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Currently learning how to use the squeeze theorem to determine a limit.

The exercise I'm working on is finding the limit of:

$\lim \ (n!)^{\frac{1}{n^2}}$

So far what I have is:

$\lim \ (n!)^{\frac{1}{n^2}} = \lim \ e^{\frac{1}{n^2}ln(n!)}$

So I know that $1 = e^0 \leq \lim \ e^{\frac{1}{n^2}\ln(n!)}$

But I need help determining what should be on the other side of the equality.

I'm trying to look at $\frac{\ln(n!)}{n^2}$ but I'm having trouble determining its limit as it approaches infinity.

Does $n^2$ increase faster than $\ln(n!)$? Could someone show me why which one increases faster?

If so the inequality would just be:

$1 = e^0 \leq \lim \ e^{\frac{1}{n^2}\ln(n!)} \leq e^0 = 1$

So our limit would just be $1$, right?

Thanks in advance.

Martin Sleziak
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Pandamonium
  • 975

2 Answers2

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Hint:
Use the inequality $$n^{\frac{n}{2}} \leqslant {n!} \leqslant {\left(\frac{n+1}{2}\right)^n}, \;\; n>1,$$ which can be proved by induction.

M. Strochyk
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Hint:

$$\log n!=\sum_{k=2}^n\log k\le n\log n\implies e^{\frac1{n^2}\sum_{k=2}^n\log k}\le e^{\frac1n\log n}=\sqrt[n]{\log n}$$

But also

$$\sqrt[n]{\log n}\le\sqrt[n]n\xrightarrow[n\to\infty]{}1$$

Timbuc
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