showing $\frac{x}{e^{x}-1} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $

Question

I want to show

$\frac{x}{e^{x}-1} = -\sum_{k=0}^{\infty} x e^{kx} = \frac{\ln(e^x)}{e^x-1} = -\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $

The first step seems trivial, what i am confused is the process of \begin{align} -\sum_{k=0}^{\infty}xe^{kx} = \frac{\ln(e^x)}{e^x-1} \end{align} and the last step \begin{align} -\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} \end{align}

Can you give me some idea for these steps?

Answer 1

In order to show the identity \begin{align*} \frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{\left(1-e^x\right)^{n-1}}{n}\qquad\qquad |1-e^x|<1\tag{1} \end{align*}

we recall that

\begin{align*} \ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\qquad\qquad\qquad |x|<1 \end{align*}

We obtain \begin{align*} x&=\ln(e^x)=\ln(1-(1-e^x))\\ &=-\sum_{n=1}^\infty \frac{1}{n} (1-e^x)^n\qquad\qquad\qquad\qquad\qquad |1-e^x|<1\\ \end{align*} and after division by $1-e^x$ with $x \neq 0$ we get (1) \begin{align*} \frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{1}{n} (1-e^x)^{n-1}\qquad\qquad\qquad\qquad x<\ln 2,x\neq 0 \end{align*}

Note, it is not necessary to consider a geometric series expansion. In fact the representation in this case is different from the series representation in (1) since the radius of convergence is different. \begin{align*} \frac{x}{e^x-1}=\sum_{n=0}^\infty xe^{kx}\qquad\qquad |e^x|<1\quad\text{resp.}\quad x<0 \end{align*}

Answer 2

As $x=\ln({e^x})$ , $$ \begin{align} -\sum_{k=0}^{\infty}xe^{kx}\end{align} =\frac{x}{e^x-1}= \frac{\ln(e^x)}{e^x-1} $$ And as $\ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n}$ , $$ -\frac{\ln[1-(1-e^x)]}{1-e^x} = -\frac{(-\sum_{n=1}^{\infty} \frac{(1-e^x)^n}{n})}{1-e^x}=\frac{1}{1-e^x}.\sum_{n=1}^{\infty} \frac{(1-e^x)^n}{n}=\sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $$