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cherrilyn

  • 5 years ago

approximate the value of the series to within an error of at most 10^-5 ?

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  1. cherrilyn
    • 5 years ago
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    \[\sum_{n=1}^{\infty} (1)^{n+1}\ln n/n! \]

  2. cherrilyn
    • 5 years ago
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    I know |Sn-S|\[\le\] An+1 = ln(N+1)/(N+1)! \[\le 10^{-5}\]

  3. watchmath
    • 5 years ago
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    Then 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.

  4. nowhereman
    • 5 years ago
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    You can take N so that \[\frac{\ln(N+1)}{(N+1)!} \leq \frac{N+1}{(N+1)!} = \frac{1}{N!}\leq 10^{-5}\]

  5. cherrilyn
    • 5 years ago
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    okayy. now how would I go about finding the exact answer?

  6. nowhereman
    • 5 years ago
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    Just add all the summands from N=1 to \(N! \leq 10^5\)

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