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
$$ \sum_{n=0}^\infty \frac{1}{(2n+1)^2} + \sum_{k=1}^\infty \frac{1}{(2k)^2}=\frac{4}{3} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$
 A: Consider the following series:
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n^{2}} &= \frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \cdots \\
&= \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}
\end{align}
which leads to
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{1}{4} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} 
\end{align}
or
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{1}{3} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}}.
\end{align}
Now,
\begin{align}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} = \frac{4}{3} \ \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}}
\end{align}
as desired.
A: HINT:
$$\zeta(n)=\sum\limits_{k=1}^{\infty}\dfrac{1}{k^n}=\dfrac{2^n}{2^n-1}\sum\limits_{k=0}^{\infty}\dfrac{1}{(2k+1)^n}.$$
where $\zeta(\cdot)$ is the Riemann zeta function.
and $\zeta(2)=\sum\limits_{k=1}^{\infty}\dfrac{1}{k^2}=\dfrac{\pi^2}{6}$
A: 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}
