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$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

We have

$$\lim_{n \to \infty} n^{1/n} = 1$$

But in my prep for a Real Analysis exam, I came across the following modification:

$$\lim_{n \to \infty} (n!)^{1/n} = ? $$

and got stumped because the usual method of taking natural logarithms does not seem to work. (Or perhaps it requires some amendment that I'm not seeing?)

Any help would be appreciated. Thanks.

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marked as duplicate by Marvis, Hagen von Eitzen, Norbert, Thomas, Emily Nov 8 '12 at 21:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Stirling approximation for $n!$ ? – vesszabo Nov 8 '12 at 17:24
up vote 7 down vote accepted

Here is a simple proof of $\lim (n!)^{\frac{1}{n}}=+\infty.$

Observe that $(n!)^{\frac{1}{n}}\geq (2\cdot2\cdots2)^{\frac{1}{n}}=(2^{n-1})^{\frac{1}{n}}=2^{1-\frac{1}{n}} \to 2$.

In general $\forall k \in \mathbb{N}$ and $n \geq k$ we have $(n!)^{\frac{1}{n}}\geq (k\cdot k\cdots k)^{\frac{1}{n}}=(k^{n-(k-1)})^{\frac{1}{n}}=k^{1-\frac{k-1}{n}} \to k$.

Therefore $\forall k \in \mathbb{N}, \ \exists n_0 \in \mathbb{N}: \forall n \geq n_0 \ \ (n!)^{\frac{1}{n}} \geq k-1$ (choose $n_0$ s.t. $\forall n \geq n_0 \ \ k^{1-\frac{k-1}{n}} \geq k-1$ ).
Hence $\lim_{n \to \infty} (n!)^{\frac{1}{n}} = \infty .$

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This is true in spirit, but false as stated. – Max Nov 10 '12 at 17:53
I believe now is better. – P.. Nov 11 '12 at 14:34
Yes, now it is. – Max Nov 11 '12 at 17:43

Note that

$$ \frac{\log n!}{n} = \log n + \sum_{k=1}^{n} \log\left( \frac{k}{n} \right)\frac{1}{n} $$

and also

$$ \int_{0}^{1} \log x \, dx \leq \sum_{k=1}^{n} \log\left( \frac{k}{n} \right)\frac{1}{n} \leq \int_{1/n}^{1} \log x \, dx $$

by comparing the area as we can see from below:

enter image description here

This shows that

$$ \frac{\log n!}{n} = \log n - 1 + o(1)$$

and hence

$$(n!)^{1/n} = \exp\left(\frac{\log n!}{n}\right) = \frac{n}{e}(1+o(1)). $$

In particular, the limit diverges. (If we are only interested in the convergence, we may argue by some much simpler arguments.)

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We have (Stirling formula) $$n!\approx n^n e^{-n}\sqrt{2\pi n},$$ hence $\sqrt[n]{n!}\approx \frac ne$ for large $n$.

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