Establishing identity: $\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^{2}} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ I need some help in trying to prove the following identity $$\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k-1)^{2}} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$$
I'm not sure where to start really. I thought it was going to be as simple as just adding the two terms together but nothing comes of that....

You may know already that $\sum_{k=1}^\infty\frac{1}{k^2} = \frac{\pi^2}{6}$ and $\sum_{k=1}^\infty\frac{1}{(2k + 1)^2} = \frac{\pi^2}{8}$. 
So
$$\frac{1}{4}\frac{\pi^2}{6} + \frac{\pi^2}{8} = \frac{\pi^2}{6}.
$$
