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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.

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

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    $\begingroup$ Forgetting the exponential and replacing every $e^a$ by $x$ simplifies this. $\endgroup$ – Did Apr 26 '17 at 18:18
  • $\begingroup$ @Did Fair point. I did initially see that simplification (well until you told me!) $\endgroup$ – Chinny84 Apr 26 '17 at 18:19
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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! $$S(1-x)^2+x=\frac{2x}{1-x}$$ $$S=\frac{x+x^2}{(1-x)^3} =\sum_{n=1}^{\infty}n^2x^n$$

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  • $\begingroup$ Thanks for your answer. Your solution is very nice. $\endgroup$ – Navin Apr 26 '17 at 19:09
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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 $x=1$ $$ \sum_{n\geq 1} n^2 e^{-n} = \color{red}{\frac{e(e+1)}{(e-1)^3}}.\tag{3}$$

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