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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}