# Does $\sum^{\infty}_{n=1} n/(n-1)!$ converge to something?

I just ran into an expression, and I would like to know what it converges to. $$\sum^{\infty}_{n=1} \frac{n}{(n-1)!}$$

Do you know if it converges to something (like $e$) or, in alternative, how to find out what it converges to?

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Thanks for fixing the equation. –  tunnuz Jun 25 '12 at 4:51

## 1 Answer

$$\sum_{n=1}^{\infty} \dfrac{n}{(n-1)!} = \sum_{n=1}^{\infty} \dfrac{n-1+1}{(n-1)!} = \sum_{n=2}^{\infty} \dfrac1{(n-2)!} + \sum_{n=1}^{\infty} \dfrac1{(n-1)!} = e + e = 2e$$

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Wow, that was fast. –  copper.hat Jun 25 '12 at 4:50
Curse you for typing faster! shakes fist –  anon Jun 25 '12 at 4:50
Damn neutrino typists. –  copper.hat Jun 25 '12 at 4:51
Or if $g(x)=xe^x$, then $g'(1)=2e$, and $g'(x)=\frac{d}{dx}\sum\limits_{n=1}^\infty\frac{1}{(n-1)!}x^n=\sum\limits_{n=1}‌​^\infty\frac{n}{(n-1)!}x^{n-1}$, so $g'(1)$ gives the series in question. –  Jonas Meyer Jun 25 '12 at 4:52
Any faster the answer would have come before the question!\ –  Arjang Jun 25 '12 at 8:11