<s>[INST] How to calculate this infinite sum? $$ \sum_ {n=0}^\infty \frac {1}{(4n+1)^2} $$
I am not sure how to calculate the value of this summation. My working so far is as follows:
Let $S=\sum_ {n=0}^\infty \frac {1}{(4n+1)^2}$.
$\Longrightarrow S=\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{9^2}+...$
$\Longrightarrow S=(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...)-(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{6^2}+...) $
$\Longrightarrow S=\zeta(2)-[(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}...)-(\frac{1}{5^2}+\frac{1}{9^2}+...)]$
$\Longrightarrow S=\zeta(2)-[(\zeta(2)-1)-(S-1)]$
$\Longrightarrow S=\zeta(2)-\zeta(2)+1+S-1$
$\Longrightarrow 0=0$
Does anyone have a better way of evaluating this that does not involve a cyclical answer as mine eventually does?
 [/INST] This has no closed form, unless you want to use Catalan's constant. Then the answer is $\frac{K}{2}+\pi^2/16$. See here: http://mathworld.wolfram.com/CatalansConstant.html
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