<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] Step 1: The sum of all the reciprocal squares is $\zeta(2)=\frac{\pi^2}{6}$.
Step 2: The sum of all the reciprocal even squares is $\frac{1}{4}\zeta(2)=\frac{\pi^2}{48}$.
Step 3: Their difference, the sum of all the reciprocal odd squares, is $\frac{\pi^2}{8}$.
Step 4: Catalan's constant is $G=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}\cdots$
Step 5: Adding the results from steps 3 and four, and dividing by 2, gives the desired series.  It also gives the sum $$\frac{1}{2}\left(\frac{\pi^2}{8}+G\right)$$
There is not much known about $G$, but it's also $\beta(2)$, the Dirichlet beta function.  Hence if you wanted an answer in terms of special functions, the sum is $$
\frac{3}{8}\zeta(2)+\frac{1}{2}\beta(2)$$
</s>