<s>[INST] Proving $1+\frac{4}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{4}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}+\frac{4}{10^2}+\frac{1}{11^2}+\cdots=\frac{\pi ^2}{4}$ Proving $$1+\frac{4}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{4}{6^2}+\frac{1}{7^2}+\frac{1}{9^2}+\frac{4}{10^2}+\frac{1}{11^2}+\cdots=\frac{\pi ^2}{4}$$
Firstly, I thought to prove it by comparison the terms with the terms of $1/n^2$ , but the problem with the missing terms, so I couldn't  reach to the proving . Can anybody help? Best regards.
 [/INST] This sum is explicitly:
$$\sum_{n=1}^\infty\frac{1}{n^2}+3\frac{1}{(2n)^2}-4\frac{1}{(4n)^2}$$
You can check for yourself that this eliminates $n$'s which are a multiple of four, while multiplying by four terms with even $n$ but which are not a multiple of four. In other terms:
$$\sum_{n=1}^\infty\frac{1}{n^2}\left(1+\frac{3}{4} - \frac{1}{4}\right)=\frac{\pi^2}{6}\cdot\frac{3}{2}=\frac{\pi^2}{4}$$
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