Basel problem: Difference between revisions

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:<math>\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots</math> .

The sum of the series is approximately equal to 1.644934.<ref>{{Cite OEIS|1=A013661}}</ref> The Basel problem asks for the ''exact'' sum of this series (in [[Closed-form expression|closed form]]), as well as a [[mathematical proof|proof]] that this sum is correct. Euler found the exact sum to be {{sfrac|{{pi}}²|6}} and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, although he was later proven correct, and it was not until 1741 that he was able to produce [[Basel problem#Euler's second proof|a truly rigorous proof]].

==Euler's approach==

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Then

:<math>\displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}+t^{2}}=\frac{\pi \left(te^{2\pi t}+t\right)-e^{2\pi t}+1}{2\left(t^{2}e^{2\pi t}-t^2\right)}=-\frac{1}{2t^{2}}+\frac{\pi}{2t}\coth(\pi t).</math>

=\frac{\pi \left(te^{2\pi t}+t\right)-e^{2\pi t}+1}{2\left(t^{2}e^{2\pi t}-t^2\right)}

=-\frac{1}{2t^{2}}+\frac{\pi}{2t}\coth(\pi t)

=\frac{\pi t \coth(\pi t)-1}{2t^2}.</math>

Now we take the [[limit (mathematics)|limit]] as <math>t</math> approaches zero and use [[l'hôpital's rule|L'Hôpital's rule]] thrice:

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:<math>\displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}=\frac{\pi ^{2}}{6}.</math>

== Euler's second proof ==

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}}} )dx\ & = & \ \ \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}}} dx\\

& & & \\

& & = & \ \ \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{14}{2t^{23}}+\frac{\pi ^{2}}{2t8} \coth( =\frac{\pi t)^{2}}{6}

</math>

==A rigorous proof using Fourier series==