TITLE: Finding the closed form for $\sum_{n=1}^{\infty }\frac{\zeta (4n)}{\beta^{4n-1}}$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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}