# 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.

• Take the shft $\ e^{-1}=x$ then you can operate on it as a power series Apr 26, 2017 at 18:14
• Hint: $$\sum_{n=1}^\infty n^2x^n=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+x\sum_{n=0}^\infty nx^{n-1}$$
– Did
Apr 26, 2017 at 18:15
• @imranfat "but taylor series of $e^x$ is going to be useful" How?
– Did
Apr 26, 2017 at 18:15
• @imranfat Yeah, with no Taylor series of the exponential at all.
– Did
Apr 26, 2017 at 18:17
• ah, forget it.... Apr 26, 2017 at 18:20

$$\frac{d}{da}\mathrm{e}^{an} = n\mathrm{e}^{an}$$ 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 $$\sum_{n=0}^\infty \lambda^n = 1 + \sum_{n=1}^\infty \lambda^n$$
• Forgetting the exponential and replacing every $e^a$ by $x$ simplifies this.
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 $x=1$ $$\sum_{n\geq 1} n^2 e^{-n} = \color{red}{\frac{e(e+1)}{(e-1)^3}}.\tag{3}$$
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! $$S(1-x)^2+x=\frac{2x}{1-x}$$ $$S=\frac{x+x^2}{(1-x)^3} =\sum_{n=1}^{\infty}n^2x^n$$