# Finding the limit of $\frac {n}{\sqrt[n]{n!}}$

I'm trying to find $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .$$

I tried couple of methods: Stolz, Squeeze, D'Alambert

Thanks!

Edit: I can't use Stirling.

• Hint: Stirling.
– Did
Mar 22, 2011 at 11:58
• @Didier: Thank you for the comment, but unless you ment the city in scotland, I didn't study stirling method yet.
– user6163
Mar 22, 2011 at 12:00
• Try taking the natural log and finding the limit of that. Mar 22, 2011 at 12:02
• Second try: Stirling formula.
– Did
Mar 22, 2011 at 12:06
• This is (after a slight modification) Problem 1.2.2 from Radulescu, Radulescu, Andreescu: Problems from Real Analysis, p.8. Oct 19, 2011 at 12:28

Let $\displaystyle{a_n=\frac{n^n}{n!}}$. Then the power series $\displaystyle{\sum_{n=1}^\infty a_n x^n}$ has radius of convergence $R$ satisfying $\displaystyle{\frac{1}{R}=\lim_{n\to \infty} \sqrt[n]{a_n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}}$, provided these limits exist. The first limit is what you're looking for, and the second limit is $\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$.

Added: I just happened upon a good reference for the equality of limits above, which gives a more general result which is proved directly without reference to power series. Theorem 3.37 of Rudin's Principles of mathematical analysis, 3rd Ed., says:

For any sequence $\{c_n\}$ of positive numbers, $$\liminf_{n\to\infty}\frac{c_{n+1}}{c_n}\leq\liminf_{n\to\infty}\sqrt[n]{c_n},$$ $$\limsup_{n\to\infty}\sqrt[n]{c_n}\leq\limsup_{n\to\infty}\frac{c_{n+1}}{c_n}.$$

In the present context, this shows that $$\liminf_{n\to\infty}\left(1+\frac{1}{n}\right)^n\leq\liminf_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\leq\limsup_{n\to\infty}\frac{n}{\sqrt[n]{n!}}\leq\limsup_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$$ Assuming you know what $\displaystyle{\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n}$ is, this shows both that the limit in question exists (in case you didn't already know by other means) and what it is.

From the comments: User9176 has pointed out that the case of the theorem above where $\displaystyle{\lim_{n\to\infty}\frac{c_{n+1}}{c_n}}$ exists follows from the Stolz–Cesàro theorem applied to finding the limit of $\displaystyle{\frac{\ln(c_n)}{n}}$. Explicitly, $$\lim_{n\to\infty}\ln(\sqrt[n]{c_n})=\lim_{n\to\infty}\frac{\ln(c_n)}{n}=\lim_{n\to\infty}\frac{\ln(c_{n+1})-\ln(c_n)}{(n+1)-n}=\lim_{n\to\infty}\ln\left(\frac{c_{n+1}}{c_n}\right),$$ provided the latter limit exists, where the second equality is by the Stolz–Cesàro theorem.

• @Theo: Thanks. @Nir: I know that this can be stated without explicit reference to power series and radii of convergence, but this answer reveals my bias toward thinking of power series. Is this what you meant by "delambre"? Do you have a reference for Delambre's theorem? (My internet searching isn't successful.) Mar 22, 2011 at 13:48
• @jonas: Yeah, this is what I ment. I really tried to look for a reference for that in english but I couldn't find anything that relates to this specifically, I'm sorry.
– user6163
Mar 22, 2011 at 13:58
• @Nir: A reference not in English would be good, too. (With probability approximately one I cannot read the reference you have, but I could use online translators.) I'm really curious because my searches for delambre and limits don't turn up anything relevant. Mar 22, 2011 at 14:04
• @lhf: Of course! Thanks. I didn't think of that in part because searching for a mathematician named Delambre turned up this guy: www-history.mcs.st-and.ac.uk/Mathematicians/Delambre.html D'Alembert is credited with the ratio test. I'd still be interested in a reference to this particular "trick" that doesn't use power series as an intermediary (not that I have any thing against power series). Mar 23, 2011 at 1:07
• Just a short comment, the mentioned Theorem is just the Stolz-Cezaro theorem applied to $\frac{\ln (a_n)}{n}$. Apr 10, 2011 at 5:53

This is going to be a bit difficult (since apparently lots of things aren't allowed). Here's how I would do it (this is far from a complete solution but just a couple of hints):

I hope you know that $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{n}$ (this is often taken as the definition of $e$).

