Proof of $\sum_{n=1,3,5,\ldots}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{96}$ 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. 
 A: You can also use the well known summation formula $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{residue of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{ poles}\right\} 
 $$ which is a consequence of the residue theorem. So it is sufficient to note that $$ \sum_{n\geq1}\frac{1}{\left(2n-1\right)^{4}}=\frac{1}{2}\sum_{n\in\mathbb{Z}}\frac{1}{\left(2n-1\right)^{4}}=-\frac{1}{2}\left\{ \textrm{residue of }\frac{\pi\cot\left(\pi z\right)}{\left(2z-1\right)^{4}}\textrm{ at }\frac{1}{2}\right\} =\color{red}{\frac{\pi^{4}}{96}}.$$
A: $$\frac{\pi^4}{90}=\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}\implies$$
$$\implies\sum_{n=1}^\infty\frac1{(2n-1)^4}=\left(1-\frac1{16}\right)\frac{\pi^4}{90}=\frac{\pi^4}{96} $$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\mrm}[1]{\,\mathrm{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}

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