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Math Help - Direct Comparison test of 1/n! to what?

  1. #1
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    Direct Comparison test of 1/n! to what?

    What should I compare \sum_{n \ge 0} \frac 1{n!} if the question requires me to prove its convergence/divergence using the direct comparison test. I know it converges. It seems like \frac 1{n!} < \frac 1{n^2} for n \ge 4, and \frac 1{n^2} converges by p-series test. The first four terms are obviously finite. Is that enough? Is there a better series to compare it to?
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  2. #2
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    Re: Direct Comparison test of 1/n! to what?

    Hey MSUMathStdnt.

    The p-series comparison should be more than sufficient.
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  3. #3
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    Re: Direct Comparison test of 1/n! to what?

    How do you prove that 1/n! < 1/n^2?
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  4. #4
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    Re: Direct Comparison test of 1/n! to what?

    Prove n! > n^2 for all appropriate n (i.e. n > 3).
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    Re: Direct Comparison test of 1/n! to what?

    which is the same as proving (n-1)! > n. induction is your friend, here.

    or you can just use "brute force":

    (n-1)! > (n-1)(n-2) = n2 - 3n + 2 > n2 - 3n = n(n - 3) ≥ n, if n ≥ 4 (so that n - 3 ≥ 1).
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  6. #6
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    Re: Direct Comparison test of 1/n! to what?

    Or, what amounts to the same thing as Deveno did, if n \ge 4,

    n! \ge n(n-1)(n-2) = n^3 - 3n^2 + 2n > n^3 - 3n^2 = n^2(n-3) \ge n^2.

    - Hollywood
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  7. #7
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    Re: Direct Comparison test of 1/n! to what?

    How come some threads have a little "thanks" link in the posts of people who reply to you and some (like this one) don't?
    P.S. Thanks for replies, guys (and gals, if any).
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  8. #8
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    Re: Direct Comparison test of 1/n! to what?

    Some guesses:

    1. You need to be logged in - in fact, I think you need to be logged in when you download the page.

    2. You can not give thanks to yourself.

    3. You have to mouse over the post.

    It's working for me on this thread now - I've never noticed that it's not working.

    - Hollywood
    Last edited by hollywood; December 21st 2012 at 07:07 AM. Reason: correction to #3
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