Limit of $n!^{1/n^2}$ using Squeeze Theorem?

Question

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 $n^{n^{1/n^2}}$.

Presumably, you mean $n^{1/n^2}\to 1$, not $=1$. – Thomas Andrews Oct 30 '15 at 17:17 — Thomas Andrews, Oct 30 '15 at 17:17
@ThomasAndrews What is the difference? – user270494 Nov 01 '15 at 00:08 — user270494, Nov 01 '15 at 00:08

score 3 · Accepted Answer · answered Oct 30 '15 at 17:19

3

Hint: $n!^{1/n^2} \le (n^n)^{1/n^2}=n^{1/n}.$

answered Oct 30 '15 at 17:19

zhw.

105,693

score 0 · Answer 2 · answered Oct 30 '15 at 17:37

0

One more way: $t=e^{log t}$ and then compare the sum over $ \log n$ to the integral. The limit is 1.

answered Oct 30 '15 at 17:37

Alex

19,262

score 0 · Answer 3 · answered Jan 13 '22 at 16:07

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

$$\lim \limits_{n \to \infty} (n!)^{\frac{1}{n^{2}}}=1$$

Limit of $n!^{1/n^2}$ using Squeeze Theorem?

3 Answers3