Divergence of $n!^{1/n}$

Question

I'm trying to find a simple proof that $n!^{1/n}$ is a divergent sequence. I have proved it using a lower bound you can get from an integral (or Stirling's approximation) that $n!^{1/n}>\ln(n)$, but I get the impression from the other questions I've done that it shouldn't need all that machinery.

Hints would be preferred to full solutions, please

@Jonas Thanks for pointing this out! – preferred_anon Sep 23 '14 at 13:04 — preferred_anon, Sep 23 '14 at 13:04

score 6 · Answer 1 · answered Sep 23 '14 at 13:01

6

Hint; At least half the terms in the product for $n!$ will be at least $n/2$, so$$n! \geq (n/2)^{n/2}$$ which shows the result.

answered Sep 23 '14 at 13:01

Matt Rigby

2,316
12
13

score 2 · Answer 2 · answered Sep 23 '14 at 13:00

2

Since $$(n+1)\log(n+1)-n\log n = n\log(1+1/n)+\log(n+1) \leq 1+\log(n+1)$$ we have: $$\log(n!) \geq 1-n+n\log n$$ hence: $$\log\left(n!^{1/n}\right) \geq \log n-1$$ and $a_n=n!^{1/n}$ satisfies $a_n\geq \frac{n}{e}.$

answered Sep 23 '14 at 13:00

Jack D'Aurizio

353,855

score 0 · Answer 3 · answered Sep 23 '14 at 13:02

0

Hint: can you write a polynomial expression for $n!$? What is its lead term?

answered Sep 23 '14 at 13:02

Jason Knapp

1,669

You can't write a polynomial expression for $n!$ (unless you use an infinite polynomial). – Akiva Weinberger Sep 23 '14 at 13:08
1

I'll admit it is early and I am tired, but for specific $n$ a positive integer I'm fairly sure that $n! = \prod_{k=0}^{n-1} (n-k)$ is a polynomial in $n$. – Jason Knapp Sep 23 '14 at 13:11
Oh, I had assumed you meant one polynomial that works for all $n$. – Akiva Weinberger Sep 23 '14 at 13:15
Ah, right. Yes, that would be bad! – Jason Knapp Sep 23 '14 at 13:15
No, that would be awesome=) – flawr Sep 23 '14 at 13:38

Divergence of $n!^{1/n}$

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