Infinite sum of squares It is known that $1 + \frac{1}{4}+\frac{1}{9}+\frac{1}{16} + \cdots = \frac{\pi^2}{6}$
. Find the sum
$1 + \frac{1}{9}+\frac{1}{25} + \frac{1}{49} + \cdots$.
What method can we use to answer this?  I tried expressing the 2nd equation into 2 fractions which contain the first summation but i couldnt find one

The usual trick: separate even and odd indices of the sum (things converge absolutely, so you can).
We have $$\frac{\pi^2}{6}
= \sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \frac{1}{(2n)^2}
+\sum_{n=0}^\infty \frac{1}{(2n+1)^2}
$$
but
$$
\sum_{n=1}^\infty \frac{1}{(2n)^2} = \sum_{n=1}^\infty \frac{1}{4n^2}
= \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{4}\frac{\pi^2}{6}
$$
so
$$
\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{8}\,.
$$
