Sum of the series $\sum_{n=1}^\infty n^2e^{-n}$ 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. 
 A: 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 $x=1$
$$ \sum_{n\geq 1} n^2 e^{-n} = \color{red}{\frac{e(e+1)}{(e-1)^3}}.\tag{3}$$
A: Just for kicks, here's the way I like to solve these without calculus:
As long as $|x|<1$, we have
$$S=\sum_{n=1}^{\infty}n^2x^n$$
$$S(1-x) = \sum_{n=1}^{\infty}n^2x^n - \sum_{n=2}^{\infty}(n-1)^2x^n = \sum_{n=1}^{\infty}(2n-1)x^n =\sum_{n=1}^{\infty}2nx^n-\sum_{n=1}^{\infty}x^n$$
$$S(1-x) + \frac{x}{1-x} = 2\sum_{n=1}^{\infty}nx^n$$
$$\big(S(1-x) + \frac{x}{1-x}\big)\frac{1-x}{2} = \sum_{n=1}^{\infty}nx^n -\sum_{n=2}^{\infty}(n-1)x^n =\sum_{n=1}^{\infty}x^n=\frac{x}{1-x}$$
Now that we have dealt with the $n$'s,  we solve for S!
$$S(1-x)^2+x=\frac{2x}{1-x}$$
$$S=\frac{x+x^2}{(1-x)^3} =\sum_{n=1}^{\infty}n^2x^n$$ 
A: $$
\frac{d}{da}\mathrm{e}^{an} = n\mathrm{e}^{an}
$$
and
$$
\frac{d^2}{da^2}\mathrm{e}^{an} = n^2\mathrm{e}^{an}
$$
so I posit that we can use
$$
\sum_{n=1}^\infty\frac{d^2}{da^2}\mathrm{e}^{an} = \sum_{n=1}^\infty n^2\mathrm{e}^{an}
$$
we can pull the derivative out of the sum to find
$$
\frac{d^2}{da^2}\sum_{n=1}^\infty\mathrm{e}^{an} =\frac{d^2}{da^2}\sum_{n=1}^\infty\lambda^{n}
$$
where $\lambda = \mathrm{e}^a$. This sum is a geometric series if we consider
$$
\sum_{n=0}^\infty \lambda^n = 1 + \sum_{n=1}^\infty \lambda^n
$$
