The equality$$\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)\tag{1}$$follows from the fact that the sum of the first series is $\dfrac\pi4$, whereas the sum of the second one is $\dfrac{\pi^2}6$.

My question is: can someone provide a proof that $(1)$ holds without using this?

**Answer**

If we put $$f(t) =\sum_{n=1}^{\infty}\frac{\sin nt} {n} $$ then $$f^{2}(t)=\sum_{n=1}^{\infty}\frac{\sin^{2}nt}{n^{2}} +\text{(terms containing }\sin nt\sin mt) $$ and integrating this term by term with respect to $t$ over $[-\pi, \pi] $ should give us $$\pi\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$ and therefore we see that the RHS of the equation in question is $$\frac{3}{8\pi}\int_{-\pi}^{\pi}f^{2}(t)\,dt$$ This needs to be proved to be equal to $f^{2}(\pi/2)$. The function $f(t) $ is given in closed form as $$f(t) = \begin{cases}

\dfrac{\pi – t} {2}, 0<t\leq \pi\\

0,t=0\\

-\dfrac{\pi +t} {2}, – \pi\leq t<0

\end{cases}$$ and $f(t+2\pi)=f(t)$. So this works out fine.

**Attribution***Source : Link , Question Author : José Carlos Santos , Answer Author : Paramanand Singh*