How to solve $\sum_{k=1}^{\infty} ke^{-k}$? I am interested in
$$\sum_{k=1}^{\infty} ke^{-k}$$
And I can get the closed form on Wolfram Alpha, $\frac{e}{(e-1)^2}$, but I am curious how to derive it. It doesn't appear to be a typical geometric series.

There are many ways to evaluate
$S 
= \sum_{k=1}^{\infty} kx^k
$.
This is possibly the simplest.
$xS 
= x\sum_{k=1}^{\infty} kx^k
= \sum_{k=1}^{\infty} kx^{k+1}
= \sum_{k=2}^{\infty} (k-1)x^{k}
$
so
$\begin{array}\\
S-xS
&= \sum_{k=1}^{\infty} kx^k-\sum_{k=2}^{\infty} (k-1)x^{k}\\
&= x+\sum_{k=2}^{\infty} kx^k-\sum_{k=2}^{\infty} (k-1)x^{k}\\
&= x+\sum_{k=2}^{\infty} (kx^k-(k-1)x^{k})\\
&= x+\sum_{k=2}^{\infty} x^k\\
&= x+\dfrac{x^2}{1-x}\\
&= \dfrac{x-x^2+x^2}{1-x}\\
&= \dfrac{x}{1-x}\\
\text{so}\\
S
&= \dfrac{x}{(1-x)^2}\\
\end{array}
$
Then put $x = e^{-1}$.
