Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?$$\sum_{n=1}^{+\infty}{\frac{1}{(2n-1)^2}} = \frac{\pi^2}{8}$$

share|cite|improve this question
The question has no answer, except that the sums one can realize are some (but not all) numbers in $[0,\pi^2/6]$. The specific instance in the second part of the question is answered below. – Did Oct 14 '12 at 9:17
The first sum is absolutely convergent, so you may reorder the terms. Consider splitting into even and odd terms and the second sum follows almost instantly. – fretty Oct 14 '12 at 9:30
By the way, $\pi^2/8$ in the RHS of the second identity should be replaced by $(\pi^2/8)-1$. – Did Oct 14 '12 at 9:57
I changed the LHS. – Cydonia7 Oct 14 '12 at 11:54
up vote 13 down vote accepted

Observe that:

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{6}$$


$$\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{24} $$

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*}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.