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  4. It is given that ∑ limitsr=1∞ (1/(2r-1)2)=(π 2/8), then ∑ limitsr=1∞ (1/r2) is equal to

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Q. It is given that $ \sum\limits_{r=1}^{\infty }{\frac{1}{{{(2r-1)}^{2}}}}=\frac{{{\pi }^{2}}}{8}, $ then $ \sum\limits_{r=1}^{\infty }{\frac{1}{{{r}^{2}}}} $ is equal to

Bihar CECEBihar CECE 2010

Solution:

We have, $ \sum\limits_{r=1}^{\infty }{\frac{1}{{{(2r-1)}^{2}}}}=\frac{{{\pi }^{2}}}{8} $
$ \Rightarrow $ $ \frac{1}{{{1}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{5}^{2}}}+...\infty =\frac{{{\pi }^{2}}}{8} $ ..... (i)

Let $ x=\sum\limits_{r=1}^{\infty }{\frac{1}{{{r}^{2}}}=\frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+...\infty } $
$ \Rightarrow $ $ x=\left( \frac{1}{{{1}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{5}^{2}}}+...\infty \right) $
$ +\left( \frac{1}{{{2}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{6}^{2}}}+...\infty \right) $
$ \Rightarrow $ $ x=\frac{{{\pi }^{2}}}{8}+\frac{1}{4}\left[ \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+...\infty \right] $
$ \Rightarrow $ $ x=\frac{{{\pi }^{2}}}{8}+\frac{1}{4}.x $
$ \Rightarrow $ $ x-\frac{x}{4}=\frac{{{\pi }^{2}}}{8} $
$ \Rightarrow $ $ \frac{3x}{4}=\frac{{{\pi }^{2}}}{8} $
$ \Rightarrow $ $ x=\frac{{{\pi }^{2}}}{6} $