Bernoulli Numbers are rationals? There is a explicit form that admit Bernoulli numbers rationals but there is another definition where the Bernoulli numbers are $B_n$, such that $\displaystyle \frac{x}{e^x-1}= \sum_{n=0}^\infty B_n \frac{x^n}{n!}$. 
How can I prove that?? or it's equivalence?

We have
$$\frac{x}{e^x-1} = \frac{x}{x + x^2/2 + x^3/6 + x^4/24 + \cdots} = \frac{1}{1 + x/2 + x^2/6 + x^3/24 + \cdots}.$$
Now, if we let $f(x) := \frac{x}{2} + \frac{x^2}{6} + \frac{x^3}{24} + \cdots$, then
$$\frac{x}{e^x-1} = \frac{1}{1 + f(x)} = 1 - f(x) + (f(x))^2 - (f(x))^3 - \cdots.$$
In this expansion, each power of $f(x)$ has rational coefficients, and since $f(x)$ is divisible by $x$, the final coefficient of $x^n$ is the same as the coefficient of $x^n$ in $1 - f(x) + (f(x))^2 - (f(x))^3 + \cdots + (-1)^n (f(x))^n$.
