Need to find $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$

Question

I've attempted to solve this $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$ via converging 1/n to zero, but somehow it's incorrect. Can someone explain what i'm doing wrong?

Thank's for your reply, but i know that it can be solved without stirling approximation. I mean, i can't rely on it every time — Ramil Ahmedov, Oct 08 '20 at 04:27
Also tried and it worked, but still there is solution via squeeze theorem. — Ramil Ahmedov, Oct 08 '20 at 04:32
Then you have elaborate it in your question well. you must be getting the answer as 1(using squeeze thm), right? — Anindya Prithvi, Oct 08 '20 at 04:33
You' re right, next time i'll write it down in description. The correct answer is 0 — Ramil Ahmedov, Oct 08 '20 at 04:35

score 3 · Accepted Answer · answered Oct 08 '20 at 05:35

3

Hint: $$0<\frac 1{(n!)^\frac 1{n}}=\sqrt[n]{\frac{1}{n!}}<\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}\to 0.$$ In general, we have: if $a_n\to a$, then $$\lim_{n\to\infty}\frac{a_1+a_2+\cdots+a_n}{n}=a.$$

answered Oct 08 '20 at 05:35

Riemann

7,203

Hi, thank you for your reply, however this approach is not obvious. How can we prove the part with < inequality? – Ramil Ahmedov Oct 09 '20 at 04:03
the inequality used is AM-GM inequality! And you can refer https://math.stackexchange.com/questions/691807/proofs-of-am-gm-inequality – Riemann Oct 09 '20 at 09:04
AM-GM inequality is an elementary inequality even for high school students! You should master it (learn to use it). – Riemann Oct 09 '20 at 09:08

Qiaochu Yuan · Answer 2 · 2020-10-08T07:06:19.807

2

The limit is $0$, and the following very hands-on argument suffices: for any positive integer $k \le n$ we have

$$n! \ge k^{n-k}$$

(we look at $1 \cdot 2 \cdot \dots \cdot n$ and just ignore the terms less than or equal to $k$ and replace all the terms bigger than $k$ with $k$), which gives

$$\sqrt[n]{n!} \ge k^{1 - \frac{k}{n}}.$$

Taking, for example, $k = \lfloor \frac{n}{2} \rfloor$ gives

$$\sqrt[n]{n!} \ge \left\lfloor \frac{n}{2} \right\rfloor^{\frac{1}{2}}$$

so $\lim_{n \to \infty} \sqrt[n]{n!} = \infty$ and $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}} = 0$. Of course much tighter bounds are available, and even quite exact asymptotics via Stirling's formula, but it's worth knowing what can be proven in a very elementary way.

edited Oct 08 '20 at 07:06

answered Oct 08 '20 at 05:29

Qiaochu Yuan

419,620

I like this one, but how can we proove that n! is actually more than k^n-k – Ramil Ahmedov Oct 09 '20 at 04:08
@Amil: I gave the argument in parentheses: you look at the product $1 \cdot 2 \cdot \dots \cdot n$, ignore the terms less than or equal to $k$, and bound the terms greater than or equal to $k$ by $k$. It's very simple. – Qiaochu Yuan Oct 09 '20 at 06:10

DatBoi · Answer 3 · 2020-10-20T10:25:00.123

1

$$y=\lim_{n\to\infty}\frac 1{(n!)^\frac 1{n}}$$

$$\ln y=\lim_{n\to\infty}-\frac{1}{n}\ln(n!)$$

$$\ln y=\lim_{n\to\infty}-\frac{1}{n}\left({\ln1+\ln2+\cdots\ln n}\right)$$

$$\ln y=\lim_{n\to\infty}-\frac{1}{n}\left({[\ln\frac{1}{n}+\ln\frac{2}{n}+\cdots\ln\frac{n}{n}]+n\ln n}\right)$$

$$\ln y=\lim_{n\to\infty}\left({\int_0^1\ln x dx-\ln n}\right)$$

$$\ln y=\lim_{n\to\infty}\left({-1-\ln n}\right)\to-\infty$$

$$y=e^{-\infty}\to0$$

edited Oct 20 '20 at 10:25

answered Oct 08 '20 at 06:13

DatBoi

4,055

How did you convert the rienman sum into integral without a $ \frac{1}{n}..?$ – tryst with freedom Oct 20 '20 at 10:21
See here – tryst with freedom Oct 20 '20 at 10:22
@Buraian Yep, an error there – DatBoi Oct 20 '20 at 10:23

score -1 · Answer 4 · edited Oct 08 '20 at 06:18

Use the common limit $$\lim_{n \to \infty} \left( \frac{n^n}{n!}\right)^\frac 1n = \lim_{n \to \infty} n \times \frac{1}{(n!)^\frac1n} = e$$ It's easy to see that $$\lim_{n \to \infty} \frac{1}{(n!)^\frac1n} = \lim_{n \to \infty} \frac{e}{n} = 0$$

And that's the final answer.

It's easier to do this via Stirling's approximation, as $n! = \sqrt{2\pi n}\left( \frac ne\right)^n$ and the nth root of $1/n!$ is equal to $e/n$, which tends to 0.

Need to find $\lim_{n\to\infty}$ $\frac 1{(n!)^\frac 1{n}}$

4 Answers4