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m The sum one calculates starts from 1, not 0 |
m The sum of odd numbers starts from 0 |
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Let's 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=
=\frac{1}{4}\sum ^{+\infty }_{n=1}\frac{1}{n^{2}} +\displaystyle \ \frac{\pi ^{2}}{8}
</math>
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