How to calculate this infinite sum?

Question

$$ \sum_ {n=0}^\infty \frac {1}{(4n+1)^2} $$

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?

Answer 1

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

Answer 2

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 $$ \frac{3}{8}\zeta(2)+\frac{1}{2}\beta(2)$$