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: 
There is an explicit formula 
$$
B\_m = \sum\_{k=0}^m\sum\_{v=0}^k(-1)^v\binom kv\frac{ v ^m}{k+1}
$$ 

**Edit.** Put $P(k,v)=\displaystyle (-1)^v\binom kv\frac{ v ^m}{k+1}.$

Then
 $B\_2=P \left( 0,0 \right) +P \left( 1,0 \right) +P \left( 1,1 \right) +P
 \left( 2,0 \right) +P \left( 2,1 \right) +P \left( 2,2 \right).$

Taking into account the calculation 
${\it P} \left( 0,0 \right) =0,{\it P} \left( 1,0 \right) =0,{\it P
} \left( 1,1 \right) =-1/2,{\it P} \left( 2,0 \right) =0,{\it P}
 \left( 2,1 \right) =-2/3,{\it P} \left( 2,2 \right) =4/3
$
you will get easilly $B\_2.$
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 $B\_2=\dfrac16$.