# Calculate sum of series $\sum \frac{n^2}{n!}$ [duplicate]

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..

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

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

• Gotta be careful in the case $n=1$, of course... – Thomas Andrews Jan 11 '15 at 0:36
• @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). – Akiva Weinberger Jan 23 '15 at 0:08
• Really? I've never seen that convention, ever, and I've done a lot of math. @columbus8myhw – Thomas Andrews Jan 23 '15 at 0:11
• Either one, but the second one was really broad. @columbus8myhw – Thomas Andrews Jan 23 '15 at 0:12
• @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). – 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.

Hint:

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 $f(x)=e^x$.