You can show easily that the sequence $c_{k} = \left(1 + \frac{1}{k}\right)^k$ is monotonically increasing and that the sequence $d_{k} = \left(1 + \frac{1}{k}\right)^{k+1}$ is monotonically decreasing. This gives the squeezing $$\displaystyle \left(1 + \frac{1}{k}\right)^k = c_k \lt e \lt d_k = \left(1 + \frac{1}{k}\right)^{k+1}.$$

By taking the products $c_{1} c_{2} \cdots c_{n}$ and $d_{1} d_{2} \cdots d_{n}$ you can then show $$\displaystyle \frac{(n+1)^n}{n!} \lt e^n \lt \frac{(n+1)^{n+1}}{n!}$$ using a few manipulations.

Now extract roots on both sides of the last inequalities and you're there.

By applying Cauchy-d'Alembert criterion we get that:

$$\lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}=\lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{\frac{1}{n}} \\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}} = \lim_{n\to\infty} \frac{(n+1)^{(n+1)}}{(n+1)!}\cdot \frac{n!}{n^n} \\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}= \lim_{n\to\infty} \frac{(n+1)^n}{n^n} \\ \lim_{n\to\infty} \frac{n}{n!^{\frac{1}{n}}}=\lim_{n\to\infty} {\left(1+\frac{1}{n}\right)^{n}}=e.$$

Q.E.D.

• I'll just add that the result used here is shown in this post: math.stackexchange.com/questions/287932/… (And many other posts on this site.) This is pointed out more explicitly in Jonas Meyer's answer. Jul 16, 2014 at 8:53

If $f(n)=\frac{n}{\sqrt[n]{n!}}$ and $g(n) = f(n)^n$ then

$$g(n) = \frac{n^n}{n!}$$

and taking the ratio of terms, removing the factorials and using $\frac{n+1}{n} = 1+\frac{1}{n}$,

$$\frac{g(n+1)}{g(n)} = \left(1 + \frac{1}{n}\right)^n$$

You may recognise this as having a limit of $e$. It implies

$$\lim_{n \to \infty} \frac{g(n+1)}{g(n)} \frac{1}{e} = 1$$

and so multiplying a string of these together

$$\lim_{n \to \infty} \frac{g(n)}{e^n h(n)} = 1$$

for some function $h(n)$ which grows more slowly than $e^n$ or decays more gently than $e^{-n}$, [not that it matters, but $h(n)$ is about $1/\sqrt{2 \pi n}$] so taking the $n$-th root

$$\lim_{n \to \infty} \frac{f(n)}{e} = \lim_{n \to \infty} h(n)^{1/n} = 1$$

and so $\lim_{n \to \infty} f(n) = e$.

• +1. Any fine answer should end with a statement which shows the answer to the OP question as you did it !!!. May 25, 2014 at 6:49

If you take the log, it is:

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

Which is a Riemann sum for $$\int_{0}^1 \log x$$.

The indefinite integral is $$F(x)=x\log x-x$$ and $$\lim_{x\to 0} x\log x -x =0$$, and $$F(1)=-1$$.

You have to deal with the fact that this integral is an improper integral, but it "just works."

Let $[x]$ denote the largest integer not exceeding $x.$ For $n\geq 1$ we have $$\log n! =\int_1^{n+1}\log [x]\; dx<\int_1^{n+1}\log x\; dx=-n+(n+1)\log (n+1)$$ and $$\log n!=\int_1^n \log (1+[x]) \;dx\geq \int_1^n\log x \;dx=1-n+n\log n.$$ So $$1/n\leq 1+\log ( (n!^{1/n}/n)<(1+1/n)\log (n+1)-\log n=\log (1+1/n)+(1/n)\log (n+1).$$ Since $(1/n)\log (n+1)\to 0$ as $n\to \infty$ we have $$\lim_{n\to \infty}\log (n!^{1/n}/n)=-1.$$

what's wrong with just logging the expression? $$\varphi (n) = \frac{n}{n!^{\frac{1}{n}}}\\ L \varphi(n) = \log \varphi(n) = \log n - \frac{\log n!}{n} = \log n -\sum_{k=1}^{n}\frac{\log k}{n} \\ \sim \log n -\frac{n \log n -n + 1 }{n} = \log n - \log n +1 + \frac{1}{n}= 1 + o(1)$$ Hence $\lim_{n \to \infty} \varphi(n) =e^1 = e$

EDIT: to make things sharper, here's the approximation using Euler-Maclaurin formula: $\sum_{k=1}^{n} \log k = \int_{1}^{n}\log x dx + O(\log n) = n \log n -n +1 +O(\log n ).$ Obviously $\lim_{n \to \infty} \frac{\log n }{n} = 0$, hence the statement above holds: $$\frac{n \log n -n -\frac{1}{2} \log n +1}{n} = \log n -1 +o(1)$$ and the result holds because $e^{o(1)} = 1$.

