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Math Help - calculating limit of a sequence

  1. #1
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    calculating limit of a sequence

    i need to calculate the limit of this sequence:

    \sqrt[n^2]{n!}

    i tried to use the sandwich rule to squeeze it between to sequences, but got stuck.

    this is what i did so far:

    \sqrt[n^2]{n!}<\sqrt[n^2]{n\cdot n\cdot \cdot \cdot \cdot \cdot n}=\sqrt[n^2]{n^n}=\sqrt[n]{n}\rightarrow 1

    but i couldn't find anything smaller then the given sequence that can be easily calculated.
    for example, i know that:

    \sqrt[n^2]{n}<\sqrt[n^2]{n!}

    and i know that \sqrt[n]{n}\rightarrow 1, but is it enough to say that if \sqrt[n]{n}\rightarrow 1 then also \sqrt[n^2]{n}\rightarrow 1?

    thanks in advanced!!!
    BTW, if there's any other way to get to the answer ( \sqrt[n^2]{n!}\rightarrow 1) i'd love to hear.
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  2. #2
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    Re: calculating limit of a sequence

    \sqrt[n^2]{n}=e^{(\ln n)/n^2}\to1 because (\ln n)/n^2\to0 as n\to\infty. In fact, without using n=e^{\ln n}, the function a^{1/x} where a > 1 monotonically decreases as x increases. Therefore, n^{1/n^2}<n^{1/n}. On the other hand, n^{1/n^2}>1 for n > 1. Therefore, the fact that n^{1/n}\to1 indeed implies that n^{1/n^2}\to1.

    You could also try using Stirling's approximation.
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  3. #3
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    Re: calculating limit of a sequence

    brilliant.
    thanks!
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