Prove the following: $$\sum_{k=1}^{\infty }\frac{1}{(2k-1)^{2}}=\frac{\pi ^{2}}{8}$$
I don't really know how to prove this, will assuming that $$cos(x)=\sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k}}{(2k)!}$$ help?
 A: If we admit that
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$
Then
$$\sum_{n=1}^\infty \frac{1}{(2n)^2}=\frac{\pi^2}{24}$$
And, finally :
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{3\pi^2}{24} = \frac{\pi^2}{8}$$
A: If you must use $\cos (x)$, you can focus on zeros of it.  $(...,-\frac{(2n-1)\pi}{2},....,-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},...,\frac{(2n-1)\pi}{2},..)$  where $n$ is a  positive integer
Hint:
 $$\cos x = \cdots\left(1+\frac{x}{\frac{3\pi}{2}}\right)\left(1+\frac{x}{\frac{\pi}{2}}\right)\left(1-\frac{x}{\frac{\pi}{2}}\right)\left(1-\frac{x}{\frac{3\pi}{2}}\right)\cdots =\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)\cdots$$
Take derivatives of both side and  divide to $\cos (x)$ .
Please let me know if you cannot go forward.
EDIT:
$$\frac{\cos' x}{\cos x} =\frac{-\sin x}{\cos x}  =\cfrac{\left(-\frac{2x}{\frac{\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{5^2\pi^2}{4}}\right)\cdots+\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)\left(-\frac{2x}{\frac{3^2\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{5^2\pi^2}{4}}\right)\cdots}{\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)\left(1-\frac{x^2}{\frac{5^2\pi^2}{4}}\right)\cdots}$$
$$\frac{\cos' x}{\cos x} =\frac{-\sin x}{\cos x}  =-\frac{8x}{\pi^2}[\cfrac{\left(\frac{1}{1}\right)}{\left(1-\frac{x^2}{\frac{\pi^2}{4}}\right)}+\cfrac{\left(\frac{1}{3^2}\right)}{\left(1-\frac{x^2}{\frac{3^2\pi^2}{4}}\right)}+....]$$
Is it clear what I mean now? I believe you can handle after that.
