# Finding $\sum_{n=1}^\infty {{1} \over ({2n-1})^2}$ given $\sum_{n=1} ^ \infty {{1} \over {n^2}} = {{\pi^2}\over {6}}$

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?

• 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
• – Martin Sleziak Jul 2 '19 at 16:24

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*}
Hint: consider $$\sum_1^{\infty}{1\over(2n)^2}$$
$$\sum_{n\geq 1}\frac{1}{n^2}=\sum_{n\geq 1}\frac{1}{(2n)^2}+\sum_{n\geq 1}\frac{1}{(2n-1)^2}$$
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 $$\sum_{1}^{n} \frac{1}{n^2}=\pi^2/6$$