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
$a_{n}<\frac{1}{4}$
 A: Here's a sketch of a direct proof that only requires calculus.  Note that for each $k$,  we have $$\frac{1}{(2k + 1)^2} \leq \int_{k-1}^k \frac{1}{(2x + 1)^2}\,dx.$$
This implies that each $a_n$ is less than $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \int_{3}^\infty \frac{1}{(2x + 1)^2}.$$
Calculating this integral directly gives the upper bound of $$\frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{14} \approx .242 < \frac{1}{4}.$$
This method is a good technique for computing an upper bound on a sum (or a corresponding lower bound by altering the integral). 
A: The $n$th term is less than $\frac{1}{4n(n+1)}$. This replaces your series with a standard series that sums to $\frac{1}{4}$.
A: $$\sum_{n\geq 1}\frac{1}{(2n+1)^2}<\sum_{n\geq 1}\frac{1}{2n(2n+2)}\stackrel{\text{telescopic!}}{=}\frac{1}{4}. $$
Since the LHS equals $\frac{\pi^2}{8}-1$ by Fourier series, this turns out to be a proof of $\pi^2<10$.
A: adding and subtracting $1 + \frac{1}{2^2} + \frac{1}{4^2} + \ldots + \frac{1}{(2n)^2}$, and using Basel problem we get
$\lim_{n\to \infty}a_n = \sum_{k=1}^{2n+1} \frac{1}{k^2} - \frac{1}{4}\left(\sum_{k=1}^{n} \frac{1}{k^2}\right) - 1 = \frac{\pi^2}{6} - \frac{1}{4}\frac{\pi^2}{6} - 1 = \frac{\pi^2}{8} - 1$
