# 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

• Thanks for the neat reference. – Andrés E. Caicedo Dec 9 '13 at 0:38
• I believe my answer here can be adapted to answer your question. Essentially, the answer shows how to obtain $\zeta(2) = \pi^2/6$, by squaring $1-\dfrac13 + \dfrac15 \mp \cdots = \dfrac{\pi}4$. I am now too lazy to write the entire thing out here. – user17762 Dec 9 '13 at 0:41
• Some things are so obvious that we never see them ! :-) – Lucian Dec 9 '13 at 3:24
• Oh, and as far as that integral is concerned, $$\int_0^\infty\frac{dx}{1+x^n}=\frac{\frac\pi n}{\sin\left(\frac\pi n\right)}$$ – Lucian Dec 9 '13 at 4:16
• Let $t=\displaystyle\frac1{1+x^n}$. It will become a Beta function, which is expressible in terms of the Gamma function. Then use the reflection formula for the latter. See my answer here. – Lucian Dec 12 '13 at 1:19

Let's have a try. $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{4n+1}=\int_{0}^{1}\frac{dx}{1+x^4},\qquad S=\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right)=\int_{0}^{1}\frac{1+x^2}{1+x^4}dx,$$ $$S = \int_{0}^{1}\frac{x+x^{-1}}{x^{-2}+x^2}\frac{dx}{x}=\int_{1}^{+\infty}\frac{z}{(2z^2-1)\sqrt{1-z^2}}\,dz = \int_{0}^{1}\frac{dt}{(2-t^2)\sqrt{1-t^2}},$$ $$S = \int_{0}^{\pi/2}\frac{d\theta}{2-\sin^2\theta}=\int_{0}^{\pi/2}\frac{d\theta}{1+\cos^2\theta}=\frac{1}{2}\int_{\mathbb{R}}\frac{du}{2+u^2},$$ where in the last integral we used the substitution $\theta=\arctan u$. This gives: $$S^2 = \frac{1}{8}\int_{\mathbb{R}^2}\frac{du\,dv}{(1+u^2)(1+v^2)}=\int_{0}^{1}\int_{0}^{+\infty}\frac{1}{(1+z^2)(1+x^2)}dx\,dz$$ On the other hand, $$\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\int_{0}^{1}\frac{\log y}{y^2-1}dy=\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2y^2)}dx\,dy,$$ where I learned the last equality from the Mike Spivey's note on the Luigi Pace's proof of $\zeta(2)=\frac{\pi^2}{6}$, just here. By setting $y=\frac{z}{x}$ in the last integral we get $S^2=\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}$, QED. So it looks like @user17762's proof-by-squaring-the-arctangent-series and Pace's proof can be combined in order to get a very short proof of your claim.
For the sake of exposing a one-line-proof of $\zeta(2)=\frac{\pi^2}{6}$: $$\zeta(2)=\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy=\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy=\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{6}.$$
• In fact, to consider the integral $\int_{\mathbb{R}^2}\frac{dx\,dy}{(1+x^2)(1+y^2)}$ (or a slight variant) is just to consider the square of the arctangent series. @user17762 manipulates it as a double sum, here it appears as a double integral. I did not know Danielle Ritelli's proof, but our approaches are almost identical, so I think I cannot take credits for this :) – Jack D'Aurizio Dec 25 '13 at 9:12
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: