Euler, Erdős

Four years ago, I wrote about my failed attempt to directly prove the following identity:

\[ \left(1 + \frac{1}{3} - \frac{1}{5} -\frac{1}{7} + \frac{1}{9} + \cdots \right)^2 = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \cdots \]

The crucial part was I wanted to show this without introducing integrals. One way to answer this question is to rewrite the left/right-hand sides as integrals and showing that the two integrals were the same. But this feels like a roundabout way to show something that seemed like it should have a straight-forward proof. To make matters even more frustrating, there was a StackExchange (SE) answer that did precisely what I wanted with a similar series. At the time, I didn't properly understand it and trying to blindly repurpose it for my identity failed. After writing the Misadventure post, I moved on and never tried to prove it again.

That was until this month. What changed? I've been auditing a computational statistics class and I'm learning how to use R. As I was testing the options for plotting data points, it occured to me to plot the partial sums of the above series and see how fast the convergence happens; this visual evidence of the identity inspired me to try proving it again. On this new attempt, I was able to both succeed and fail in my original objective: I succeeded in proving the identity but I failed because the proof is nearly identical to the SE answer; this couldn't really count as a new proof for \(\zeta(2) = \frac{\pi^2}{6}\). On a related note, I found this document with numerous proofs that \(\zeta(2)=\frac{\pi^2}{6}\). The SE answer and my argument below are based on Proof 13 in that document.

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: \[\left(\sum_{i=0}^\infty a_i\right)^2 = \sum_{i=0}^\infty b_i\] Or to rephrase the problem, I want to show: \[ \lim_{n \to \infty} \left[ \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i \right]= 0 \] The key observation that I missed on my previous attempts 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 Proof 13: \[ \begin{aligned} \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i &= \left(\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1}\right)^2 - \sum_{i=-n-1}^n \frac{1}{(4i+1)^2} \\ &= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^i}{4i+1}\frac{(-1)^j}{4j+1} \\ &= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{4j-4i}\left(\frac{1}{4i+1}-\frac{1}{4j+1} \right) \\ &= \sum_{\substack{i,j=-n-1 \\ i \neq j}}^n \frac{(-1)^{i+j}}{2j-2i} \cdot \frac{1}{4i+1} \\ &= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i}{4i+1} \sum_{\substack{j=-n-1 \\ i \neq j}}^n \frac{(-1)^j}{j-i} \\ &= \frac{1}{2}\sum_{i=-n-1}^n \frac{(-1)^i }{4i+1}c_{i,n} \\ &= \frac{1}{2}\sum_{i=0}^n a_i \,c_{i,n} \end{aligned} \] 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: \[ \begin{aligned}\left| \left(\sum_{i=0}^n a_i\right)^2 - \sum_{i=0}^n b_i \right| &\le \frac{1}{2} \sum_{i=0}^n \frac{2}{4i+1} \cdot \frac{1}{n-i+1} \\ &\le \sum_{i=0}^n \frac{1}{4n+5}\left( \frac{4}{4i+1} + \frac{1}{n-i+1} \right) \\ &\le \frac{1}{4n+5}\left( 4+ \int_0^{n} \frac{4}{4x+1}\,dx + 1+ \int_0^n \frac{1}{x+1}\, dx \right) \\ &\le \frac{1}{4n+5}\left( 5 + \ln(4n+1) +\ln(n+1)\right) \\ & \to 0 ~\text{ as }~ n \to \infty \end{aligned} \] This concludes the proof. In fact, with the same idea, I can prove a general family of identities: fix an integer \(m \ge 2\) and \(0 < k < m\), then \[ \begin{aligned} & \left( \frac{1}{k} + \frac{1}{m-k} - \frac{1}{m+k} - \frac{1}{2m-k} + \frac{1}{2m+k} + \frac{1}{3m-k} - \cdots \right)^2 \\ =& ~ \left(\sum_{i=-\infty}^\infty \frac{(-1)^i}{im+k}\right)^2 \\ =& ~\sum_{i=-\infty}^\infty \frac{1}{(im+k)^2} \\ =& ~ \frac{1}{k^2} + \frac{1}{(m-k)^2} + \frac{1}{(m+k)^2} + \frac{1}{(2m-k)^2} + \frac{1}{(2m+k)^2} + \frac{1}{(3m-k)^2} + \cdots \end{aligned} \]

Happy Black History Month

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