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Math Help - size order sequences,

  1. #1
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    size order sequences,

    Can someone please explain how I would approach this question, other thanplugging in numbers and trying to figure it out from there?


    Is the last one the biggest ?


    thank you.
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  2. #2
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    Re: size order sequences,

    Quote Originally Posted by Tweety View Post
    Can someone please explain how I would approach this question, other thanplugging in numbers and trying to figure it out from there? Is the last one the biggest ?
    You need a lot of experience with these functions.

    For example, the first is an increasing sequence bounded above by e^2 its limit.

    You need to know that for F>1 then \log(F)< F.

    So \log(n^{2^{-1}}\log(n))< n^{2^{-1}}\log(n)

    Here is a hint on the last one: \frac{n!}{n^n}\le 1~. WHY?
    Last edited by Plato; January 16th 2014 at 02:40 PM.
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