Question: How is the Bernoulli numbers? For example, found that in internet 
$$\sum\_{n=0}^\infty\frac{B\_nx^n}{n!}=\frac x{e^x-1}$$
but if I want to find $B\_2$ then
$$B\_0+B\_1x+\frac{B\_2x^2}{2}+\sum\_{n=3}^\infty\frac{B\_nx^n}{n!}=\frac x{e^x-1}$$
and I believe this is not much help. 
I want to learn how to calculate Bernoulli numbers to learn how to calculate $\zeta(2n)$.

Answer: 
As suggested by anon, develop $\dfrac{x }{e^x - 1}$ as a Taylor series. You then obtain for the rhs 

$$1 - \frac x 2 + \frac{x^2}{12} - \frac{x^4}{ 720} + \frac{x^6}{ 30240} - \frac{x^8}{ 1209600} + \cdots$$

The lhs write 

$$B\_0 + B\_1 x + B\_2\frac{x^2}{2!} + B\_3 \frac{x^3}{3!} + B\_4\frac{x^4}{4!} +\cdots$$ 

Then $$B\_0 = 1, B\_1 = -1/2, B\_2 = 1/6, B\_3 = 0, B\_4 = -1/30 ....$$
Compare
$$\sum\_{n=0}^{\infty} \frac{B\_nx^n}{n!}=B\_0+B\_1x+\frac{B\_2x^2}2+\frac{B\_3x^3}6+\cdots\left(=\frac{x}{\mathrm e^x-1}\right)
$$
with
$$
\sum\_{n=0}^{\infty} \frac{B\_2x^n}{2}=\frac{B\_2}2+\frac{B\_2x}2+\frac{B\_2x^2}2+\frac{B\_2x^3}2+\cdots\left(=\frac{B\_2}{2(1-x)}\right)$$
To find $B\_2$, either compute $f''(0)$ for $f:x\mapsto\dfrac{x}{\mathrm e^x-1}$, or expand $f(x)$ into powers of $x$ up to the power $x^2$ (this will involve expanding $\mathrm e^x$ into powers of $x$ up to the power $x^3$, but not more). And indeed, the result is $B\_2=\dfrac16$.