$ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} $ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $

Question

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?

Answer 1

$$\sum_{n=1}^\infty \frac1{n^2} =\sum_{n=1}^\infty \frac1{(2n)^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}=\frac14 \sum_{n=1}^\infty \frac1{n^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}$$

So, $\sum\limits_{n=1}^\infty \frac1{(2n-1)^2}= \frac34 \sum\limits_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{8}$.

Answer 2

you can use the relation $$\sum_{n=1}^\infty \frac{1}{(2n-1)^k}=(1-\frac{1}{2^k})\sum_{n=1}^\infty\frac{1}{n^k}$$ when the $k$ even integer number($k\geq2) $