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Find the sum of the series $ \sum_{n=1}^{\infty} \frac{n^2}{n!} $

$My \ attempt :$ $$ \sum_{n=1}^{\infty} \frac{n^2}{n!} \\ = \sum_{n=1}^{\infty} \frac{n}{(n-1)!} \\ = \sum_{n=0}^{\infty} \frac{n+1}{n!} = \sum_{n=0}^{\infty} \frac{1}{(n-1)!} + \frac{1}{n!} $$ But I don't know if this will take me anywhere.

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    $\begingroup$ it may be useful to know that $\sum 1/n!=e$(notice that $1/(n-1)!$ is not defined in $n=0$, it would be more correct to split the series and let $n$ start from $1$ in the first one). $\endgroup$
    – Kandinskij
    Commented Dec 7, 2020 at 17:12
  • $\begingroup$ I just noticed that the sum of one part is e but the other part isn't defined for n = 0...okay I'll do that...and if I can do it correctly I'll post it as an answer. $\endgroup$
    – Itachi
    Commented Dec 7, 2020 at 17:16
  • $\begingroup$ See math.stackexchange.com/questions/576976/… $\endgroup$ Commented Dec 7, 2020 at 17:20
  • $\begingroup$ @eureka I tried to split the series first but there the same problem is occurring because there'll be a term $\frac{1}{(n-2)!}$ and in that I can't run $n$ from 1. $\endgroup$
    – Itachi
    Commented Dec 7, 2020 at 17:24
  • $\begingroup$ @Pritam I'll explain myself better, your initial mistake was that you could simply simplify $\frac{n}{n!}$ into $\frac{1}{(n-1)!}$ if and only if $n\neq 0$, so more correctly $\sum_{n=0}^{\infty} \frac{n}{n!}=\frac{0}{0!}+\sum_{n=1}^{\infty} \frac{n}{n!}=\sum_{n=1}^{\infty} \frac{1}{(n-1)!}$ $\endgroup$
    – Kandinskij
    Commented Dec 7, 2020 at 17:27

2 Answers 2

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Since

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

We can differentiate to get

$$\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!}=e^x\implies \sum_{n=1}^{\infty}\frac{nx^{n}}{n!}=xe^x$$

differentiating again, we get that

$$\sum_{n=1}^{\infty}\frac{n^2x^{n-1}}{n!}=xe^x+e^x=(1+x)e^x$$

plugging in $x=1$, we get that

$$\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$$

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$$S=\sum_{n=1}^{\infty} \frac{n^2}{n!}=\sum_{n=1}^{\infty} \frac{n(n-1)+n}{n!}=\sum_{n=1}^{\infty} \left(\frac{1}{(n-2)!}+\frac{1}{(n=1)!} \right)$$ $$=\sum_{p=-1}^{\infty} \frac{1}{p!}+\sum_{m=0}^{\infty} \frac{1}{m!}=e+e=2e$$ Note that $(-1)!=\infty$.

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  • $\begingroup$ How factorial is defined for negative integers? And if so then what I wrote as my attempt then from there also I can split and do as you did here. But as far as I know that factorial is not defined like this. $\endgroup$
    – Itachi
    Commented Dec 7, 2020 at 17:26
  • $\begingroup$ That's pretty informal $\endgroup$
    – Kandinskij
    Commented Dec 9, 2020 at 11:21

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