5

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..

lksz43
  • 81

3 Answers3

13

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

kingW3
  • 13,496
  • Gotta be careful in the case $n=1$, of course... – Thomas Andrews Jan 11 '15 at 00:36
  • @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 00:08
  • Really? I've never seen that convention, ever, and I've done a lot of math. @columbus8myhw – Thomas Andrews Jan 23 '15 at 00:11
  • Either one, but the second one was really broad. @columbus8myhw – Thomas Andrews Jan 23 '15 at 00:12
  • @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 00:16
  • @ThomasAndrews In addition, it gives the right answer in this context (i.e. kingW3's post). $\frac{1^2}{1!}=0+\frac1{(1-1)!}$. – Akiva Weinberger Jan 23 '15 at 00:17
  • @ThomasAndrews Also, I recently-ish ran into a sum involving logarithms, where one of the terms is undefined (though the limit as $n$ went to its value was defined). If I interpreted that term to mean it's limit, I got the right answer. EDIT: This thing. To avoid confusion I wrote $\rho\neq1$ under the sum sign. – Akiva Weinberger Jan 23 '15 at 00:18
  • If the limit isn't zero, then the sum is undefined, always. You'd still have to given an argument for why that works. You can't simply assert it. That's bad math. You can easily get right answers for wrong reasons, develop bad habits, and start to get wrong answers without knowing it later. @columbus8myhw – Thomas Andrews Jan 23 '15 at 00:20
  • @ThomasAndrews http://www.wolframalpha.com/input/?i=1%2F%28-1%29%21 – Akiva Weinberger Jan 23 '15 at 00:32
8

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.

Community
  • 1
Mhenni Benghorbal
  • 47,431
2

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 $f(x)=e^x$.

lhf
  • 216,483