Prove that $a^n

Question

Prove that $a^n <n! < n^n$ when $a>1$ and n always big enough.

How can I solve that? I need to prove that $$\lim\limits_{n\to\infty} \frac{a^n}{n!} = 0$$ and $$\lim\limits_{n\to\infty} \frac{n!}{n^n} = 0.$$

Is it possible to solve by another way? Some like this :

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What else do we know about $a$? If $a \geq n$, this is clearly not true — MacCab, Jul 08 '15 at 18:43
This isn't true, take $a = n$, then your inequality says $n^n < n^n$. What are the restrictions on $a$? — Zain Patel, Jul 08 '15 at 18:44
Do you mean to say for $a>1$ there exists $N\in\mathbb N$ with $a^n<n!<n^n$ for all $n\geq N$? — Hirshy, Jul 08 '15 at 18:46
Yes! Thank you about the sugestions and for the visual treatment of my question. The n is always big enough. Sorry my bad english too. I speak portuguese. — Leandro, Jul 08 '15 at 18:53
To me it is still not clear what you want; do you want to prove $a^n<n!<n^n$ or do you just want to calculate the limits? — Hirshy, Jul 08 '15 at 19:11
possible duplicate of Which has a higher order of growth, n! or n^n? — , Jul 08 '15 at 19:17
Also of Do factorials really grow faster than exponential functions? for the other part. — , Jul 08 '15 at 19:18
Sorry, I can't express myself very well in english. I will post a photo. By the time I just need to calculate the limits. — Leandro, Jul 08 '15 at 19:23

Hirshy · Answer 1 · 2015-07-08T18:55:13.530

4

If you just want to calculate the limits, you could look at the series $$\sum\limits_{n=0}^\infty \frac{a^n}{n!}~\text{and}~\sum\limits_{n=1}^\infty \frac{n!}{n^n}$$ and check for convergence of these series.

edited Jul 08 '15 at 18:55

answered Jul 08 '15 at 18:48

Hirshy

5,040

2

Convergence of those sums is stronger than the original problem no? – anon Jul 08 '15 at 19:32
The convergence of these sums will not tell you wether $a^n<n!<n^n$, but as OP said in a comment, that he only wants to calculate $$\lim\limits_{n\to\infty} \frac{a^n}{n!}=0=\lim\limits_{n\to\infty}\frac{n!}{n^n},$$ this is a fast and easy way to get there. – Hirshy Jul 09 '15 at 08:13

score 2 · Answer 2 · answered Jul 08 '15 at 18:54

2

Also, regarding the first Statement, I don't think this is actually true, consider a=3 and n=4, then $a^n$ = 81 > n! = 24.

answered Jul 08 '15 at 18:54

Fluffy12

152

Sure! Sorry! I already edited it. Thanks. – Leandro Jul 08 '15 at 18:57

Mark Viola · Answer 3 · 2016-03-23T19:40:45.910

Let's look at the ratio

$$\frac{a^{n+1}/(n+1)!}{a^n/n!}=\frac{a}{n+1}<\frac12$$

for $n>2a-1$. Thus, for $n>2a-1$, the terms of the sequence decay faster than the geometric progression $\frac{1}{2^n}$. By comparison, the sequence tends to zero.

Now let's look at $\lim_{n\to \infty}\frac{n!}{n^n}$. Recall Stirling's formula

$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)$$

Then, we have

$$\begin{align} \lim_{n\to \infty}\frac{n!}{n^n}&=\lim_{n\to \infty}\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right)}{n^n} \\\\ &=\lim_{n\to \infty}\sqrt{2\pi n}e^{-n}\left(1+O\left(\frac{1}{n}\right)\right)\\\ &=0 \end{align}$$

score 2 · Answer 4 · answered Jul 08 '15 at 19:18

$$0<\left| \frac { { a }^{ n } }{ n! } \right| =\frac { \left| a \right| }{ 1 } \frac { \left| a \right| }{ 2 } \frac { \left| a \right| }{ 3 } ...\frac { \left| a \right| }{ m } \quad \frac { \left| a \right| }{ (m+1) } ...\frac { \left| a \right| }{ n } \le \frac { \left| { a }^{ m } \right| }{ m! } { \left( \frac { \left| a \right| }{ m+1 } \right) }^{ n-m }<\varepsilon \\ $$ for every $\forall \varepsilon >0$ and $m+1>\quad \left| a \right| $

score 2 · Answer 5 · edited Apr 13 '17 at 12:20

The first limit is evaluable using the Stolz-Cesàro theorem. It is equal to $$\lim_{n\to\infty}\frac{a^{n+1}-a^n}{(n+1)!-n!}=\lim_{n\to\infty}\frac{a^{n}(a-1)}{n\cdot n!}=0,$$ where the rightmost equality follows from the upperboundedness of $\displaystyle\frac{a^n}{n!}$, which in turn is a consequence of its obvious decreasingness for large enough $n$.

As for the second one, we show by induction that for $n>1$ $$n!<n^n.\tag{1}$$ We have $$2=2!<2^2=4,$$and $$\frac{(n+1)!}{n!}<\frac{(n+1)^{n+1}}{n^n} \\ n+1<(n+1)\left(1+\frac{1}{n}\right)^n \\ \left(1+\frac{1}{n}\right)^n>1,$$ therefore $(1)$ does hold for $n>1$. If we suppose $$\lim_{n\to\infty}\frac{n!}{n^n}\ne0,$$ then from $$\min\left({\frac{n!}{n^n}}\right)\le\frac{n!}{n^n}<1 $$ follows $$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=1,$$ and again, by Stolz-Cesàro, $$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=1.$$ But combining $$\sqrt{2}-1<1 $$ and the decreasingness of $$\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}, $$ we get a contradiction.

Prove that $a^n

5 Answers5