Evaluating $\sum_{n=0}^\infty \frac 1 {(2n+1)^4}$ using Mittag-Leffler's expansion I am trying to evaluate the following series using Mittag-Leffler's expansion theorem. What function would be useful?
$$\sum_{n=0}^\infty \frac 1 {(2n+1)^4} = \frac{\pi^4}{96}$$
I considered differentiating the following relation twice but it did not help. And I managed to prove $0 = 0$ :D
$$\text{tanh}(z) = 2z \sum_{n=0}^\infty \frac{1}{z^2 + ((2n+1)\pi/2)^2}$$
The above expansion gives $\displaystyle \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$. I need something of $4-th$ order. Which function's expansion will give me something like $\displaystyle F(z) = \sum_{n=0}^\infty \frac{1}{(2n+1)^4 + f(z)}$

\begin{align*}
\frac 1z \frac{d}{dz}\left( \frac{\text{tanh}(z)}{z} \right ) 
&= \frac{z \; \text{sech}(z)^2 -\text{tanh}(z)}{z^3} \\
&= \frac{z (1 - z^2 + O(z^3)) - (z  - z^3/3 + O(z^2))}{z^3} \\
&= \frac{-\frac 2 3 z^3 + O(z^5)}{z^3 } \\ 
&= -2/3  \\
&= \frac 1z \frac{d}{dz}\left(  2 \sum_{n=0}^\infty \frac{1}{z^2 + ((2n+1)\pi/2)^2}\right )\\
&= \frac 1z 2 \sum_{n=0}^\infty - \frac{2z}{(z^2 + ((2n+1)\pi/2)^2)^2} \\
&= - 4\sum_{n=0}^\infty \frac{1}{(z^2 + ((2n+1)\pi/2)^2)^2}
\end{align*}

You should evaluate:
$$
     \lim_{z \to 0} -\frac{(\pi/2)^4}{2 z} \frac{\mathrm{d}}{\mathrm{d}z} \frac{\tanh(z)}{2z}
$$
