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I'm working through the following exercise which asks me to use the squeeze theorem to determine the limit of: $$n^{(\frac{1}{n^2})}$$ and $$n!^{(\frac{1}{n^2})}$$

I figured that both limits approach $1$ as $(\frac{1}{n^2})$ approaches $0$. But, I'm having trouble determining the functions to apply the squeeze theorem. Is there a general method as to figuring out which functions to squeeze a sequence between?

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  • $\begingroup$ Do you know Stirling's approximation? I think that will help. $\endgroup$ – Vim Jul 13 '15 at 4:38
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    $\begingroup$ For the second, it will help to note that $n!\le n^n$ if $n\gt 0$. $\endgroup$ – André Nicolas Jul 13 '15 at 4:41
  • $\begingroup$ For the first one, you might find $n\leq n^n$ for $n>1$ to be useful. $\endgroup$ – Michael Burr Jul 13 '15 at 4:42
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Hint we don't even need Stirling here, just note that with $n$ very large $$1<n<n^n$$ $$1<n!<n^n$$

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If you know that $n^{1/n} \to 1$ as $n \to \infty$, you can use the fact that $x^{1/x}$ is decreasing for $x > e$ and squeeze: $$1 \leq n^{1/n^2} \leq n!^{1/n^2} \leq n^{1/n}.$$ Re: your question about how to choose the squeezing bounds. Usually this comes from doing some dirty work (i.e. computing for some values) or knowing some bounds by practice. Here, I saw that $n^{1/n^2}$ looks a lot like $n^{1/n}$, and it just so happens that it works. The idea is similar to that of direct/limit comparison tests for series -- finding the right series to compare to is hard, but after seeing a several approaches and tricks, you get the idea and can solve (textbook) problems.

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