TITLE: Sum of the series $\sum_{n=1}^\infty n^2e^{-n}$
QUESTION [7 upvotes]: The question is to find out the sum of the series $$\sum_{n=1}^\infty n^2 e^{-n}$$
I tried to bring the summation in some form of telescoping series but failed. I then tried approximating the sum by the corresponding integral(which I am not sure about) to get the value as $2/e$ indicating that the sum converges. Any help shall be highly appreciated. Thanks.

REPLY [6 votes]: We have
$$ \sum_{n\geq 0} e^{-nx} = \frac{1}{1-e^{-x}}\tag{1} $$
hence by applying $\frac{d^2}{dx^2}$ to both sides
$$ \sum_{n\geq 0} n^2 e^{-nx} = \frac{e^x(e^x+1)}{(e^x-1)^3}\tag{2} $$
and by evaluating at $x=1$
$$ \sum_{n\geq 1} n^2 e^{-n} = \color{red}{\frac{e(e+1)}{(e-1)^3}}.\tag{3}$$