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and this completes the proof.
== A proof with Wallis' integrals ==
It is noticeable that <math>\displaystyle \int ^{\frac{\pi }{2}}_{0} Arcsin( sin\ t) \ dt\ =\int ^{\frac{\pi }{2}}_{0} t\ dt\
=\frac{\pi ^{2}}{8}
</math>
Now let's use the power series of Arcsin :
<math>\begin{array}{l}
\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{array}
</math>
The term-by-term integration theorem was used as well as the values of Wallis' integrals.
Then,
<math>\displaystyle \sum ^{+\infty }_{n=0}\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=0}\frac{1}{n^{2}} +\displaystyle \ \frac{\pi ^{2}}{8}
</math>
Finally,
<math>\displaystyle \sum ^{+\infty }_{n=0}\frac{1}{n^{2}}
=\frac{4}{3}\frac{\pi ^{2}}{8} \ =\frac{\pi ^{2}}{6}
</math>
==Other identities==
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