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watchmath

  • 5 years ago

Calculate \[\sum_{n=1}^\infty \frac{n-1}{n!}\]

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  1. anonymous
    • 5 years ago
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    0

  2. anonymous
    • 5 years ago
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    it has to be positive though

  3. watchmath
    • 5 years ago
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    Can't be zero since it is a positive series starting from \(n=2\)

  4. anonymous
    • 5 years ago
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    I can't prove it, but the sum seems to approach one, in that every successive sum is (n!-1)/n!

  5. watchmath
    • 5 years ago
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    Well observation is half of the proof thebestpig :). You just need to push a little bit more :D.

  6. anonymous
    • 5 years ago
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    gimme a second - i'll do an induction thing

  7. anonymous
    • 5 years ago
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    the base step is \[(1!-1)/1! = 0\] which is in fact the first term, and \[\frac{(n!-1)}{n!}+\frac{n}{(n+1)!}=\frac{(n!-1)(n+1)+n}{(n+1)!}=\frac{(n+1)!-n-1+n}{(n+1)!}\] \[=\frac{(n+1)!-1}{(n+1)!}\] and it looks like we're done.

  8. anonymous
    • 5 years ago
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    oh wait one more thing

  9. anonymous
    • 5 years ago
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    \[\lim_{n \rightarrow \infty} \frac{n!-1}{n!}=1\]

  10. watchmath
    • 5 years ago
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    Awesome! :)

  11. watchmath
    • 5 years ago
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    We can also using telescoping sum. Note that \[\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}\]

  12. anonymous
    • 5 years ago
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    ha, /facepalm/

  13. anonymous
    • 5 years ago
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    well, I enjoy using induction :)

  14. anonymous
    • 5 years ago
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    Here's another way: a little relabeling gives (check it!) \[ \sum_{n=1}^\infty \frac{n-1}{n!} = \sum_{n=0}^{\infty}\frac{1}{(n+2)n!} \] Now \[ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}, \] so \[ xe^x = \sum_{n=0}^\infty \frac{x^{n+1}}{n!}, \] so \[ \int xe^x = xe^x-e^x +C = \sum_{n=0}^\infty \frac{x^{n+2}}{(n+2)n!} \] Plug in \(x=0\) to find that \(C =1\), so \[ 1 e^1 -e^1 +1 = 1 = \sum_{n=0}^\infty \frac{1}{(n+2)n!}. \]

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