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Can we prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier series?

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marked as duplicate by user147263, Andrew D. Hwang, Jack D'Aurizio, user98602, Mathmo123 Aug 10 '14 at 23:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Yes. The most common way to do this is attributed to Euler. It does still require Maclaurin series, however.

Consider the Maclaurin polynomial for $\frac{\sin x}{x}$:

$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$

However, note that this is a polynomial $p(x)$ with zeroes $\{\pm k\pi\;|\;k \in \Bbb N\}$, and for which $p(0) = 1$. These two properties mean that

$$\frac{\sin x}{x} = \left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{2\pi}\right)\cdots$$

And by multiplying adjacent terms,

$$\frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right)\cdots$$

Equating the $x^2$ terms in the Maclaurin polynomial and its factored form yields

$$-\frac{x^2}{3!} = -x^2\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots\right)$$

And multiplying both sides by $-\frac{pi^2}{x^2}$ gives us

$$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \cdots$$

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    $\begingroup$ The modern objections to this argument should probably be mentioned. One cannot say "an infinite series is just like a polynomial" until convergence issues have been resolved. In this case, Taylor series arguments give us that $f_n(x) = \sum_{k=0}^n \frac{(-1)^k x^{2k}}{(2k+1)!}$ converges to $f(x) = \frac{\sin x}{x}$ as $n \to \infty$. But to use the argument above, one must justify that $g_n(x) = \prod_{k=1}^n 1 - \frac{x^2}{k^2 \pi^2}$ also converges to $f(x)$ as $n \to \infty$. One method would be to apply Weierstrass's factorization theorem (since $\frac{\sin z}{z}$ is holomorphic). $\endgroup$ – Ian Aug 10 '14 at 22:55
  • $\begingroup$ Right, the objections are valid, and though Euler's method of proof here was eventually vindicated, it should be noted that infinite series can't necessarily be treated in the same way as finite polynomials. $\endgroup$ – Fargle Aug 10 '14 at 22:57
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Yes. You are hereby awarded this Wikipedia article: http://en.wikipedia.org/wiki/Basel_problem.

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    $\begingroup$ You are hereby awarded a downvote for posting link-only answer. $\endgroup$ – user147263 Aug 10 '14 at 22:52
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This question Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ already contains proofs that does not rely on Fourier series. Also Robin Chapman has a collection of proofs on his homepage.

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