How do I show $\sum\limits_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots$?

Question

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 $$\sum_{j=0}^{\infty}e^{-wj}=\frac{1}{w}+\frac{1}{2}+\frac{w}{12}-\frac{w^3}{720}+\frac{w^5}{30240}+\cdots?$$

Answer 1

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