<s>[INST] The equality $\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}$ I was reading an article and the author immediately states that the following identity is clear
 $$\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}.$$
I verified that $\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2}$, but I am having trouble seeing how it is obvious that $\sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}$ equates as well. 
I tried expanding terms, but it did not get me anywhere. Am I missing something obvious?
 [/INST] It's because
$$\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{\underset{n\,even}{n = 1}}^\infty \frac{1}{n^2} = \sum_{\underset{n\, odd}{n = 1}}^\infty \frac{1}{n^2} = \sum_{n = 0}^\infty \frac{1}{(2n+1)^2}.$$
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