Formula for $\zeta(3)$ -verification By simple manipulating with some series I have found the following formula for $\zeta(3)$:
$$\zeta(3)=\frac27\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k+2}}{(2k+2)!},$$
where $b_k$ are Bernoulli numbers, defined from the equations:
$$
B_0=1,\quad B_k=-\frac{1}{k+1}\sum_{i=0}^{k-1}\binom{k+1}{i} B_i,\quad k=1,2,3,\dots
$$
Questions: Is this formula for $\zeta(3)$ correct? Can somebody check it numerically? If it's correct, is this well known identity or not? Thanks for your help.
Added. By usimg my identities another interesting formulas follows:
$$\ln 2=\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k}}{(2k+1)!}$$
$$\zeta(3)=\frac45\sum_{k=0}^{\infty}(-1)^kB_{2k}\frac{\pi^{2k+2}}{(2k+3)!}$$
 A: So,I put here my derivation, but I'm not sure, if all my steps are fully correct. 
The key idea is very simple. We just put $x=-i\pi$ in this formula:
\begin{equation} 2\zeta(3)+\zeta(2)x+\frac{x^3}{12}+\sum_{k=0}^{\infty}B_{2k}\frac{x^{2k+2}}{(2k+2)!}=2\sum_{k=1}^{\infty}\frac{e^{kx}}{k^3}-x\sum_{k=1}^{\infty}\frac{e^{kx}}{k^2},\quad x\in(-2\pi,0\,\rangle\end{equation}
plus some simplifacations. That's all.
(Formula can be easily obtaided from two expansions:
$$\frac{x}{e^x-1}=-\frac{x}2+\sum_{k=0}^{\infty}B_{2k}\frac{x^{2k}}{(2k)!},\quad x\in(-2\pi,2\pi),$$
$$\frac{x}{e^x-1}=-x\frac{1}{1-e^x}=-x\sum_{k=0}^{\infty}e^{kx},\quad x\in(-\infty,0),$$
so formal double integration gives:
$$\int_0^x\!\!\int_0^x\frac{x}{e^x-1} dx dx=-\frac{x^3}{12}+\sum_{k=0}^{\infty}B_{2k}\frac{x^{2k+2}}{(2k+2)!}$$
$$\int_0^x\!\!\int_0^x\frac{x}{e^x-1} dx dx=2\sum_{k=1}^{\infty}\frac{e^{kx}}{k^3}-x\sum_{k=1}^{\infty}\frac{e^{kx}}{k^2}-\frac{x^{3}}{6}-\zeta(2)x-2\zeta(3)$$
which implies the desired identity.)
Now back to my derivation of formula for $\zeta(3)$.
Putting $x=-i\pi$ we have:
$$
2\zeta(3)-i\pi\zeta(2)+\frac{(-i\pi)^3}{12}+\sum_{k=0}^{\infty}B_{2k}
\frac{(-i\pi)^{2k+2}}{(2k+2)!}=2\sum_{k=1}^{\infty}\frac{e^{(-i\pi)k}}{k^3}+i\pi\sum_{k=1}^{\infty}\frac{e^{(-i\pi)k}}{k^2}
$$
and because 
$$e^{-i\pi}=\cos(-\pi)+i\sin(-\pi)=-1$$
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}=-\frac12\zeta(2)$$
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}=-\frac34\zeta(3),$$
we have step by step:
\begin{align} 2\zeta(3)-i\pi\zeta(2)+\frac{i\pi^3}{12}-\sum_{k=0}^{\infty}B_{2k}\frac{(-1)^{k}\pi^{2k+2}}{(2k+2)!}&=2\sum_{k=1}^{\infty}\frac{(-1)^k}{k^3}+i\pi\sum_{k=1}^{\infty}\frac{(-1)^k}{k^2}\\
2\zeta(3)-i\pi\zeta(2)+\frac{i\pi^3}{12}-\sum_{k=0}^{\infty}B_{2k}\frac{(-1)^{k}\pi^{2k+2}}{(2k+2)!}&=-\frac32\zeta(3)-i\pi\frac12\zeta(2)\\
\frac72\zeta(3)-\frac{i\pi}2\zeta(2)&=\sum_{k=0}^{\infty}B_{2k}\frac{(-1)^{k}\pi^{2k+2}}{(2k+2)!}-\frac{i\pi^3}{12},
\end{align}
so we immediately obtain even two formulas:
$$\zeta(2)=\frac{\pi^2}6$$
$$\zeta(3)=\frac{2}7\sum_{k=0}^{\infty}(-1)^{k}B_{2k}\frac{\pi^{2k+2}}{(2k+2)!}.$$
