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Thread: 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:

    $\displaystyle \sqrt[n^2]{n!}$

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

    this is what i did so far:

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

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

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

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

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

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

    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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