Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Thanks for fixing the equation. – tunnuz Jun 25 '12 at 4:51
up vote 14 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.