Given the sum of this series, find the sum of another series.

Question

I've stumbled across this problem that asks, given this series whose sum is that:

$$\sum_{n=1}^∞ \frac{1}{n^2} = \frac{{\pi^2}}{6}$$

Find the sum of this series: $$\sum_{n=1}^∞ \frac{1}{(2n-1)^2} $$ The second function is just the odd function of the first function, however, I do not know where to proceed from there.

Answer 1

Hint:

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

Answer 2

Hint: $\sum \frac{1}{n^2}= \sum \frac{1}{(2n)^2}+\sum \frac{1}{(2n-1)^2}= \frac{1}{4}\sum \frac{1}{n^2}+\sum \frac{1}{(2n-1)^2}$