Infinite Series with Pi I have this homework problem, that I'm stuck on.
We know that:$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
I have to find the sum of: $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots$$
I came up with this equation:
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}\right)^2$$
I know that the answer is $\frac{\pi^2}{8}$ I found $a=1$, but can't seem to figure it out...
 A: Of more historical than mathematical interest, Euler's original approach:
$\cos x = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+...$
So $\cos x$ may be regarded as a polynomial with roots $\pm \pi/2, \pm 3\pi/2, \pm 5\pi/2,...$ 
That is, 
$$\cos x = (x - \frac{\pi}{2})(x + \frac{\pi}{2})(x - \frac{3\pi}{2})(x + \frac{3\pi}{2})...$$
$$= (x^2-\frac{\pi^2}{4})(x^2-\frac{3^2\pi^2}{4})(x^2-\frac{5^2\pi^2}{4})... $$
which we can write as 
$$(*) \hspace{10mm}1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+...  = A(1-\frac{4x^2}{\pi^2})(1-\frac{4x^2}{3^2\pi^2})(1-\frac{4x^2}{5^2\pi^2})...$$ 
for some constant A; and since $\cos x \to 1 ~~\text{as}~~ x\to 0,$ A must equal $1$.
Then from (*) we can equate coefficients of $x^2:$
$$-\frac{1}{2!} = -\frac{4}{\pi^2}- \frac{4}{3^2\pi^2}- \frac{4}{5^2\pi^2}-... $$
so finally 
$$\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}+... $$
A: Hint:
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$
