Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$ 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?
 A: 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. 
A: I found a proof relying on some results from number theory, which hopefully fits your requirements

Claim:
$$
\left(\sum_{n \geq 1}\frac{(-1)^n}{2n+1}\right)^2=\frac{3}{8}\sum_{n\geq1}\frac{1}{n^2}
$$
We define 


*

*$\chi_l(n)$ as the l'th Dirichlet character $\text{mod}\,4$

*$L(\chi,s)\equiv\sum_{n\geq1}\tfrac{\chi(n)}{n^s}$ is a Dirichlet-$L$ sum

*$\zeta(s,q)\equiv\sum_{n\geq0}\tfrac{1}{(n+q)^s}$ is a Hurwitz zeta function ($q=1$ gives Riemanns $\zeta(s)$)


Now we recoginze that the orignal problem can be reformulated as follows  
$$
L^2(1,\chi_1)=\frac{1}{2}L(2,\chi_0) \quad(*)
$$

Proof:
By virtue of the identity (this holds for general characters $\text{mod}\,\,a $)$  L(\chi,s)=\sum_{b\leq a}\chi(b)\zeta(s,\frac ba)$ and the explict values of the character table we get
 for the left hand side of $(*)$
$$
\frac{1}{16}\left(\zeta(1,\tfrac{1}{4})-\zeta(1,\tfrac{3}{4})\right)^2
$$
where the limit $s\rightarrow 1$ is implicitly taken. By the Stieltjes expansion of the Hurwitz Zeta function this equal to
$$
\frac{1}{16}(\psi_0(1/4)-\psi_0(3/4))^2=\frac{\pi^2}{16}\cot^2(\frac{\pi}{4}) \quad(**)
$$
where the equality is a consequence of the reflection formula for the Polygamma function $\psi_n(z)$. 
On the other hand, for the rhs of $(*)$ we can write by the nearly the same token (here no limiting procedure is necessary)
$$
\frac{1}{2\cdot16}\left(\psi_1(1/4)+\psi_1(3/4)\right)=\frac{\pi^2}{2\cdot16}\frac{1}{\sin^2(\frac{\pi}4)}\quad(***)
$$
Now since $\cos^2(\frac{\pi}{4})=\frac{1}{2}$, $(**)=(***)$ and therefore $(*)$ is proven

Since i'm a theoretical physicist my knowledge of number theory is close to zero. I guess a more experienced person could conclude here much faster using some general theorems of $L$-functions or Dirichlet convolution. 
