Calculating the limit $\lim((n!)^{1/n})$

Question

Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm, $$\lim_{n\to\infty} \ln((n!)^{1/n}) = \lim_{n\to\infty} (1/n)\ln(n!) = \lim_{n\to\infty} (\ln(n!)/n)$$ Applying L'hopital's rule: $$\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))= \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k)) = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k) = \lim_{n\to\infty} (\ln(n))$$ I proceeded to expand the $\ln(n)$ out into Maclaurin form $$\lim_{n\to\infty} (n + (n^2/2)+...) = \infty$$ Since I $\ln$'ed in the beginning, I proceeded to e the infinity $$= e^\infty = \infty$$

So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it.

Answer 1

By rearranging terms, we can see that $$(n!)^2=[1\cdot n][2\cdot (n-1)][3\cdot (n-2)] \cdots [(n-1)\cdot 2][n\cdot 1].$$ Each of the $n$ products $(k+1)\cdot (n-k)$, for $0\le k<n$, is $\ge n$. Thus $$(n!)^2 \ge n^{n} \quad\text{and therefore}\quad (n!)^{1/n}\ge \sqrt{n}.$$

Answer 2

$$ \begin{aligned} \lim_{n\to\infty} (n!)^{1/n} &=\lim_{n\to\infty} \exp(\tfrac{1}{n} \ln n!)\\ &= \lim_{n\to\infty} \exp[\tfrac{1}{n} (\ln 1+\ln 2+\cdots + \ln n)]\\ &\ge \lim_{n\to\infty}\exp \left[ \frac{1}{n} \int_1 ^n \ln x dx\right]\\ &=\lim_{n\to\infty} \exp \frac{n\ln n -n+1}{n} \end{aligned} $$

and last side of above inequality diverges.

Answer 3

Taking $\log$ and using Stolz-Cesaro: $$ \log\lim_{n\to\infty}n!^{1/n}= \lim_{n\to\infty}\log n!^{1/n}= \lim_{n\to\infty}{\log 1+\cdots+\log n\over n}= \lim_{n\to\infty}{\log(n+1)\over(n+1)-n}= \lim_{n\to\infty}\log(n+1)=\infty,$$ so $\lim n!^{1/n}=\infty$.

Answer 4

Assume WLOG that $n$ is even.

Clearly, $$n!=1\cdot2\cdots n > \left(\frac{n}{2}+1\right)\left(\frac{n}{2}+2\right)\cdots n> \left(\frac{n}{2}\right)^{\frac{n}{2}}$$

Can you take it from here?

Answer 5

As suggested by Martín-Blas Pérez Pinilla, using Stirling approximation $$n!\simeq\sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ helps a lot. Raising to power $\frac{1}{n}$ then leads to $$(n!)^{\frac{1}{n}}\simeq\ (2 \pi n)^{\frac{1}{n}} \frac{n}{e}$$ For large values of $n$, the first term goes to $1$ and so $(n!)^{\frac{1}{n}}$ behaves as $\frac{n}{e}$

A better approximation can be obtained using Taylor series; writing the beginning of the expansion as $$(n!)^{\frac{1}{n}}\simeq\frac{n}{e}+\frac{\log (2 \pi n)}{2 e}$$ which shows how would behave $$\frac{(n!)^{\frac{1}{n}}}{n}$$ for large values of $n$.

Answer 6

The inequalities I often find useful because of their simplicity are $(n/e)^n < n! < (n/e)^{n+1} $. These give $\frac{(n!)^{1/n}}{n} \to 1/e $.

These follow from $(1+1/n)^n < e < (1+1/n)^{n+1} $ (which have been proven here many times).

Answer 7

Let a $n\in \Bbb N$. By definition $$[\frac n2]\leq \frac n2<[\frac n2]+1.$$ Then $n!=1\cdot 2\cdot ...\cdot[\frac n2]\cdot ([\frac n2]+1)\cdot...\cdot n>(\frac n2)^{n-[\frac n2]+a}>(\frac n2)^{\frac n2 +a}$, so $(n!)^{\frac 1n}>(\frac n2)^{\frac 12 + \frac an}\to \infty$, thus $(n!)^{\frac 1n}\to \infty.$ We set $a:=0$, if $n$ is even, and $a:=1$, if it is odd.