Neglect $1/2 \ln(2\pi n)$ in Stirlings approximation formula, but this term is not bounded or gets smaller, but larger One form of Stirlings approximation reads
$$
 \ln(n!) \approx n\ln(n) - n + \frac{1}{2} \ln(2\pi n)
$$
another
$$
 \ln(n!) \approx n\ln n - n.
$$
But thats makes me wonder, for the difference of both is $\frac{1}{2}\ln(2\pi n)$, which gets arbitrary large (surely very slow, but it is not bounded...), so the error between both approximations gets larger and larger for $n \to \infty$, but it is not the point of an approximation formula to give a lower error for $n \to \infty$?
So, in what sense is the second approximation valid, when the difference between both terms nonetheless becomes larger and larger for $n\to \infty$? Could anybody please explain this to me?

This is called an asymptotic expansion.
Since you have $$\frac{\ln(n!)}{n\ln(n)}\xrightarrow[n\to\infty]{} 1,$$
$n\ln(n)$ is a valid approximation for $\ln(n!)$.
You actually have that the error goes to $0$ at the speed $\frac 1{\ln(n)}$ which is very slow.
So if you want a better approximation, you can notice that 
$$\frac{\ln(n!)}{n\ln(n)-n}\xrightarrow[n\to\infty]{} 1,$$ 
and the error goes to $0$ at the speed $\frac {1/2 \ln(2\pi n)}{n\ln(n)-n}$ which is better already.
And so on... you will always have better approximations as the error goes to $0$ faster and faster.
