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If $$s_*=\sum_1^\infty \frac{1}{n^2}$$ show $$\frac{3}{4}s_*=\sum_0^\infty\frac{1}{(2k+1)^2}$$

This is homework and i have no idea how to get into the right direction. I was looking for a rearrangement of $s_*$ from which i could show the desired result but i had no luck. I would be happy about any tips - no full solutions please.

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HINT: The second sum ranges over all odd numbers. How does this relate to the sum over all even numbers?

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$$s_*=\sum_1^\infty \frac{1}{n^2}=\sum_1^\infty \frac{1}{(2n)^2}+\sum_0^\infty \frac{1}{(2n+1)^2}= \frac{1}{4}\sum_1^\infty\frac{1}{n^2}+\sum_0^\infty \frac{1}{(2n+1)^2}$$ Hence $$s_*=\frac{1}{4}s_*+\sum_0^\infty \frac{1}{(2n+1)^2}$$

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