<s>[INST] Sum of a sequence of reciprocals of square of odd natural numbers If $$a_{n}=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+....+\frac{1}{(2n+1)^2}$$where $n\in N$.
Then prove that
$a_{n}<\frac{1}{4}$
 [/INST] adding and subtracting $1 + \frac{1}{2^2} + \frac{1}{4^2} + \ldots + \frac{1}{(2n)^2}$, and using Basel problem we get
$\lim_{n\to \infty}a_n = \sum_{k=1}^{2n+1} \frac{1}{k^2} - \frac{1}{4}\left(\sum_{k=1}^{n} \frac{1}{k^2}\right) - 1 = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} - 1 = \frac{\pi^2}{8} - 1$
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