Question: I am not to find an upper bound to this succession. A lower bound seem to be $\log(n)$. Any ideas?

Answer: 
Hint. By the [Stolz-Cesàro Theorem](https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem),
$$\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \frac{\sum\_{k=1}^n\log{k}}{n}=\lim\limits\_{n\to\infty} \frac{\log (n+1)}{(n+1)-n}.$$
We have:

$$n! = n\cdot (n-1) \cdots 2 \cdot 1 > \underbrace{\frac{n}2 \cdot \frac{n}2 \cdot \cdots \cdot \frac{n}2}\_{\frac{n}2 \text{ times}} \underbrace{\cdot 1 \cdot \cdots \cdot 1}\_{\frac{n}2 \text{ times}} = \left(\frac{n}2\right)^{\frac{n}2}$$

$$n! = n\cdot (n-1) \cdots 2 \cdot 1 < n \cdot n \cdot \cdots \cdot n = n^n$$

So:

$$\frac12 \cdot \log \frac{n}2= \frac{n}2 \cdot \frac{\log \frac{n}2}{n}= \frac{\log \left(\frac{n}2\right)^{\frac{n}2}}{n} <\frac{\log n!}{n} < \frac{\log n^n}{n} = \frac{n\log n}{n} = \log n$$

This means the sequence diverges to $+\infty$.