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Deacon Vorbis, I think you did not read the references carefully enough. The proof that I add is exactly the proof published by Euler. I just rewrote it using modern mathematical tools so that readers can understand it more easily and skip certain details that would seem tedious nowadays. I did not write that it is Euler's proof. I just wrote one historical fact : Euler published this proof. Please mention sources on the talk page to prove what you assert. |
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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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:<math>\displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}=\frac{\pi ^{2}}{6}.</math>
== The first truly rigorous proof published by Euler ==
In 1741, Euler published a second proof. <ref>{{Cite book|title=Opera Omnia, series 1, volume 14|last=Euler|first=Leonhard|publisher=|year=1741|isbn=|___location=|pages=177-186}}</ref><ref>{{Cite web|url=http://eulerarchive.maa.org/hedi/HEDI-2004-03.pdf|title=How Euler did it|last=Sandifer|first=Ed|date=March 2004|website=MAA Online|url-status=live|archive-url=http://archive.wikiwix.com/cache/?url=http%3A%2F%2Feulerarchive.maa.org%2Fhedi%2FHEDI-2004-03.pdf|archive-date=|access-date=}}</ref><ref>{{Cite web|url=https://www.apmep.fr/IMG/pdf/Article_probleme_Bale.pdf|title=Euler and Basel problem (page 17-19)|last=Association of Professors of mathematics of public education|first=Apmep (France)|date=|website=apmep.fr|url-status=live|archive-url=|archive-date=|access-date=}}</ref><ref>{{Cite journal|last=|first=|date=October 2011|title=How Euler could have used information & communication technology|url=https://pdfs.semanticscholar.org/5a8a/5a18e10917d364b61282eb76fd57024bdc0d.pdf|journal=National institute of education Singapore|volume=|pages=3|via=}}</ref>, six years after his first proof :
First of all, <math>\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}{\rm d}x=\left[\frac{(\arcsin x)^2}2\right]_0^1=\frac{\pi^2}8</math>
Now one can use the [[List of mathematical series|power series of Arcsin]] :
<math>\begin{aligned}
& \int ^{1}_{0} \ (\sum ^{+\infty }_{n=0}\frac{(2n)!}{\left( n!\ 2^{n}\right)^{2} (2n+1)\ } \ \frac{x^{2n+1}}{\sqrt{1-x^{2}}} ){\rm d}x\ & = & \ \ \sum ^{+\infty }_{n=0}\frac{(2n)!}{\left( n!\ 2^{n}\right)^{2} (2n+1)\ } \ \int ^{1}_{0}\frac{x^{2n+1}}{\sqrt{1-x^{2}}} {\rm d}x\\
& & & \\
& & = & \ \ \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{aligned}
</math>
The sum and the integral signs were interchanged thanks to Beppo Levi's [[monotone convergence theorem]] for Lebesgue integral. Euler wrote that <math>\int ^{1}_{0}\frac{x^{2n+1}}{\sqrt{1-x^{2}}} dx = \frac{\left( n!\ 2^{n}\right)^{2}}{(2n+1)!}
</math> using integration by parts. But on can also use the integration by substitution <math>x=sin(t)
</math> and recognize [[Wallis' integrals]].
One can separate even and odd numbers since all the terms are positive :
<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>
==A rigorous proof using Fourier series==
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