# Math Help - What does the series converge to?

1. ## What does the series converge to?

What does the series converge to?
$\sum_{k=0}^\infty \frac{k+1}{k!}$

This was on my final tonight, I don't know (or don't remember) how to figure it out.

2. Originally Posted by angel.white
What does the series converge to?
$\sum_{k=0}^\infty \frac{k+1}{k!}$

This was on my final tonight, I don't know (or don't remember) how to figure it out.
Rewriting it as $\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty }\frac{1}{n!}=2e$

3. Originally Posted by Mathstud28
Rewriting it as $\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty }\frac{1}{n!}=2e$
Sorry I did not really explain that well

$\sum_{n=0}^{\infty}\frac{n}{n!}=\sum_{n=0}^{\infty }\frac{1}{(n-1)!}=\frac{D[\sum_{n=0}^{\infty}\frac{1^n}{n!}]}{dx}=e^1$

and $\sum_{n=0}^{\infty}\frac{1}{n!}=\sum_{n=0}^{\infty }\frac{1^n}{n!}=e^1$

so $\sum_{n=0}^{\infty}\frac{n+1}{n!}=\sum_{n=0}^{\inf ty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}=e+ e=2e$

4. Originally Posted by angel.white
What does the series converge to?
$\sum_{k=0}^\infty \frac{k+1}{k!}$

This was on my final tonight, I don't know (or don't remember) how to figure it out.

Alternately

$xe^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!}$

$(xe^x)' = xe^x + e^x = \left(\sum_{k=0}^\infty \frac{x^{k+1}}{k!}\right)' = \sum_{k=0}^\infty \frac{(k+1)x^{k}}{k!}$

Substituting x=1, we get:
$(xe^x + e^x)\bigg{|}_{x=1} = 2e = \sum_{k=0}^\infty \frac{(k+1)}{k!}$

5. Originally Posted by Mathstud28
Sorry I did not really explain that well

$\sum_{n=0}^{\infty}\frac{n}{n!}=\sum_{n=0}^{\infty }\frac{1}{(n-1)!}={\color{red}\frac{D[\sum_{n=0}^{\infty}\frac{1^n}{n!}]}{dx}}=e^1$
The derivative of a constant is 0...

All you have to know is that $\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$

$\sum_{n=0}^\infty \frac{n}{n!}=0+\sum_{n={\color{red}1}}^\infty \frac{n}{n!}=\sum_{n=1}^\infty \frac{1}{(n-1)!}$
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

$=\sum_{n=0}^\infty \frac{1}{n!}=e$

Added to the other term, it yields what the previous posters wrote.

6. Thanks guys, I'm a little annoyed that I forgot that >.< It's one of those things I picked up outside of class and never used again and so I forgot it. If I were more astute, I could have figured that out on the final, but I wasn't

7. Originally Posted by Moo
The derivative of a constant is 0...

All you have to know is that $\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$

$\sum_{n=0}^\infty \frac{n}{n!}=0+\sum_{n={\color{red}1}}^\infty \frac{n}{n!}=\sum_{n=1}^\infty \frac{1}{(n-1)!}$
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

$=\sum_{n=0}^\infty \frac{1}{n!}=e$

Added to the other term, it yields what the previous posters wrote.
It was 2:30 over here...cut me some slack...I was tired

So Techincally what I meant to say was $\frac{d}{dx}\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\bigg]\bigg|_{x=1}$

8. Originally Posted by Mathstud28
It was 2:30 over here...cut me some slack...I was tired

So Techincally what I meant to say was $\frac{d}{dx}\bigg[\sum_{n=0}^{\infty}\frac{x^n}{n!}\bigg]\bigg|_{x=1}$
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

9. Originally Posted by CaptainBlack
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
So...I am unclear are you saying this because my correction was wrong or because I needed to make a correction?

And secondly...thats a little harsh...everybody makes mistakes...