MHB Prove the Famous Result: \(\sum_{k = 1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}\)

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Hi MHB Community,

I'm sorry I haven't been around. For several months I've been very sick. I wish you all a Happy New Year! In respect of the MHB equations above, here's a good problem to start the new year:

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Prove the famous result \[\sum_{k = 1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}\] Use any method(s) you like.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one answered this week's problem. You can read my solution below.

Spoiler

Apply the Parseval identity to the function $f(x) = x$, $-\pi \le x \le \pi$:
\[\sum_{-\infty}^\infty \lvert c_n\rvert^2 = \frac{1}{2\pi} \int_{-\pi}^\pi \lvert f(x)\rvert^2\, dx\]
where the $c_n$ are the coefficients of the complex Fourier series of $f$. We have \(c_n = \frac{1}{2\pi}\int_{-\pi}^\pi x e^{-inx}\, dx\), so that $c_0 = 0$ and for $n\neq 0$, integration by parts yields \[c_n = \frac{1}{2\pi}\left\{x\left(-\frac{1}{in}e^{-inx}\right) - \left(-\frac{1}{n^2}e^{-inx}\right)\right\}\Bigl|_{x = -L}^L = \frac{(-1)^{n+1}}{in}\] The Parseval equation above reduces to \[2\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{1}{2\pi}\int_{-\pi}^\pi x^2\, dx\] or \[2\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{3}\] Therefore, \[\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\]