Summation of series. If $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}.$$
Then find the value of $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+..$$
The answer given in the book is $\frac{\pi^2}{8}$.
How can I find this and also how to find summation of fractions? Thank you!
 A: $$\sum_{1\le r<\infty}\frac1{r^2}=\sum_{1\le r<\infty}\frac1{(2r)^2}+\sum_{1\le r<\infty}\frac1{(2r-1)^2}=\frac14\sum_{1\le r<\infty}\frac1{r^2}+\sum_{1\le r<\infty}\frac1{(2r-1)^2}$$
$$\implies \sum_{1\le r<\infty}\frac1{(2r-1)^2}=\left(1-\frac14\right)\sum_{1\le r<\infty}\frac1{r^2} $$
A: Riemann's zeta function is given by, $$ \zeta (s)=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+.... $$
Group even and odd number containing terms of this function as shown below, 
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+(\frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}....) $$
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+\frac{1}{2^s}(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+.... ) $$
The second bracket is nothing but the function itself, hence we can write,
$$ \zeta (s)=(\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....)+\frac{1}{2^s}\zeta (s) $$
So, finally we get, 
$$ \frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+....=\zeta (s)(1-\frac{1}{2^s})$$
Hope this cleared your doubt.
Now, for you case s=2 . So, we get
$$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....=\zeta (2)(1-\frac{1}{2^2})$$
$$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....=\frac{\pi^2}{6}(1-\frac{1}{4})$$
$$ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....=\frac{\pi^2}{8}$$
