7
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ Take the shft $\ e^{-1}=x$ then you can operate on it as a power series $\endgroup$
    – CTSnake
    Apr 26, 2017 at 18:14
  • 6
    $\begingroup$ 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, 2017 at 18:15
  • $\begingroup$ @imranfat "but taylor series of $e^x$ is going to be useful" How? $\endgroup$
    – Did
    Apr 26, 2017 at 18:15
  • $\begingroup$ @imranfat Yeah, with no Taylor series of the exponential at all. $\endgroup$
    – Did
    Apr 26, 2017 at 18:17
  • $\begingroup$ ah, forget it.... $\endgroup$
    – imranfat
    Apr 26, 2017 at 18:20

3 Answers 3

10
$\begingroup$

$$ \frac{d}{da}\mathrm{e}^{an} = n\mathrm{e}^{an} $$ 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 $$ \sum_{n=0}^\infty \lambda^n = 1 + \sum_{n=1}^\infty \lambda^n $$

$\endgroup$
2
  • 2
    $\begingroup$ Forgetting the exponential and replacing every $e^a$ by $x$ simplifies this. $\endgroup$
    – Did
    Apr 26, 2017 at 18:18
  • $\begingroup$ @Did Fair point. I did initially see that simplification (well until you told me!) $\endgroup$
    – Chinny84
    Apr 26, 2017 at 18:19
6
$\begingroup$

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 $x=1$ $$ \sum_{n\geq 1} n^2 e^{-n} = \color{red}{\frac{e(e+1)}{(e-1)^3}}.\tag{3}$$

$\endgroup$
5
$\begingroup$

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! $$S(1-x)^2+x=\frac{2x}{1-x}$$ $$S=\frac{x+x^2}{(1-x)^3} =\sum_{n=1}^{\infty}n^2x^n$$

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for your answer. Your solution is very nice. $\endgroup$
    – Navin
    Apr 26, 2017 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.