Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Assuming we know that : $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}$$
How do you find the sum of a series where all terms are in this one ?
For instance, how do you prove that ?$$\sum_{n=1}^{+\infty}{\frac{1}{(2n-1)^2}} = \frac{\pi^2}{8}$$
Observe that:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{6}$$
and
$$\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{24} $$
therefore \begin{eqnarray*} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} & = & \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\ & = & \frac{\pi^2}{6} - \frac{\pi^2}{24} \\ & = & \frac{\pi^2}{8} \end{eqnarray*}
