# Infinite sum, not arithemtic nor geometric progression.

How could you compute this:

$$\sum_{n}^\infty\frac{1}{\left( 2n+1\right)^4}$$

I know the solution is $\pi^4/96$, according to Maple, but I have no idea how to do this or how to start.

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Can you first show what $\sum \frac{1}{n^4}$ is? –  Gautam Shenoy Apr 21 '13 at 15:36
What mathematics do you know? –  Thomas Andrews Apr 21 '13 at 15:37

$$\frac{\pi^4}{90}=\sum_{n=1}^\infty \frac{1}{n^4}$$
\begin{align}\sum_{n=0}^\infty \frac{1}{(2n+1)^4} &= \sum_{n=1}^\infty\frac{1}{n^4} - \sum_{n=1}^\infty\frac{1}{(2n)^4} \\&= \left(1-\frac{1}{16}\right)\sum_{n=1}^\infty \frac{1}{n^4} \end{align}
You might also see the answers at: Nice proofs of $\zeta(4) = \pi^4/90$? for why the first statement is true.