Establishing identity: $\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^{2}} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$

Question

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....

Answer 1

$$\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^{2}} = \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$$

Note that every natural number $n$ is either even which is $n=2k$ or it is odd, that is $n=2k+1$

So we can partition the set of $$\left\{\frac {1}{n^2} : n\in \mathbb {N}\right\} $$ into two sets namely $$\left\{\frac {1}{(2k+1)^2} : k\in \mathbb {N}\right\} $$ and $$\left\{\frac {1}{(2k)^2} :k\in \mathbb {N}\right\} $$ Thus $$\sum_{k = 1}^{\infty} \frac{1}{(2k)^{2}} + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^{2}} = \sum_{n= 1}^{\infty} \frac{1}{n^{2}} = \sum_{k= 1}^{\infty} \frac{1}{k^{2}} $$

Answer 2

Just split the sum in RHS into the part where $k$ is even and the part where $k$ is odd.

Answer 3

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 $$\frac{1}{4}\frac{\pi^2}{6} + \frac{\pi^2}{8} = \frac{\pi^2}{6}. $$