Question

Prove that:
[\sum_{n=1}^{\infty}\frac{1}{(2n - 1)^2}] - [\sum_{n=1}^{\infty}\frac{1}{4n^2} =
[\sum_{n=1}^{\infty}\frac{1}{n^2}]

Answer 1

\[\huge [\sum_{n=1}^{\infty}\frac{1}{(2n - 1)^2}]-[\sum_{n=1}^{\infty}\frac{1}{4n^2}]=[\sum_{n=1}^{\infty}\frac{1}{n^2}]\]