Value of $\sum\limits_{n=1}^{\infty}n^2e^{-n}$

I was solving this question and as you can see when we expand the sum we get,

$$\frac{1}{e}+\frac{4}{e^2}+\frac{9}{e^3}+++..$$ It's not an GP neither an AP , how shall I find the sum then?

  • 9
    $\begingroup$ $$\sum_{n=1}^\infty n^2e^{-nx}=\frac{d^2}{dx^2}\left(\sum_{n=1}^\infty e^{-nx}\right)=\frac{d^2}{dx^2}\left(\frac1{e^x-1}\right)=\cdots$$ $\endgroup$ – Did May 11 '17 at 12:50
  • $\begingroup$ do you know taylor series ? if so take the derivative of the function $ (1-x)^{-1}= 1+\sum_{n=1}^{\infty}x^{n} $ $\endgroup$ – Jose Garcia May 11 '17 at 12:51
  • $\begingroup$ Actually I don't but I'll go and check more about it. $\endgroup$ – Iti Shree May 11 '17 at 12:52
  • 3
    $\begingroup$ See math.stackexchange.com/questions/338852/… $\endgroup$ – lab bhattacharjee May 11 '17 at 13:03
  • $\begingroup$ You could as well notice that if you call that sum $S$ and multiply the series by $\frac1e$ you get $S-\frac1e S=\frac{1}{e}+\frac{3}{e^2}+\frac{5}{e^3}+\cdots$,now you can similarly solve the sum you found ($\sum_{n=1}^\infty(2n-1)e^{-n}$) $\endgroup$ – kingW3 May 11 '17 at 13:05

So, here what I did, thanks to everyone who commented for helping me. $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$

after differentiating we get, $\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}nx^{n-1}$

multiply both side with x we get, $\frac{x}{(1-x)^2}=\sum_{n=0}^{\infty}nx^{n}$ again differentiate and multiply with x we get $\frac{x(1+x)}{(1-x)^3}=\sum_{n=0}^{\infty}n^2x^{n}$

now when we replace $x$ with $\frac{1}{e}$ we get required result i.e $$\frac{e^2+e}{(e-1)^3}$$

  • $\begingroup$ Yes, thanks for correction. $\endgroup$ – Iti Shree May 11 '17 at 13:28

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.