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cherrilyn
 5 years ago
approximate the value of the series to within an error of at most 10^5 ?
cherrilyn
 5 years ago
approximate the value of the series to within an error of at most 10^5 ?

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cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=1}^{\infty} (1)^{n+1}\ln n/n! \]

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0I know SnS\[\le\] An+1 = ln(N+1)/(N+1)! \[\le 10^{5}\]

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Then from there we just make a guess what is the smallest n that will do the job. So you pick some n if it doesn't work, try n+1 an so on.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0You can take N so that \[\frac{\ln(N+1)}{(N+1)!} \leq \frac{N+1}{(N+1)!} = \frac{1}{N!}\leq 10^{5}\]

cherrilyn
 5 years ago
Best ResponseYou've already chosen the best response.0okayy. now how would I go about finding the exact answer?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Just add all the summands from N=1 to \(N! \leq 10^5\)
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