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

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?

• $$\sum_{n=1}^\infty n^2e^{-nx}=\frac{d^2}{dx^2}\left(\sum_{n=1}^\infty e^{-nx}\right)=\frac{d^2}{dx^2}\left(\frac1{e^x-1}\right)=\cdots$$ – Did May 11 '17 at 12:50
• do you know taylor series ? if so take the derivative of the function $(1-x)^{-1}= 1+\sum_{n=1}^{\infty}x^{n}$ – Jose Garcia May 11 '17 at 12:51
• Actually I don't but I'll go and check more about it. – Iti Shree May 11 '17 at 12:52
• – lab bhattacharjee May 11 '17 at 13:03
• 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 ($\sum_{n=1}^\infty(2n-1)e^{-n}$) – kingW3 May 11 '17 at 13:05

## 1 Answer

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 $$\frac{e^2+e}{(e-1)^3}$$

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