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I have to calculate sum of series $\sum \frac{n^2}{n!}$. I know that $\sum \frac{1}{n!}=e$ but I dont know how can I use that fact here..


marked as duplicate by Martin Sleziak, Claude Leibovici, Namaste, JonMark Perry, Dando18 Jul 21 '17 at 13:06

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  • $\begingroup$ Ok I know. Its 2e $\endgroup$ – lksz43 Jan 10 '15 at 13:11

HINT $$\frac{n^2}{n!}=\frac{n^2}{n\cdot (n-1)!}=\frac{n}{(n-1)!}=\frac{n-1+1}{(n-1)!}=\frac{n-1}{(n-1)!}+\frac{1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$$

  • $\begingroup$ Gotta be careful in the case $n=1$, of course... $\endgroup$ – Thomas Andrews Jan 11 '15 at 0:36
  • $\begingroup$ @ThomasAndrews I think it's a fairly common convention to let $\frac1{n!}$ equal $0$ when $n$ is a negative integer (even though, strictly speaking, it's not true). $\endgroup$ – Akiva Weinberger Jan 23 '15 at 0:08
  • $\begingroup$ Really? I've never seen that convention, ever, and I've done a lot of math. @columbus8myhw $\endgroup$ – Thomas Andrews Jan 23 '15 at 0:11
  • $\begingroup$ Either one, but the second one was really broad. @columbus8myhw $\endgroup$ – Thomas Andrews Jan 23 '15 at 0:12
  • $\begingroup$ @ThomasAndrews You know how $\binom nk$ is, by convention, $0$ when $n<k$? Also, $\binom nk=\frac{n!}{k!}\cdot\frac1{(n-k)!}$. Taking the case where $n<k$, this kind of implies that $\frac1{(n-k)!}=0$ (note that $n-k$ is negative here). $\endgroup$ – Akiva Weinberger Jan 23 '15 at 0:16

Apply the operator $DxD$, where $D=\frac{d}{dx} $, to

$$\sum_{n=0}^{\infty} \frac{x^n} {n!} = e^x $$

and then substitute $x=1$. See similar techniques.



Let $f(x)=\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!}$. Express $ g(x)=\displaystyle \sum_{n=0}^{\infty} \frac{n^2x^{n}}{n!} $ in terms of $f'(x)$ and $f''(x)$.

Interpret the result using $f(x)=e^x$.


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