How do I show $\sum\limits_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots$? Since $$\sum_{k=0}^{\infty}x^k=\frac{1}{1-x},$$we have
$$\sum_{j=0}^{\infty}e^{-wj}=\sum_{j=0}^{\infty} (e^{-w})^j=\frac{1}{1-e^{-w}}=\frac{e^w}{e^w-1}$$
Also, since $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},$$we have
$$\sum_{j=0}^{\infty}e^{-wj}=\frac{1+w+\frac{w^2}{2}\cdots}{w+\frac{w^2}{2}+\frac{w^3}{6}+\cdots}$$ 
How can I obtain $$\sum_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots?$$

The Bernoulli numbers are defined by
$$\frac{t}{e^t-1}=\sum_{n=0}^{\infty} B_n \frac{t^n}{n!}$$
We have
$$\begin{split}
\sum_{j=0}^{\infty}e^{-wj} &=\frac{1}{1-e^{-w}}\\
&=\frac 1 w \frac{(-w)}{e^{-w}-1}\\
&= \frac 1 w \sum_{n=0}^{\infty} B_n \frac{(-w)^n}{n!}\\
&= \frac 1 w + \sum_{n=0}^{\infty}  \frac{(-1)^{n+1}B_{n+1}}{(n+1)!}w^n \text{ (as } B_0=1)
\end{split}$$