• thanks, I fixed the first part. I guess Euler-Maclaurin approximation should be enough to show that the remainder terms $\to 0$ and hence when exponentiated I get $e^{1+o(1)} = e^1$
– Alex
Jun 9, 2014 at 18:07
• Euler-Maclaurin is more than enough. You get the necessary bounds by more elementary means already (but of course, if you have Euler-Maclaurin, why not use it?). Jun 9, 2014 at 18:14
– Alex
Jun 9, 2014 at 18:34

$$\lim\frac{n}{\sqrt[^n]{n!}}=\lim \sqrt[n]{\frac{n^n}{n!}}=\lim \sqrt[n]{\frac{n}{1}\frac{n}{2}\dots\frac{n}{n}}=\lim \exp\bigg({\cfrac{\ln\frac{n}{1}+\ln\frac{n}{2}+\dots+\ln\frac{n}{n}}{n}}\bigg)=$$$$=\lim \exp\bigg({\frac{1}{n}\sum_{k=1}^n\ln\frac{n}{k}}\bigg)=\exp\bigg(\lim{\frac{1}{n}\sum_{k=1}^n\ln\frac{n}{k}}\bigg)=$$$$=\exp\bigg(\int_0^1(-\ln x)dx\bigg)=\exp(1)=e \ \ \ \ \square$$

By the way, you can calculate $$\int_0^1(-\ln x)dx$$ easily by integrating by parts.

One other method is to also derive an assymptote for the factorial function on the fly. Here's a short proof. Take, $$\ln(n!)=\sum_{i=1}^{n} ln(i) \sim \int_{1}^{n}\ln(t) dt=n\ln(n)-n+1$$ Exponentiation each side, we get, $$n!\sim n^ne^{-n}e=e\left(\frac{n}{e}\right)^n \sim \left(\frac{n}{e}\right)^n$$ Substituting this in the question, we get $$\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{\left(\frac{n}{e}\right)^n}}=e$$

Edit, by the way you have used the Stirling's approximation, indirectly, here's the Stirling's approximation used for practical purposes; $$n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ In the above case, we won't need to worry about the $$\sqrt{2\pi n}$$ thing. As $$(n/e)^n$$ is the dominating factor here.

\begin{align*} \lim_{n\to +\infty}\frac{n}{\sqrt[n]{n}} &=\lim_{n\to \infty}\sqrt[n]{\frac{n^n}{n!}}\\ &=\lim_{n\to \infty}\frac{n}{n+1}\sqrt[n]{\prod_{k=1}^n(\frac{k+1}{k})^k}\\ &=\lim_{n\to \infty} \frac{n}{n+1}\lim_{n\to \infty}(\frac{n+1}{n})^n\\ &=e \end{align*} We only need to know that, if $$a_n>0$$ and $$\lim_{n\to \infty} a_n=a$$ exists, then $$\lim_{n\to \infty} \sqrt[n]{a_1a_2\cdots a_n}=\lim_{n\to \infty}a_n=a$$ In fact, for any $$\epsilon>0$$, there exists $$N$$ such that if $$n>N$$, then $$|a_n-a|<\epsilon$$(Assume that $$a>0$$). Then we can observe that

$$\sqrt[n]{a_1\cdots a_N}(a-\epsilon)^{n-N}\le \sqrt[n]{a_1a_2\cdots a_n}\le \sqrt[n]{a_1\cdots a_N}(a+\epsilon)^{n-N}$$ Let $$n\to \infty$$, we can get that $$a-\epsilon\le \lim_{n\to \infty}\sqrt[n]{a_1a_2\cdots a_n}\le a+\epsilon$$ Thus, $$\lim_{n\to \infty} \sqrt[n]{a_1a_2\cdots a_n}=a$$