calculate the limit $\lim_{n \to \infty} n!^{\frac1n}$ 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 ?
 A: Are you sure about $$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n} \to 0 ?$$
A: 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 $n\to\infty$.
A: 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
A: 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 
 $$\frac{\ln 2}{n} + \frac{\ln 3}{n} + \cdots+\frac{\ln n}{n}\to 0$$
A: 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 $n\to\infty$.
