Why is $$\lim_{n\rightarrow \infty}\sqrt[n]{n!} = \infty$$
if : $$\lim_{n\rightarrow \infty}\frac{n!}{n^n} = 0 \implies n^n >n! \quad\text{for "big" n}$$
I don't understand it.
Asked
Active
Viewed 136 times
0
-
3$\lim_\limits{n\to\infty} \sqrt [n] {n^n} =\infty$ – Doug M Jan 07 '21 at 20:38
-
@DougM I fixed it. – VLC Jan 07 '21 at 20:39
-
3$n^n > n!$ always. No fancy limit argument needed. And the $n$th root of $n^n$ is just $n$. – Ethan Bolker Jan 07 '21 at 20:40
-
4But $\sqrt[n]{n^n}\to\infty$ as well, so $n^n>n!$ proves nothing about the limit of $\sqrt[n]{n!}$. – Arthur Jan 07 '21 at 20:40
-
2If $f(n) \geq g(n) \geq 0$, then $f(n) \to 0$ implies that $g(n) \to 0$. However, it is not true that $f(n) \to \infty$ implies that $g(n) \to \infty.$ – Ben Grossmann Jan 07 '21 at 20:42
-
I thought that any exponent less than $n$ gives the limit of n-th root equal to $1$ – VLC Jan 07 '21 at 20:42
-
3It is true that $n! > (n/e)^n$. Might be easier to show that $ n! > (n/3)^n$. – Calvin Lin Jan 07 '21 at 20:43
1 Answers
2
Let $a_n = n!$.
The limit $\lim \sqrt[n]{a_n}$ is related to the root test for series convergence. In this context, it is well known that if $\lim \dfrac{a_{n+1}}{a_n}$ exists, then so does $\lim \sqrt[n]{a_n}$ and they are equal. Let's compute the ratios then: $$ \lim \frac{a_{n+1}}{a_n} = \lim (n+1) = \infty $$ Therefore, $\lim \sqrt[n]{a_n}= \infty$.
lhf
- 216,483