Why the limit of the following sequence is like this:
$$\frac{2 \sqrt[n]{n!}}{n}=\frac{2}{e}$$
Asked
Active
Viewed 82 times
3 Answers
4
We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see this question).
We have that $a_n=n!/n^n$ and $$ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(n+1)!n^n}{(n+1)^{n+1}n!}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\frac1{\lim_{n\to\infty}(1+1/n)^n}=\frac1e. $$
1
Hint: We have $$\frac{\sqrt[n]{n!}}{n}=(\frac{n!}{n^n})^{\frac{1}{n}}=(\frac{1}{n}\cdot\frac{2}{n}\cdots\frac{n}{n})^{\frac{1}{n}}$$ which implies that $$\log\frac{\sqrt[n]{n!}}{n}=\frac{1}{n}\sum_{i=1}^n\log\frac{i}{n}.$$ By Riemann sum, we have $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\log\frac{i}{n}=\int_0^1\log x dx.$$ Using integration by parts, we can evaluate the last integral, which I just leave it to you.
Paul
- 19,140
0
Are you allowed to use Stirling's approximation?
For large values of $n$, $n!$ goes like $\sqrt{2\pi n} \left( \frac{n}{e} \right)^n$, so
$$\lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2\pi n} \left( \frac{n}{e} \right)^n}}{n} = \ldots$$
StackTD
- 27,903
- 34
- 63
-
no i even do not know the stirling approximation – kurkowski Feb 19 '15 at 10:35
-
Too bad, because it's quite easy (not to say trivial) if you could use it :-). – StackTD Feb 19 '15 at 10:36
-
see that, thank you :) – kurkowski Feb 19 '15 at 10:39