Basel problem: Difference between revisions

Revision as of 09:09, 6 December 2019

Revision as of 12:52, 6 December 2019

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Let's separate even and odd numbers :

<math>\displaystyle \sum ^{+\infty }_{n=01}\frac{1}{n^{2}} =\sum ^{+\infty }_{n=1}\frac{1}{( 2n)^{2}} \ +\sum ^{+\infty }_{n=01}\frac{1}{( 2n+1)^{2}} \

=\frac{1}{4}\sum ^{+\infty }_{n=01}\frac{1}{n^{2}} +\displaystyle \ \frac{\pi ^{2}}{8}

</math>

Finally,

<math>\displaystyle \sum ^{+\infty }_{n=01}\frac{1}{n^{2}}

=\frac{4}{3}\frac{\pi ^{2}}{8} \ =\frac{\pi ^{2}}{6}

</math>