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How to prove

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

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

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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.

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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 $\zeta(2)=\sum\limits_{k=1}^{\infty}\dfrac{1}{k^2}=\dfrac{\pi^2}{6}$

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