# 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.

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Hint: Stirling. –  Did Mar 22 '11 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 '11 at 12:00
Try taking the natural log and finding the limit of that. –  Becca Winarski Mar 22 '11 at 12:02
Second try: Stirling formula. –  Did Mar 22 '11 at 12:06
This is (after a slight modification) Problem 1.2.2 from Radulescu, Radulescu, Andreescu: Problems from Real Analysis, p.8. –  Martin Sleziak Oct 19 '11 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.

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+1: Very nice. Didn't know that trick (obviously). –  t.b. Mar 22 '11 at 13:33
I tried to use that but I wasnt smart enough to change the original series. thanks jonas. –  user6163 Mar 22 '11 at 13:35
@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.) –  Jonas Meyer Mar 22 '11 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 '11 at 13:58
Just a short comment, the mentioned Theorem is just the Stolz-Cezaro theorem applied to $\frac{\ln (a_n)}{n}$. –  N. S. Apr 10 '11 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.

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

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

Q.E.D.

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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. –  Martin Sleziak Jul 16 '14 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 $\lim_{n \to \infty} f(n) = e$.

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+1. Any fine answer should end with a statement which shows the answer to the OP question as you did it !!!. –  Felix Marin May 25 '14 at 6:49

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 $e^{o(1)} = 1$.

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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 '14 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?). –  Daniel Fischer Jun 9 '14 at 18:14
please see the edit. –  Alex Jun 9 '14 at 18:34

If you take the log, it is:

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

Which is a Reimann 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."

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