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Finding the closed-form

$$\sum_{n=1}^{\infty }\frac{\zeta (4n)}{\beta^{4n-1}}$$ for $\beta\in(1,+\infty)$.

I learned from this site many many important things but I till need more, so I need to know if there is a closed-form of this series . Thanks for any body devised me to learn from any answer.

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    $\begingroup$ Integer number? Then why don't you do even the most basic substitution to replace $a+1$ to simplify the question? Not to mention $-2$ in the denominator is also completely superfluous. 0 effort from you, -1 from me. $\endgroup$ – user2345215 Jan 9 '15 at 20:03
  • $\begingroup$ I slightly modified the title and the question body, I hope it is an improvement (especially about readability). $\endgroup$ – Jack D'Aurizio Jan 9 '15 at 20:22
  • $\begingroup$ Expand $\zeta(4n)$ into its well-known infinite series, then reverse the order of the two summations. $\endgroup$ – Lucian Jan 9 '15 at 21:21
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Since: $$\zeta(4n)=\frac{1}{(4n-1)!}\int_{0}^{+\infty}\frac{z^{4n-1}}{e^{z}-1}\,dz \tag{1}$$ and: $$ \sum_{n\geq 1}\frac{w^{4n-1}}{(4n-1)!}=\frac{\sinh w-\sin w}{2}\tag{2}$$ it follows that, for any $\beta>1$, $$ \sum_{n=1}^{+\infty}\frac{\zeta(4n)}{\beta^{4n-1}} = \frac{1}{2}\int_{0}^{+\infty}\frac{\sinh(w/\beta)-\sin(w/\beta)}{e^w-1}\,dw.\tag{3}$$ For special values of $\beta$ ($\beta\in\mathbb{N}$, for instance) we can compute the last integral through the residue theorem. In general, by exploiting the inverse Laplace transform, we have: $$ \sum_{n=1}^{+\infty}\frac{\zeta(4n)}{\beta^{4n-1}} = \frac{1}{2}\left(\beta - \frac{\pi}{2}\cot\frac{\pi}{\beta}-\frac{\pi}{2}\coth\frac{\pi}{\beta}\right).\tag{4}$$ We can achieve the same by considering the Taylor series of $x\cot x$ and $x\coth x$ in a neighbourhood of $z=0$.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n} \over \beta^{4n - 1}}:\ {\large ?}. \qquad\beta\in\pars{1,\infty}}$.


\begin{align}&\color{#66f}{\large% \sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n} \over \beta^{4n - 1}}} =\sum_{n\ =\ 4}^{\infty}\ {\zeta\pars{n} \over \beta^{n - 1}} -\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n + 1} \over \beta^{4n}} -\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n + 2} \over \beta^{4n + 1}} -\sum_{n\ =\ 1}^{\infty}\ {\zeta\pars{4n + 3} \over \beta^{4n + 2}} \\[5mm]&=\sum_{n\ =\ 1}^{\infty}\bracks{{\zeta\pars{n + 3} \over \beta^{n + 2}} - {\zeta\pars{4n + 1} \over \beta^{4n}} - {\zeta\pars{4n + 2} \over \beta^{4n + 1}}- {\zeta\pars{4n + 3} \over \beta^{4n + 2}}} \\[5mm]&=\beta\sum_{k\ =\ 1}^{\infty}\ \sum_{n\ =\ 1}^{\infty}\bracks{{1 \over \pars{\beta k}^{n + 3}} - {1 \over \pars{\beta k}^{4n + 1}} -{1 \over \pars{\beta k}^{4n + 2}}- {1 \over \pars{\beta k}^{4n + 3}}} \\[5mm]&=\beta\sum_{k\ =\ 1}^{\infty}{1 \over \pars{\beta k}^{4} - 1} =\color{#66f}{\large{1 \over 4}\bracks{2\beta - \pi\,\cot\pars{\pi \over \beta} -\pi\coth\pars{\pi \over \beta}}} \end{align}

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    $\begingroup$ Wouldn't it be simpler just to use $$\sum_{n=1}^\infty\left(\sum_{k=1}^\infty{1\over k^{4n}} \right){1\over\beta^{4n-1}}=\beta\sum_{k=1}^\infty\left(\sum_{n=1}^\infty{1\over(\beta k)^{4n}}\right)$$ $\endgroup$ – Barry Cipra Jan 18 '16 at 22:12

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