Show $(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots$ Last month I was calculating $\displaystyle \int_0^\infty \frac{1}{1+x^4}\, dx$ when I stumbled on the surprising identity:
$$\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right) = \frac{\pi}{\sqrt8}$$
and I knew
$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}$$
So if I could find a proof that $$\left(\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)\right)^2 = \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$ then this could be a new proof that $\zeta(2)=\frac{\pi^2}{6}$. I've thought over this for almost a month and I'm no closer on showing this identity.
Note: Article on the multiplication of conditionally convergent series: http://www.jstor.org/stable/2369519

Let $a_k = (-1)^k \left(\frac{1}{4k+1} + \frac{1}{4k+3}\right)$ and $b_k = \frac{1}{(4k+1)^2} + \frac{1}{(4k+3)^2}$. The goal is to show that: $$ \left(\sum_{i=0}^\infty a_i\right)^2 = \sum_{i=0}^\infty b_i $$
The key observation that I missed on my previous attempt is that: $$ \sum_{i=0}^n a_i = \sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} $$ This transformation allows me to then mimic the proof that was suggested in the comments by @user17762.
\begin{align*}  \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i 
&= \left(\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1}\right)^2 - \sum_{i=-n-1}^n \frac{1}{(4i+1)^2}   \\
&= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^i}{4i+1}\frac{(-1)^j}{4j+1} \\
&= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{4j-4i}\left(\frac{1}{4i+1}-\frac{1}{4j+1} \right) \\
&= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{2j-2i} \cdot \frac{1}{4i+1} \\
&= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} \sum_{\substack{j=-n-1 \\ i \neq j}}^n \frac{(-1)^j}{j-i} \\
&= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i }{4i+1}c_{i,n} \\
&= \frac{1}{2}\sum_{i=0}^n a_i \,c_{i,n}
\end{align*}
Where the last equality follows from $c_{i,n} = c_{-i-1, n}$. Since $c_{i,n}$ is a partial alternating harmonic sum, it is bounded by its largest entry in the sum: $\left| c_{i,n} \right| \le \frac{1}{n-i+1}$. We also know that $\left|a_i\right| \le \frac{2}{4i+1}$. Apply these two inequalities to get:
\begin{align*}\left| \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i \right| &\le \frac{1}{2} \sum_{i=0}^n \frac{2}{4i+1} \cdot \frac{1}{n-i+1} \\ &\le \sum_{i=0}^n \frac{1}{4n+5}\left( \frac{4}{4i+1} + \frac{1}{n-i+1} \right) \\ &\le \frac{1}{4n+5}\left( 5 + \ln(4n+1) +\ln(n+1)\right) \\
& \to 0 ~\text{ as }~ n \to \infty
\end{align*}
This concludes the proof.  In fact, with the same idea, you can prove this general family of identities: Fix an integer $m \ge 3$, then:
\begin{align*} & \left( 1 + \frac{1}{m-1} - \frac{1}{m+1} - \frac{1}{2m-1} + \frac{1}{2m+1} + \frac{1}{3m-1} - \cdots \right)^2  \\
=& ~ \left(\sum_{i=-\infty}^\infty \frac{(-1)^i}{im+1}\right)^2 \\ 
=& ~ \sum_{i=-\infty}^\infty \frac{1}{(im+1)^2} \\
=& ~ 1 +  \frac{1}{(m-1)^2} + \frac{1}{(m+1)^2} + \frac{1}{(2m-1)^2} + \frac{1}{(2m+1)^2} + \frac{1}{(3m-1)^2} + \cdots \\
=& ~ \left(\frac{\frac{\pi}{m}}{\sin\frac{\pi}{m}}\right)^2 \end{align*}
The last equality follows from the comment by @Lucian.
