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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?
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 $$\sum_{1}^{n} \frac{1}{n^2}=\pi^2/6$$
