## watchmath 5 years ago Calculate $\sum_{n=1}^\infty \frac{n-1}{n!}$

1. anonymous

0

2. anonymous

it has to be positive though

3. watchmath

Can't be zero since it is a positive series starting from $$n=2$$

4. anonymous

I can't prove it, but the sum seems to approach one, in that every successive sum is (n!-1)/n!

5. watchmath

Well observation is half of the proof thebestpig :). You just need to push a little bit more :D.

6. anonymous

gimme a second - i'll do an induction thing

7. anonymous

the base step is $(1!-1)/1! = 0$ which is in fact the first term, and $\frac{(n!-1)}{n!}+\frac{n}{(n+1)!}=\frac{(n!-1)(n+1)+n}{(n+1)!}=\frac{(n+1)!-n-1+n}{(n+1)!}$ $=\frac{(n+1)!-1}{(n+1)!}$ and it looks like we're done.

8. anonymous

oh wait one more thing

9. anonymous

$\lim_{n \rightarrow \infty} \frac{n!-1}{n!}=1$

10. watchmath

Awesome! :)

11. watchmath

We can also using telescoping sum. Note that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$

12. anonymous

ha, /facepalm/

13. anonymous

well, I enjoy using induction :)

14. anonymous

Here's another way: a little relabeling gives (check it!) $\sum_{n=1}^\infty \frac{n-1}{n!} = \sum_{n=0}^{\infty}\frac{1}{(n+2)n!}$ Now $e^x = \sum_{n=0}^\infty \frac{x^n}{n!},$ so $xe^x = \sum_{n=0}^\infty \frac{x^{n+1}}{n!},$ so $\int xe^x = xe^x-e^x +C = \sum_{n=0}^\infty \frac{x^{n+2}}{(n+2)n!}$ Plug in $$x=0$$ to find that $$C =1$$, so $1 e^1 -e^1 +1 = 1 = \sum_{n=0}^\infty \frac{1}{(n+2)n!}.$