About the strange identity $\frac{\pi}{4}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}=\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\cdot\frac{\zeta(4n+2)}{2^{2n+1}}$ I just wonder if this series 
$${\pi\over 4}=\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\cdot{\zeta(4n+2)\over 2^{2n+1}}$$
Is equivalent to
$${\pi \over 4}=\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}?$$
Can anyone have a method to verify these sums?
 A: The generating function for the values of the $\zeta$ function at even, positive integers is well-known:
$$ \frac{1-\pi x \cot(\pi x)}{2}=\sum_{n\geq 1}x^{2n}\zeta(2n) \tag{1}$$
and that is easy to prove by considering the logarithmic derivative of the Weierstrass product for the sine function. By applying a discrete Fourier transform to $(1)$ we get:
$$ \sum_{n\geq 1}x^{4n+2}\zeta(4n+2) = -\frac{\pi x}{12}\left(2\pi x+3\cot(\pi x)-3\coth(\pi x)\right)\tag{2} $$
and by termwise integration:
$$\sum_{n\geq 1}\frac{x^{4n+2}}{4n+2}\zeta(4n+2)=\frac{1}{12}\left(-\pi^2 x^2+3\log\frac{\pi x}{\sin(\pi x)}-3\log\frac{\pi x}{\sinh(\pi x)}\right)\tag{4} $$
as well as:
$$\sum_{n\geq 1}\frac{x^{2n+1}}{4n+2}\zeta(4n+2)=\frac{1}{12}\left(-\pi^2 x+3\log\frac{\pi\sqrt{x}}{\sin(\pi\sqrt{x})}-3\log\frac{\pi\sqrt{x}}{\sinh(\pi\sqrt{x})}\right)\tag{5} $$
To compute the first series given by the OP, it is enough to evaluate $(5)$ at $x=\frac{i}{2}$.
That ensures
$$ \sum_{n\geq 0}\frac{(-1)^n}{2n+1} = \frac{\pi}{4} = \sum_{n\geq 0}\frac{(-1)^n}{2n+1}\cdot\frac{\zeta(4n+2)}{2^{2n+1}}\tag{6}$$
as wanted.
