# How does one get the Bernoulli numbers via the generating function?

Here is the definition:

Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$

I've tried to naively expand $\frac{x}{e^x-1}$ around $x_0=0$ and didn't quite understand what should be done with all the $(e^x-1)^{k+1}$ in the denominators of $f^{(k)}(0)$?

How does one get to $$1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \ ...$$ directly from the definition (without rearranging the whole thing into the recursive formula)?

Thanks a lot!

-
This is the exponential generating function of Bernoulli numbers – Mhenni Benghorbal May 4 '13 at 12:39

Expand $e^{x}$ around $x=0$ and then reexpand \begin{align}\frac{1}{1+\frac{x}{2!}+\frac{x^2}{3!}+\ldots}=1-\left(\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\ldots\right)+\left(\frac{x}{2!}+\frac{x^2}{3!}+\ldots\right)^2-\left(\frac{x}{2!}+\ldots\right)^3+\ldots=\\ =1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\ldots\right)+\frac{x^2}{4}\left(1+\frac{x}{3}+\ldots\right)^2-\frac{x^3}{8}\left(1+\ldots\right)^3+\ldots=\\ =1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}\right)+\frac{x^2}{4}\left(1+\frac{2x}{3}\right)-\frac{x^3}{8}+O\left(x^4\right)=\\ =1-\frac{x}{2}+\left(-\frac{1}{6}+\frac14\right)x^2+\left(-\frac{1}{24}+\frac{1}{6}-\frac{1}{8}\right)x^3+O\left(x^4\right)=\\ =1-\frac{x}{2}+\frac{x^2}{12}+O\left(x^4\right).\end{align}
Could you please elaborate how to get from your RHS to $1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \ ...$? – Leo May 4 '13 at 12:42
I know that the question asks for a method not using recursion, but this is just so much easier and cleaner than computing the series for $\dfrac{x}{e^x-1}$.
Given $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k\frac{x^k}{k!}\tag{1}$$ we can rewrite $(1)$ as \begin{align} 1 &=\sum_{j=0}^\infty\frac{x^j}{(j+1)!}\sum_{k=0}^\infty B_k\frac{x^k}{k!}\\ &=\sum_{k=0}^\infty x^k\sum_{j=0}^k\frac{B_{k-j}}{(k-j)!\,(j+1)!}\\ &=\sum_{k=0}^\infty\frac{x^k}{(k+1)!}\sum_{j=0}^k\binom{k+1}{j+1}B_{k-j}\tag{2} \end{align} which yields $B_0=1$ and the recurrence $$B_k=-\frac1{k+1}\sum_{j=1}^k\binom{k+1}{j+1}B_{k-j}\tag{3}$$ for $k\ge1$.