Limit of $n!^{1/n^2}$ using Squeeze Theorem?
I'm trying to find the limit of $n!^{1/n^2}$ as $n$ goes to infinity.
What I've done so far is:
I know that $n < n^2 < n! < n^n$ for large $n$,
and I know that $n^{1/n^2} = 1$, but I'm not sure how to find the limit of $n^{n^{1/n^2}}$.
$\endgroup$ 23 Answers
$\begingroup$Hint: $n!^{1/n^2} \le (n^n)^{1/n^2}=n^{1/n}.$
$\endgroup$ $\begingroup$One more way: $t=e^{log t}$ and then compare the sum over $ \log n$ to the integral. The limit is 1.
$\endgroup$ $\begingroup$It is clear that $(n!)^{\frac{1}{n^{2}}}≤(n^{n})^{\frac{1}{n^{2}}}$ since $n!≤n^{n}$, the above implies that$$(n!)^{\frac{1}{n^{2}}}≤(n^{n})^{\frac{1}{n^{2}}}=n^{\frac{1}{n}}$$
As in turn $1≤n!$, then
$$1≤(n!)^{\frac{1}{n^{2}}}≤n^{\frac{1}{n}}$$
And as $\lim \limits_{n \to \infty} (1) = \lim \limits_{n \to \infty} (n^{\frac{1}{n}}) = 1$ then by Squeeze Theorem
$$\lim \limits_{n \to \infty} (n!)^{\frac{1}{n^{2}}}=1$$
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