<s>[INST] $ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} $ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $ If $ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} 
 $ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $
Dont know what kind of series is this. Please educate. 
How to do such problems?
 [/INST] $$\sum_{n=1}^\infty \frac1{n^2} =\sum_{n=1}^\infty \frac1{(2n)^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}=\frac14 \sum_{n=1}^\infty \frac1{n^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}$$
So, $\sum\limits_{n=1}^\infty \frac1{(2n-1)^2}= \frac34 \sum\limits_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{8}$.
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