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Math Help - Sum to infinity of series

  1. #1
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    Sum to infinity of series

    Show that n(n+1)\equiv(n-1)(n-2)+4(n-1)+2 and hence, that for n\geq 3,
    \frac{n(n+1)}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!}
    Deduce or find otherwise the sum to infinity of the series
    1.2+\frac{2.3}{1!}+\frac{3.4}{2!}+...+\frac{n(n+1)  }{(n-1)!}+...

    I've done the first two parts correctly, but I don't know how to do this last part.
    \sum^{n}_{r=1} \frac{r(r+1)}{(r-1)!}=\sum^{n}_{r=1}\frac{1}{(r-3)!}+\frac{4}{(r-2)!}+\frac{2}{(r-1)!}
    This is the furthest I got, where n<3, I can't find values, so I don't know how else to find the sum to n, or to infinity.
    Thanks!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by arze View Post
    Show that n(n+1)\equiv(n-1)(n-2)+4(n-1)+2 and hence, that for n\geq 3,
    \frac{n(n+1)}{(n-1)!}=\frac{1}{(n-3)!}+\frac{4}{(n-2)!}+\frac{2}{(n-1)!}
    Deduce or find otherwise the sum to infinity of the series
    1.2+\frac{2.3}{1!}+\frac{3.4}{2!}+...+\frac{n(n+1)  }{(n-1)!}+...

    I've done the first two parts correctly, but I don't know how to do this last part.
    \sum^{n}_{r=1} \frac{r(r+1)}{(r-1)!}=\sum^{n}_{r=1}\frac{1}{(r-3)!}+\frac{4}{(r-2)!}+\frac{2}{(r-1)!}
    This is the furthest I got, where n<3, I can't find values, so I don't know how else to find the sum to n, or to infinity.
    Thanks!
     \displaystyle \sum_{n=1}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum_{n=3}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum^{\infty}_{n=3}\left(\frac1{(n-3)!}+\frac4{(n-2)!}+\frac2{(n-1)!}\right)

    Now use the fact that  \displaystyle e=\sum_{n=0}^\infty \frac1{n!}
    Last edited by chiph588@; August 12th 2010 at 09:24 PM.
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  3. #3
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    Quote Originally Posted by chiph588@ View Post
     \displaystyle \sum_{n=1}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum_{n=3}^\infty \frac{n(n+1)}{(n-1)!} = 8+\sum^{\infty}_{n=3}\left(\frac1{(n-3)!}+\frac4{(n-2)!}+\frac2{(n-1)!}\right)

    Now use the fact that  \displaystyle e=\sum_{n=0}^\infty \frac1{n!}
    so i have
    \displaystyle 8+(\sum_{n=0}^{\infty}\frac{1}{n!}-1-\frac{1}{n!}-\frac{1}{(n-1)!}-\frac{1}{(n-2)!})+2(\sum_{n=0}^{\infty}\frac{1}{n!}-1-\frac{1}{n!}-\frac{1}{(n-1)!})+4(\sum_{n=0}^{\infty}\frac{1}{n!}-1-\frac{1}{n!})
    8+(e-1-0-0-0)+2(e-1-0-0)+4(e-1-0)=7e+1
    Answer is 7e, the eight is the problem it think
    Thanks
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  4. #4
    MHF Contributor chiph588@'s Avatar
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     \displaystyle \sum_{n=3}^\infty \frac1{(n-3)!} = \sum_{n=0}^\infty \frac1{n!} = e

     \displaystyle \sum_{n=3}^\infty \frac4{(n-2)!} = 4\sum_{n=1}^\infty \frac1{n!} = 4\sum_{n=0}^\infty \frac1{n!}-4 = 4e-4

     \displaystyle \sum_{n=3}^\infty \frac2{(n-1)!} = 2\sum_{n=2}^\infty \frac1{n!} = 2\sum_{n=0}^\infty \frac1{n!}-2-2 = 2e-4
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