How can I prove $\pi ^2=\sum_{n=0}^{\infty }\frac{1}{(2n+1+\frac{a}{3})^2}+\frac{1}{(2n+1-\frac{a}{3})^2}$ Proving this formula 
$$
\pi^{2}
=\sum_{n\ =\ 0}^{\infty}\left[\,{1 \over \left(\,2n + 1 + a/3\,\right)^{2}}
+{1 \over \left(\, 2n + 1 - a/3\,\right)^{2}}\,\right]
$$
if $a$ an even integer number so that
$$
a \geq 4\quad\mbox{and}\quad{\rm gcd}\left(\,a,3\,\right) = 1
$$

Start with the well known? expansion of $\cot z$ and differentiate,
$$\cot z = \sum_{n=-\infty}^\infty \frac{1}{z - n\pi}
\implies \frac{1}{\sin(z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z - n\pi)^2}
\implies \frac{\pi^2}{\sin(\pi z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z - n)^2}
$$
Substitute $z$ by $-\frac{1+\alpha}{2}$, we get
$$\frac{\pi^2}{4\cos(\frac{\pi\alpha}{2})^2} = \sum_{n=-\infty}^\infty\frac{1}{(2n+1+\alpha)^2}\tag{*1}$$
In the RHS of above sum, if we index those negative $n$ as $-(m+1)$, we have
$$\frac{1}{(2n+1 + \alpha)^2} = \frac{1}{(2m+1 - \alpha)^2}$$
This means one can rewrite $(*1)$ as
$$\frac{\pi^2}{4\cos(\frac{\pi\alpha}{2})^2} = \sum_{n=0}^\infty \left(\frac{1}{(2n+1 + \alpha)^2} + \frac{1}{(2n+1 - \alpha)^2}\right)\tag{*2}$$
When $\alpha = \frac{a}{3}$ where $a$ is an even integer with $\gcd(a,3) = 1$,
$\cos(\frac{\pi\alpha}{2}) = \pm \frac12$.
RHS$(*2)$ reduces to $\frac{\pi^2}{4\left(\frac12\right)^2} = \pi^2$ as desired.
