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Math Help - Infinite series of e^x?

  1. #1
    Super Member Anonymous1's Avatar
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    Infinite series of e^x?

    Hi,

    So, I'm trying to find the sum of:

    \sum_{n=0}^{\infty}\frac{x^n}{(n+1)!}

    I know that \sum_{n=0}^{\infty}\frac{x^n}{n!} = e^x....

    But How can I use this to find my sum?

    Thanks
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  2. #2
    Super Member Deadstar's Avatar
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    Right, compare the two series...

    \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{1}{2!}x^2  + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \ldots

    \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = 1 + \frac{1}{2!}x  + \frac{1}{3!}x^2 + \frac{1}{4!}x^3 + \frac{1}{5!}x^4 + \ldots

    Compare coefficients and you can see if you divide the first series by x you get the second with the exception of an extra \frac{1}{x} term.

    Hence just subtract that and you get...

    \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = \frac{e^x - 1}{x}
    Last edited by Deadstar; April 12th 2010 at 03:42 PM.
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  3. #3
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Deadstar View Post
    Posting solution hold on...

    Right, compare the two series...

    \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{1}{2!}x^2  + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \ldots

    \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = 1 + \frac{1}{2!}x  + \frac{1}{3!}x^2 + \frac{1}{4!}x^3 + \frac{1}{5!}x^4 + \ldots

    Compare coefficients and you can see if you divide the first series by x you get the second with the exception of an extra \frac{1}{x} term.

    Hence just subtract that and you get...

    \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!} = \frac{e^x - 1}{x}
    Why thank you sir.
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  4. #4
    Math Engineering Student
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    \sum\limits_{n\ge 0}{\frac{x^{n}}{(n+1)!}}=\frac{1}{x}\sum\limits_{n  \ge 1}{\frac{x^{n}}{n!}}=\frac{1}{x}\left( e^{x}-1 \right).
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