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.
 A: In fact it is the derivative of a geometric series because
$$
\left(e^{-kx}\right)'=-ke^{-kx}
$$
You can for example calculate for $x \in \mathbb{N}^{*}$
$$
\sum_{k=0}^{+\infty}e^{-kx}=\frac{1}{1-e^{-x}}
$$
Then differenciating
$$
\sum_{k=0}^{+\infty}ke^{-kx}=-\left(\frac{1}{1-e^{-x}}\right)'=\frac{e^x}{\left(e^x-1\right)^2}
$$
Taking $x=1$ gives you the expected result.
A: For each real $x $, $e^{-x}<1$ and
$$e^{-x}+e^{-2x}+....=\frac {1}{1-e^{-x}} $$
differentiation gives
$$-(e^{-x}+2e^{-2x}+....)=\frac {-e^{-x}}{(1-e^{-x})^2} $$
$$=\frac {-e^x}{(e^x-1)^2} $$
For $x=1$, you get the result.
A: 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 $x = e^{-1}$.
A: By elementary means:
$$S:=a+a^2+a^2+a^3+a^3+a^3+a^4+a^4+a^4+a^4+\cdots\\
=(a+a^2+a^3+a^4+\cdots)+a(a+a^2+a^2+a^3+a^3+a^3+\cdots),$$
hence for $|a|<1$,
$$S=\frac a{1-a}+aS.$$
