Limit of factorial function: $\lim\limits_{n\to\infty}\frac{n^n}{n!}.$

Question

I am studying for a test and I am given this problem:

$$\lim_{n\to\infty}\frac{n^n}{n!}.$$

How do I go about solving this limit? Intuitively I see how the numerator is growing much faster, but how do I express this precisely? Thanks!

Note that $\frac{n\cdot n\cdot n\cdots\cdot n}{1\cdot 2\cdot 3\cdots \cdot n}\ge n\cdot 1\cdot 1\cdots \cdot 1$. Of course the ratio grows much faster than $n$, but what we have is enough to determine the limit. — André Nicolas, Jul 24 '14 at 21:25
As a sidenote: another intuitive way to see that this goes to zero is that this is the same as the probability of picking $n$ numbers out of ${1,\ldots,n}$ and having them all be different. (Not much help on a calculus test, but persuasive nonetheless.) — Semiclassical, Jul 24 '14 at 21:38

score 7 · Answer 1 · answered Jul 24 '14 at 21:36

7

Hint: Note that $$ \left.\frac{(n+1)^{n+1}}{(n+1)!}\middle/\frac{n^n}{n!}\right.=\left(1+\frac1n\right)^n $$ Thus, the ratio of the terms tends to $e$.

answered Jul 24 '14 at 21:36

robjohn

345,667

I am having trouble with the ration test here. I cannot seem to reduce my solution to what you have. Would you mind helping me? Particularly, I am stuck with the (n+1)^(n+1) in the numerator, and n^n in the denominator. – ianks Jul 24 '14 at 21:47
@ianks: It might help to note that $$\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^n}{n!}$$ Then divide that by $$\frac{n^n}{n!}$$ – robjohn Jul 24 '14 at 22:50

Hakim · Answer 2 · 2014-09-08T12:21:24.557

Hint:
Use Stirling's approximation: $$ n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n,$$ then after simplification you'll get: $$\lim_{n\to\infty}\dfrac{e^n}{\sqrt{2\pi n}}.$$

For the intuitive part (we're just working with integers to simplify things) we have: $$\dfrac{n^n}{n!}=\dfrac{\color{#C00}n\cdot \color{green}n\cdot \color{royalblue}n\cdots\color{brown}n}{\color{#C00}1\cdot\color{green}2\cdot\color{royalblue}3\cdots \color{brown}n}\xrightarrow[n\to\infty]{}\infty.$$

score 2 · Answer 3 · answered Jul 24 '14 at 23:15

$\ln\left(\sqrt[n]{\dfrac{n!}{n^n}}\right)$

$=\dfrac{1}{n}\ln\left(\prod\limits_{k=1}^{n}\dfrac{k}{n}\right)$

$=\dfrac{1}{n}\sum\limits_{k=1}^{n}\ln\left({0+k\dfrac{1-0}{n}}\right)$

By Riemann sum:

$\lim\limits_{n\to +\infty}\ln\left(\sqrt[n]{\dfrac{n!}{n^n}}\right) =\int\limits_0^1\ln x\,dx=\left[x\ln x-x\right]_0^1=-1-\lim\limits_{x\to 0^+}(x\ln x-x)=-1$

And so:

$\lim\limits_{n\to +\infty}\sqrt[n]{\dfrac{n!}{n^n}}=e^{-1}$

While $0<e^{-1}<1$ we have then:

$\lim\limits_{n\to +\infty}\dfrac{n!}{n^n}=0^+$

And so:

$\lim\limits_{n\to +\infty}\dfrac{n^n}{n!}=+\infty$

score 1 · Answer 4 · answered Jul 24 '14 at 21:50

"Intuitively I see how the numerator is growing much faster, but how do I express this precisely?": use little-o notation.

Using AM-GM we get (after a slight technicality with $n$ even or odd):

$n!=(1n)(2(n-1))\ldots\leq((\frac{n+1}{2})^{2})^{\frac{n}{2}}=\frac{(n+1)^{n}}{2^{n}}$ so $n!=o((n+1)^{n})$.

Since $\lim\limits_{n\rightarrow\infty}\frac{(n+1)^{n}}{n^{n}}=e$ this show that $n^{n}=\Theta((n+1)^{n})$.

Hence $n!=o(n^{n})$.

Limit of factorial function: $\lim\limits_{n\to\infty}\frac{n^n}{n!}.$

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