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Math Help - What does the series converge to?

  1. #1
    Super Member angel.white's Avatar
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    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.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by angel.white View Post
    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
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    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
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    Quote Originally Posted by angel.white View Post
    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!}
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  5. #5
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    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.
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  6. #6
    Super Member angel.white's Avatar
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    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
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    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}
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    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    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
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    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...
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