Zeta in summations

Question

Does anyone know if there is any reference that discuss the following kind of summations?

$$S = \displaystyle\sum_{n=0}^\infty a_n \zeta(n+2)$$

I have read Srivastava's article https://www.sciencedirect.com/science/article/pii/0022247X88900133. But none deals with such series. Especially when:

$a_n = \frac{2}{(n+2)(n+3)}$. Which leads to

$S = \frac{\zeta(2)}{3} + \frac{\zeta(3)}{6} + \frac{\zeta(4)}{10} + \cdots$

Have you seen this one before?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} S & \equiv \bbox[5px,#ffd]{2\sum_{n = 0}^{\infty} {\zeta\pars{n + 2} \over \pars{n + 2}\pars{n + 3}}} = 2\sum_{n = 2}^{\infty}{\zeta\pars{n} \over n\pars{n + 1}} \\[5mm] & = 2\sum_{n = 2}^{\infty}\zeta\pars{n} \pars{\int_{0}^{1}x^{n - 1}\,\,\dd x} \pars{\int_{0}^{1}y^{n}\,\dd y} \\[5mm] & = 2\int_{0}^{1}\int_{0}^{1} \bracks{\sum_{n = 2}^{\infty}\zeta\pars{n} \pars{xy}^{n - 1}}y\,\dd x\,\dd y \\[5mm] & = 2\int_{0}^{1}\int_{0}^{1} \bracks{-\gamma - \Psi\pars{1 - xy}}y\,\dd x\,\dd y \end{align} $\ds{\gamma}$ is the Euler-Mascheroni Constant and $\ds{\Psi}$ is the Digamma Function.

The relation between $\ds{\zeta\ \mbox{and}\ \Psi}$ can be seen in $\ds{\color{black}{\bf 6.3.14}}$ of $\mbox{A & S Table}$.

Then, \begin{align} S & \equiv \bbox[5px,#ffd]{2\sum_{n = 0}^{\infty} {\zeta\pars{n + 2} \over \pars{n + 2}\pars{n + 3}}} \\[5mm] & = 2\int_{0}^{1} \bracks{-\gamma - {\ln\pars{\Gamma\pars{1 - y}} \over - y}}y\,\dd y \\[5mm] & = 2\bracks{-\,{1 \over 2}\,\gamma + \int_{0}^{1}\ln\pars{\Gamma\pars{1 - y}}\dd y} \\[5mm] & = \bbx{\ln\pars{2\pi} - \gamma} \approx 1.2607 \\ & \end{align}

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I used the well known result $\ds{\int_{0}^{1}\ln\pars{\Gamma\pars{1 - y}}\,\dd y = {\ln\pars{2\pi} \over 2}}$

Answer 2

Thanks to a CAS, $$S_m=\sum_{n=0}^\infty \frac{2 _,\zeta(n+m)}{(n+m)(n+m+1)}$$ have closed forms $$S_2=\log (2 \pi )-\gamma$$ $$S_3=\log (2 \pi )-\gamma -\frac{\zeta(2)}{3}$$ $$S_4=\log (2 \pi )-\gamma -\frac{\zeta(2)}{3}-\frac{\zeta (3)}{6}$$ $$S_5=\log (2 \pi )-\gamma -\frac{\zeta(2)}{3}-\frac{\zeta (3)}{6} -\frac{\zeta (4)}{10}$$ $$S_6=\log (2 \pi )-\gamma-\frac{\zeta(2)}{3}-\frac{\zeta (3)}{6} -\frac{\zeta (4)}{10}-\frac{\zeta (5)}{15} $$

Answer 3

Since $$ \zeta (s) = {1 \over {\Gamma (s)}}\int_0^\infty {{{x^{\,s - 1} } \over {e^{\,x} - 1}}dx} $$ then $$ \eqalign{ & \sum\limits_{n = 0}^\infty {a_{\,n} \zeta (n + 2)} = \sum\limits_{n = 0}^\infty {a_{\,n} {1 \over {\left( {n + 1} \right)!}}\int_0^\infty {{{x^{\,n + 1} } \over {e^{\,x} - 1}}dx} } = \cr & = \int_0^\infty {\left( {\sum\limits_{n = 0}^\infty {a_{\,n} } {{x^{\,n + 1} } \over {\left( {n + 1} \right)!}}} \right){1 \over {e^{\,x} - 1}}dx} = \cr & = \int_0^\infty {{{I(x)} \over {e^{\,x} - 1}}dx} \cr} $$ if we admit that the integral and sum can be exchanged.

Here $$ I(x) = \sum\limits_{n = 0}^\infty {a_{\,n} } {{x^{\,n + 1} } \over {\left( {n + 1} \right)!}} = \int_0^x {\left( {\sum\limits_{n = 0}^\infty {a_{\,n} } {{t^{\,n} } \over {n!}}} \right)dt} $$ is the integral of the e.g.f of the $\{a_n\}$ sequence.