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watchmath
 5 years ago
Calculate
\[\sum_{n=1}^\infty \frac{n1}{n!}\]
watchmath
 5 years ago
Calculate \[\sum_{n=1}^\infty \frac{n1}{n!}\]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it has to be positive though

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Can't be zero since it is a positive series starting from \(n=2\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can't prove it, but the sum seems to approach one, in that every successive sum is (n!1)/n!

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Well observation is half of the proof thebestpig :). You just need to push a little bit more :D.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gimme a second  i'll do an induction thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the base step is \[(1!1)/1! = 0\] which is in fact the first term, and \[\frac{(n!1)}{n!}+\frac{n}{(n+1)!}=\frac{(n!1)(n+1)+n}{(n+1)!}=\frac{(n+1)!n1+n}{(n+1)!}\] \[=\frac{(n+1)!1}{(n+1)!}\] and it looks like we're done.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait one more thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{n \rightarrow \infty} \frac{n!1}{n!}=1\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0We can also using telescoping sum. Note that \[\frac{n1}{n!}=\frac{1}{(n1)!}\frac{1}{n!}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, I enjoy using induction :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here's another way: a little relabeling gives (check it!) \[ \sum_{n=1}^\infty \frac{n1}{n!} = \sum_{n=0}^{\infty}\frac{1}{(n+2)n!} \] Now \[ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}, \] so \[ xe^x = \sum_{n=0}^\infty \frac{x^{n+1}}{n!}, \] so \[ \int xe^x = xe^xe^x +C = \sum_{n=0}^\infty \frac{x^{n+2}}{(n+2)n!} \] 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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