watchmath
  • watchmath
Calculate \[\sum_{n=1}^\infty \frac{n-1}{n!}\]
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
0
anonymous
  • anonymous
it has to be positive though
watchmath
  • watchmath
Can't be zero since it is a positive series starting from \(n=2\)

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anonymous
  • anonymous
I can't prove it, but the sum seems to approach one, in that every successive sum is (n!-1)/n!
watchmath
  • watchmath
Well observation is half of the proof thebestpig :). You just need to push a little bit more :D.
anonymous
  • anonymous
gimme a second - i'll do an induction thing
anonymous
  • anonymous
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.
anonymous
  • anonymous
oh wait one more thing
anonymous
  • anonymous
\[\lim_{n \rightarrow \infty} \frac{n!-1}{n!}=1\]
watchmath
  • watchmath
Awesome! :)
watchmath
  • watchmath
We can also using telescoping sum. Note that \[\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}\]
anonymous
  • anonymous
ha, /facepalm/
anonymous
  • anonymous
well, I enjoy using induction :)
anonymous
  • anonymous
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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