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
$a_{n}<\frac{1}{4}$

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). 
