Prove that $\lim\limits_{n \to\infty}\sqrt[n]{n!}=\infty$

Question

My attempt:$$\lim\limits_{n \to\infty}\sqrt[n]{n!}=\lim\limits_{n\to\infty}\sqrt[n]{1\cdot 2 \cdot 3 \cdot \ldots \cdot n}=\lim\limits_{n\to\infty}\sqrt[n]{1}\cdot \sqrt[n]{2}\cdot \sqrt[n]{3}\cdot \ldots \cdot \sqrt[n]{n}=1\cdot 1\cdot 1 \cdot \ldots \cdot 1=1$$ Why is this not correct?

Answer 1

$n! = \prod_{k=1}^nk>\prod_{k=n/2}^n k>(n/2)^{n/2}$ so $(n!)^{1/n}>(n/2)^{1/2} \to \infty$.

Answer 2

As the comments have pointed out, you are taking the limit of parts of the problem while ignoring the bigger picture, namely that you have to take the limit of the amount of parts as well.

Intuitively, we are saying the geometric mean of all positive integers is $\infty$. A simple proof may be done with ratio $\Rightarrow$ root:

$$\frac{(n+1)!}{n!}=n+1\to\infty\Rightarrow\sqrt[n]{n!}\to\infty$$

Answer 3

It is incorrect to multiply the limits of each factor because the number of factors is not fixed, aech factor, which indeed tends to $1$ , is not equal to $1$, but slighly greater.

Hint: to prove the limit is $\infty$, you can try to determine the limit of the log, observing that $$\log(\sqrt[n]{n!})=\frac1n\bigl(\log 2+\dots+\log n\bigr),$$ which is an upper Riemann sum for the integral $\;\displaystyle\int_1^n \log x\,\mathrm dx$

Answer 4

By using the result that I proved here (https://math.stackexchange.com/a/3159844/629594), we have that $\lim\limits_{n\to \infty}\sqrt[n]{n!}=\lim\limits_{n\to \infty}\frac{(n+1)!}{n!}=\lim\limits_{n\to \infty}(n+1)=\infty$