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

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.

• Better use Stirling. Mar 10, 2014 at 8:17
• You could use the fact that for a sequence $(a_n)$ of positive terms, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}\root n\of{a_n}$ and the two limits are equal. Apply this to $a_n=n!$ Mar 10, 2014 at 8:19
• This is utterly illegible. Please use latex mode to write formulas, or at the very least fix those linebreaks...
– fgp
Mar 10, 2014 at 8:20
• See also $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite and other posts linked there. Feb 25, 2017 at 1:52
• @MaximilianJanisch I have mentioned your suggestion in CRUDE chatroom, perhaps somebody responds there. Dec 23, 2019 at 10:20

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 $$(n!)^2 \ge n^{n} \quad\text{and therefore}\quad (n!)^{1/n}\ge \sqrt{n}.$$

• this is by far the simplest answer +1 Jun 13, 2015 at 10:22
• @AndréNicolas With $n/2$ no justification is required as one of the factors will be at least $n/2$. It's actually true that $0\le k<n$ implies $(k+1)(n-k)>n$, but why not using an obvious inequality instead? Mar 3, 2016 at 13:24
• Because the original inequality is stronger (though far short of the truth) and prettier. Mar 3, 2016 at 13:27
• @AndréNicolas can you please show me how to prove $\ (k+1)\cdot (n-k)\ge n$? Apr 22, 2020 at 8:12
• @ManjoyDas If $k=0$, it's obvious. Now for $1\leq k\leq\frac n2$, you have $(k+1)\cdot (n-k)\geq (k+1)\frac n2\geq n.$ Similarly, complete the other case.
– cqfd
Apr 22, 2020 at 14:35

\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.

• How the sum of logarithms becomes greater than the integration?
– MAS
Jan 3, 2021 at 15:22
• @BijanDatta Replace the sum to the integration by using a step function and compare the step function with $\ln x$. Jan 3, 2021 at 15:38

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 $\lim n!^{1/n}=\infty$.

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?

• how does the 1st inequality hold? Apr 22, 2020 at 8:08
• all of the factors exist in the LHS product plus more Apr 22, 2020 at 8:58
• oh yess. I missed that n is even Apr 22, 2020 at 9:34
• The theorem is correct for all $n$ of course, but writing it would just involve some floors or ceils - not sure... Apr 22, 2020 at 10:34
• I think $1\cdot2\cdots n < \left(\frac{n}{2}+1\right)\left(\frac{n}{2}+2\right)\cdots \left(\frac{n}{2}+n\right)$ would be right Jun 3, 2021 at 16:41

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 $n$.

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).

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.