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Reverted 4 edits by Contribute.Math (talk): Bulk of the concerns still not addressed...will elaborate on talk page in a moment (TW) |
I mentionned the source of the proof this time. The source is the French page of the same wikipedia article. The latter refer to French sources (L. Euler, « Démonstration de la somme de cette suite 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + etc. », Journal lit. d'Allemagne, de Suisse et du Nord, vol. 2, 1743, p. 115-127 (lire en ligne [archive]) (E63, Opera Omnia, I.14, p. 177-186), écrit en 1741. Voir aussi sa lettre d'avril 1742 [archive] (OO396) à Clairaut.) |
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Using the [[Weierstrass factorization theorem]], it can also be shown that the left-hand side is the product of linear factors given by its roots, just as we do for finite polynomials (which Euler assumed as a [[heuristic]] for expanding an infinite degree [[polynomial]] in terms of its roots, but is in general not always true for general <math>P(x)</math>):<ref>A priori, since the left-hand-side is a [[polynomial]] (of infinite degree) we can write it as a product of its roots as
:<math>\begin{align}
\sin(x) & = x (x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2) \cdots \\
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and this completes the proof.
== Euler's second proof ==
This is the first accepted proof of the theorem. It was made by Euler in 1741<ref>{{Cite web|url=https://fr.wikipedia.org/wiki/Probl%C3%A8me_de_B%C3%A2le|title=Problème de Bâle|last=|first=|date=|website=|url-status=live|archive-url=|archive-date=|access-date=}}</ref>, six years after his first proof. The rigor of the latter was challenged at that time because the Weierstrass factorization theorem had not been discovered yet.
First of all, <math>\displaystyle \int ^{\frac{\pi }{2}}_{0} \operatorname{Arcsin}( \sin\ t) \ dt\ =\int ^{\frac{\pi }{2}}_{0} t\ dt\
=\frac{\pi ^{2}}{8}
</math>
Now one can use the [[List of mathematical series|power series of Arcsin]] :
<math>\begin{align}
& \int ^{\frac{\pi }{2}}_{0} \ (\sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \sin( t)^{2n+1}) dt\ & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \int ^{\frac{\pi }{2}}_{0} \sin( t)^{2n+1} dt\\
\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{( 2n) !}{\left( n!\ 2^{n}\right)^{2}( 2n+1) \ } \ \frac{\left( n!\ 2^{n}\right)^{2}}{( 2n+1) !}\\
& & = & \ \ \sum ^{+\infty }_{n=0}\frac{1}{( 2n+1)^{2}} \\
& & = & \ \ \frac{\pi ^{2}}{8}
\end{align}
</math>
The sum and the integral signs were interchanged thanks to Beppo Levi's [[monotone convergence theorem]] for Lebesgue integral. Then, [[Wallis' integrals]] enabled us to integrate the powers of sine.
One can separate even and odd numbers :
<math>\displaystyle \sum ^{+\infty }_{n=1}\frac{1}{n^{2}} =\sum ^{+\infty }_{n=1}\frac{1}{( 2n)^{2}} \ +\sum ^{+\infty }_{n=0}\frac{1}{( 2n+1)^{2}} \
=\frac{1}{4}\sum ^{+\infty }_{n=1}\frac{1}{n^{2}} +\displaystyle \ \frac{\pi ^{2}}{8}
</math>
Finally,
<math>\displaystyle \sum ^{+\infty }_{n=1}\frac{1}{n^{2}}
=\frac{4}{3}\frac{\pi ^{2}}{8} \ =\frac{\pi ^{2}}{6}
</math>
==Other identities==
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