Possible Duplicate:
$ \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.
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 $\lim_{n \to \infty} (n!)^{\frac{1}{n}} = \infty .$
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:
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.)
We have (Stirling formula) $$n!\approx n^n e^{-n}\sqrt{2\pi n},$$ hence $\sqrt[n]{n!}\approx \frac ne$ for large $n$.
