How many ways to evaluate $1 + e^{-x} + e^{-2x} + e^{-3x} + \ldots$ to $(e^x - 1)\cdot(e^{-x} + 2e^{-2x} +3e^{-3x} + \ldots)$

Question

How many ways to evaluate

$1 + e^{-x} + e^{-2x} + e^{-3x} + \ldots$

to

$(e^x - 1)\cdot(e^{-x} + 2e^{-2x} +3e^{-3x} + \ldots)$

Answer 1

Geometric series:

$$1+e^{-x}+e^{-2x}+\ldots=\sum_{n=0}^\infty\left(e^{-x}\right)^n=\frac1{1-e^{-x}}=\frac{e^x}{e^x-1}$$

as long as $\;e^{-x}<1\iff e^x>1\iff x>0\;$

The series above converges absolutely for positive $\;x\;$ so you can't "rearrange" it to be anything else.

Added on request: Within the radius of convergence we can differentiate elementwise:

$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies \frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

Thus, in our case:

$$\left(\frac{e^x}{e^x-1}\right)'=-\frac{e^x}{(e^x-1)^2}=-\sum_{n=1}^\infty ne^{-nx} $$

Now just watch closely the signs...