real analysis - 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

Answer

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.