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Math Help - Alternating Series Estimation Theorem

  1. #1
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    Alternating Series Estimation Theorem

    Use the alternating series estimation theorem to approximate the sum of  (-1)^n/(n!) from n=0 to infinity with an error of .000005.

    I know that after summing a number of terms, the remainder(error) will be less than the first omitted term, so

    I've set up the problem like this:

    .000005 < 1/(n+1)!

    Is this right, and if so, what's next?
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  2. #2
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    Quote Originally Posted by JewelsofHearts View Post
    Use the alternating series estimation theorem to approximate the sum of  (-1)^n/(n!) from n=0 to infinity with an error of .000005.

    I know that after summing a number of terms, the remainder(error) will be less than the first omitted term, so

    I've set up the problem like this:

    .000005 < 1/(n+1)!

    Is this right, and if so, what's next?
    So you want the least integer $$n such that:

    (n+1)!>20000

    To proceed you need either to use trial and error or some suitable approximation for the factorial (and then a numerical solution of the resulting equation).

    (look up $$7! and $$8!)

    CB
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  3. #3
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    Shouldn't it be (n+1)!<200000?
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  4. #4
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    Quote Originally Posted by JewelsofHearts View Post
    Shouldn't it be (n+1)!<200000?
    Probably counting zeros is not something I am particularly careful about
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  5. #5
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    Not only the zeros, but I think the direction of the inequality sign is supposed to be different. I really don't know, though. I don't understand the theorem.

    Help, please!
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  6. #6
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    error, E < \dfrac{1}{(n+1)!} < 5 \times 10^{-6}

    (n+1)! > \dfrac{1}{5 \times 10^{-6}} = 2 \times 10^5

    note the following factorial values ...

    8! = 40320

    9! = 362880

    so, how many terms of the series would you need to add to get an error less than 5 \times 10^{-6} ?

    fyi, to check your work, note ...

    \displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = \frac{1}{e}
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