Proving interesting formula involving zeta function I calculated fourier series of $x^s$ where $s$ is even parameter grater than zero and then I let $x=\pi$. After a few manipulations I've derived following formula:
$$\frac{s}{(2s+1)!}=\sum_{k=1}^s \frac{(-1)^{k+1} \ \zeta(2k)}{(2s-2k+1)!\ \pi^{2k}}, \ s \in \Bbb{N}_+$$
Where $\zeta(s)$ denotes Riemann zeta function.
 I was wondering, if it is possible to derive this formula in other ways. Please let me know, if you know how can this formula be proven. Solutions involving some theorems about zeta functions are highly requested.
 Thanks for all the help.
 A: Here we seek to prove that
$$\frac{n}{(2n+1)!} = \sum_{k=1}^n
\frac{(-1)^{k+1} \zeta(2k)}{(2n-2k+1)! \pi^{2k}}.$$
Using the closed form of the zeta function values at positive even 
integers this  becomes 
$$\frac{n}{(2n+1)!} = \sum_{k=1}^n
\frac{(-1)^{k+1}}{(2n-2k+1)! \pi^{2k}}
(-1)^{k+1} \frac{B_{2k} (2\pi)^{2k}}{2(2k)!}$$
or
$$n = \sum_{k=1}^n {2n+1\choose 2k} B_{2k} 2^{2k-1}.$$
We use $z/(\exp(z)-1)$ as the EGF of the Bernoulli numbers appearing
here. Now the odd index Bernoulli numbers are zero except for $B_1 =
-1/2$ so we may write this as
$$n = - {2n+1\choose 0} \times 1 \times \frac{1}{2}
-{2n+1\choose 1} \times \left(-\frac{1}{2}\right) \times 1 +
\sum_{k=0}^{2n+1} {2n+1\choose k} B_{k} 2^{k-1}$$
or
$$0 = \sum_{k=0}^{2n+1} {2n+1\choose k} B_{k} 2^{k-1}.$$
Viewing this as the convolution of two EGFs we find for the RHS
$$\frac{1}{2} (2n+1)! [z^{2n+1}]
\frac{2z}{\exp(2z)-1} \exp(z)
\\ = (2n+1)! [z^{2n+1}] \frac{z}{\exp(z)-\exp(-z)}.$$
Now with
$$f(z) = \frac{z}{\exp(z)-\exp(-z)}$$
we get
$$f(-z) = \frac{-z}{\exp(-z)-\exp(z)} = f(z)$$
so $f(z)$ is even and all the odd index terms of its Taylor series
are zero, thus proving the claim. 
 Remarks. The singularity at zero is removable. We thus obtain for 
the odd  index terms by the Cauchy Coefficient Formula
$$f_{2n+1} =
\frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{1}{z^{2n+2}} f(z) \; dz $$
which on setting $z=-w$ becomes
$$\frac{1}{2\pi i} \int_{|e^{i\pi} w|=\epsilon}
\frac{(-1)^{2n+2}}{w^{2n+2}} f(-w) \; (-dw) 
\\ = - \frac{1}{2\pi i} \int_{|w|=\epsilon}
\frac{1}{w^{2n+2}} f(w) \; dw = -f_{2n+1}.$$
Hence $f_{2n+1} = -f_{2n+1}$ or $2f_{2n+1} = 0$ or $f_{2n+1} = 0.$
