calculate the limit $\lim_{n \to \infty} n!^{\frac1n}$

Question

using Stirling's formula or the fact that $e^n \geq \frac{n^n}{n!}$.

one finds that the limit goes to $+\infty$

however, I found another result (probably false) using this method :

$$n!^{\frac1n} = e^{\frac1n\ln n!} = e^{\frac1n \sum_{k=1}^{n}\ln k} = \exp(\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n} ) $$

$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n} \to 0$

and $x \mapsto e^x$ is continuous. "Hence" $\lim_{n \to \infty} n!^{\frac1n} = 1$

I can't spot the mistake.

what did I do wrong ?

Answer 1

Are you sure about $$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n} \to 0 ?$$

Answer 2

We have that $$ \frac1n\sum_{k=2}^n\ln k\ge\frac1n\int_1^n\ln xdx=\frac1n\Bigl[x(\ln x-1)\Bigr]_1^n\to\infty $$ as $n\to\infty$.

Answer 3

You can't conclude that

$$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n} \to 0$$

since you are adding up infinitely many quantities (think to $\sum \frac1n$ which diverges).

Indeed by Stolz-Cesaro

$$\lim_{n\to \infty} \frac{\sum_{k=2}^{n} \log k}{n}=\lim_{n\to \infty} \frac{\sum_{k=2}^{n+1} \log k-\sum_{k=2}^{n} \log k}{n+1-n}=\lim_{n\to \infty}\log (n+1)=+\infty$$

For a derivation without Stirling take a look here

finding limit of a sequence square root of n factorial

Answer 4

Note that $$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n} =(1/n)[\ln 2+\ln 3+...+\ln n]$$

The integral $$\int _e^n (lnx) dx = n(ln( n) -1)$$

approximates $$\ln2+\ln 3+...+\ln n$$ which contradicts your assumption of $$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n}\to 0$$

Answer 5

Let $$a _n=n!$$ Note that $$ \frac{a_{n+1}}{a_n}=n+1\to\infty $$ as $n\to \infty$. By this MSE post which proves that, $$ \liminf(b_{n+1}/b_n) \leq \liminf(b_n^{1/n}) \leq \limsup(b_n^{1/n}) \leq \limsup(b_{n+1}/b_n) $$ for any sequence $(b_n)$ such that $b_n>0$, it follows that $$ (n!)^{1/n}\to \infty $$ as $n\to\infty$.