What does the series converge to?
$\displaystyle \sum_{k=0}^\infty \frac{k+1}{k!}$
This was on my final tonight, I don't know (or don't remember) how to figure it out.
Sorry I did not really explain that well
$\displaystyle \sum_{n=0}^{\infty}\frac{n}{n!}=\sum_{n=0}^{\infty }\frac{1}{(n-1)!}=\frac{D[\sum_{n=0}^{\infty}\frac{1^n}{n!}]}{dx}=e^1$
and $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}=\sum_{n=0}^{\infty }\frac{1^n}{n!}=e^1$
so $\displaystyle \sum_{n=0}^{\infty}\frac{n+1}{n!}=\sum_{n=0}^{\inf ty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=e+ e=2e$
Alternately
$\displaystyle xe^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!}$
$\displaystyle (xe^x)' = xe^x + e^x = \left(\sum_{k=0}^\infty \frac{x^{k+1}}{k!}\right)' = \sum_{k=0}^\infty \frac{(k+1)x^{k}}{k!}$
Substituting x=1, we get:
$\displaystyle (xe^x + e^x)\bigg{|}_{x=1} = 2e = \sum_{k=0}^\infty \frac{(k+1)}{k!}$
The derivative of a constant is 0...
All you have to know is that $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}=e^x$
$\displaystyle \sum_{n=0}^\infty \frac{n}{n!}=0+\sum_{n={\color{red}1}}^\infty \frac{n}{n!}=\sum_{n=1}^\infty \frac{1}{(n-1)!}$
I did it because I hate negative factorial, and I'm not sure it's correct to leave this n=0 while we have a (n-1)!
Change : n-1 -> n, so n=1 -> n=0
$\displaystyle =\sum_{n=0}^\infty \frac{1}{n!}=e$
Added to the other term, it yields what the previous posters wrote.
It appears that too many of your posts are composed in the small hours given the high proportion of your posts are less than satisfactory.
It would help the site if you took more time and care over your posts. As it is the low quality of your posts is damaging to the reputation of this site.
RonL