Calculating the infinite sum of one over the odd numbers squared: $\sum_{i \ge 0} \frac{1}{(2i+1)^2}$ Can someone tell me how to calculate the following infinite sum?
$$
(1/1^2)+(1/3^2)+(1/5^2)+(1/7^2)+(1/9^2)+(1/11^2)+ \cdots
$$
Don't give me the answer. Can you tell me if this is a geometric series? 

To test if it's geometric, compare the ratio of adjacent terms. For example, $\frac{1/3^2}{1/1^2}$ and $\frac{1/5^2}{1/3^2}$. Are these two quantities equal?
Hint:
$$
\left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \cdots\right) \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \frac{1}{16^2} + \cdots \right) \\
= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \cdots
$$
