# How to prove that $\sum_{n=0}^\infty \frac{1}{(2n+1)^2} + \sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{4}{3} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$

How to prove

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} + \sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{4}{3} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$

• Because $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\dfrac{\pi^2}{8}.$$
– Jika
Commented May 31, 2014 at 14:21

Consider the series $$\sum_{n=1}^\infty\frac1{n^2}=\underbrace{\sum_{n=0}^\infty\frac1{(2n+1)^2}}_{\large\text{odd parts}}+\underbrace{\sum_{n=1}^\infty\frac1{(2n)^2}}_{\large\text{even parts}}.$$ Hence \begin{align} \sum_{n=1}^\infty\frac1{n^2}-\sum_{n=1}^\infty\frac1{(2n)^2}&=\sum_{n=0}^\infty\frac1{(2n+1)^2}\\ \sum_{n=1}^\infty\frac1{n^2}-\frac14\sum_{n=1}^\infty\frac1{n^2}&=\sum_{n=0}^\infty\frac1{(2n+1)^2}\\ \frac34\sum_{n=1}^\infty\frac1{n^2}&=\sum_{n=0}^\infty\frac1{(2n+1)^2}\\ \sum_{n=1}^\infty\frac1{n^2}&=\frac43\sum_{n=1}^\infty\frac1{(2n+1)^2}\\ \color{blue}{\sum_{n=0}^\infty\frac1{(2n+1)^2}+\sum_{n=1}^\infty\frac1{(2n)^2}}&\color{blue}{=\frac43\sum_{n=1}^\infty\frac1{(2n+1)^2}}.\qquad\qquad\blacksquare \end{align}
$$\zeta(n)=\sum\limits_{k=1}^{\infty}\dfrac{1}{k^n}=\dfrac{2^n}{2^n-1}\sum\limits_{k=0}^{\infty}\dfrac{1}{(2k+1)^n}.$$
where $\zeta(\cdot)$ is the Riemann zeta function.
and $\zeta(2)=\sum\limits_{k=1}^{\infty}\dfrac{1}{k^2}=\dfrac{\pi^2}{6}$