TITLE: Proof of $\sum_{n=1,3,5,\ldots}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{96}$
QUESTION [12 upvotes]: I came across with the infinite series
$$\sum_{n=1,3,5,\ldots}^{\infty} \frac{1}{n^4}= \frac{\pi^4}{96}$$
when calculating a problem about an infinite deep square well in quantum mechanics.
Mathematica gives the result in the title, which is enough for a physics problem. But I just want to find how to evaluate the series. I think this sum should be connected to $\zeta(4)=\pi^4/90$, but can't figure out their relation.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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As shown by $\texttt{@Timbuc}$, you just need to know the sum
$\ds{\sum_{n = 1}^{\infty}{1 \over n^{4}}}$. It has an interesting evaluation by starting from the $\ds{\cot\pars{z}}$ expansion
$\ds{\pars{~\mbox{with}\ z \not= 0, \pm\pi,\pm 2\pi,\ldots~}}$:
\begin{align}
\cot\pars{z} & =
{1 \over z} + 2z\sum_{n = 1}^{\infty}{1 \over z^{2} - n^{2}\pi^{2}} =
{1 \over z} - {2z \over \pi^{2}}\sum_{n = 1}^{\infty}{1 \over n^{2}}
{1 \over 1 - z^{2}/ \pars{n^{2}\pi^{2}}}
\\[5mm] & =
{1 \over z} - {2z \over \pi^{2}}\sum_{n = 1}^{\infty}{1 \over n^{2}}
\pars{1 + {z^{2} \over n^{2}\pi^{2}} + \cdots} =
{1 \over z} - {2z \over \pi^{2}}\sum_{n = 1}^{\infty}{1 \over n^{2}}
-{2z^{3} \over \pi^{4}}\color{#f00}{\sum_{n = 1}^{\infty}{1 \over n^{4}}} - \cdots
\end{align}

\begin{align}
z\cot\pars{z} & =
1 - {2z^{2} \over \pi^{2}}\sum_{n = 1}^{\infty}{1 \over n^{2}}
-\bracks{{48 \over \pi^{4}}\color{#f00}{\sum_{n = 1}^{\infty}{1 \over n^{4}}}}{z^{4} \over 4!} - \cdots
\end{align}

$$
\color{#f00}{\sum_{n = 1}^{\infty}{1 \over n^{4}}} =
\left.-\,{\pi^{4} \over 48}\,\totald[4]{\bracks{z\cot\pars{z}}}{z}
\right\vert_{\ z\ \to\ 0} = \pars{-\,{\pi^{4} \over 48}}\pars{-\,{8 \over 15}} =
\color{#f00}{\pi^{4} \over 90}
$$