frac{1}{6}\right)}{6\Gamma \left(\frac{2}{3}\right)}\left\{6+\sqrt{3}\pi +12ln2-9ln3\right\}.\hfill \end{array}$$

5. Concluding Comments

By introducing three classes of harmonic-like elementary symmetric functions $\sigma (m,n)$, $\tau (m,n)$, and $\rho (m,n;\alpha )$, we have succeeded in evaluating, in closed form, twelve classes of infinite series containing tails of Riemann’s zeta and Dirichlet’s lambda functions. This is realized by integral representations of zeta tails together with the telescopic approach. However, if the above elementary functions $\sigma (m,n)$, $\tau (m,n)$ and $\rho (m,n;\alpha )$ are replaced by the corresponding complete symmetric functions, the related infinite series would become more difficult to evaluate. Interested readers are enthusiastically encouraged to make further attempts.