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Math Help - finding the max and min value of n

  1. #1
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    finding the max and min value of n

    Hi, what can be the limiting values of n for which n!>a^n for any given a?
    Thanks.
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  2. #2
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    Or in other words for which minimal n the next inequality always follows for any a>0:
    n!^{1/n}>a

    just pick n!^{1/n}=[a]+1.
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  3. #3
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    Thank. but how do i get n from that,and moreover it gives a upper limit to n ,i suppose, how to get the lower limit
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    Quote Originally Posted by pranay View Post
    Thank. but how do i get n from that,and moreover it gives a upper limit to n ,i suppose, how to get the lower limit
    I haven't worked this out completely, but I suspect you might find Stirling's approximation useful:
    \displaystyle n! \approx \sqrt{2 \pi n} \left ( \frac{n}{e} \right ) ^n

    That would give your problem the following form:
    \displaystyle \sqrt{2 \pi} n^{n + 1/2} > (ea)^n

    -Dan
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  5. #5
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    Quote Originally Posted by pranay View Post
    Thank. but how do i get n from that,and moreover it gives a upper limit to n ,i suppose, how to get the lower limit
    Well first thing first, what I wrote is the lowest n possible (there isn't a maximum cause you can always pick another one greater by adding 1 to the preceding number).

    I am unaware of a closed analytical formula for what I wrote, I guess only Stirling approximation can work here (if you want to find some closed formula though still an approximation).
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