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

If $$ \sum_{n=1} ^ \infty {{1} \over {n^2}} = {{\pi^2}\over {6}},$$ find $$ \sum_{n=1}^\infty {{1} \over ({{2n-1}})^2}. $$

I tried an approach using partial fractions and tried to transform ${{1} \over ({{2n-1}})^2} $ into something in terms of $ {{1} \over {n^2}}$ , but so far I haven't had any luck.

Is there some other approach I can use?

share|cite|improve this question
I believe that your $i$’s should all be $1$, not $i$. – Brian M. Scott Mar 20 '13 at 3:13
and $\pi^2/6$, not $\pi/6$. – Jonathan Mar 20 '13 at 3:15
@Jonathan , yes it is indeed pi^2/6, also the summation is over n and not k. – donvoldy666 Mar 20 '13 at 3:27
up vote 11 down vote accepted

HINT: $$\begin{align*} \frac{\pi^2}6&=\sum_{n\ge 1}\frac1{n^2}\\ &=\sum_{n\ge 1}\frac1{(2n-1)^2}+\sum_{n\ge 1}\frac1{(2n)^2}\\ &=\sum_{n\ge 1}\frac1{(2n-1)^2}+\frac14\sum_{n\ge 1}\frac1{n^2} \end{align*}$$

share|cite|improve this answer

Hint: consider $$\sum_1^{\infty}{1\over(2n)^2}$$

share|cite|improve this answer

$$ \sum_{n\geq 1}\frac{1}{n^2}=\sum_{n\geq 1}\frac{1}{(2n)^2}+\sum_{n\geq 1}\frac{1}{(2n-1)^2} $$

share|cite|improve this answer

You could try to find $\sum_{1}^{\infty} \frac{1}{(2n)^2}$ and then decompose a partial sum of $\sum_{1}^{n} \frac{1}{n^2}$ into odd and even numbers and take the limit. By the way $$\sum_{1}^{n} \frac{1}{n^2}=\pi^2/6$$

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.