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?

The exact formula should read as $\;\ln(n!) \color{red}{\sim_\infty} n\ln(n) - n + \frac{1}{2} \ln(2\pi n)$ or $\;\ln(n!) \color{red}{\sim_\infty} n\ln(n) - n $, in the sense of equivalence of functions, not of numerical approximation. 
It simply means the ratio of the functions tends to $1$ as $n$ tends to $\infty$, not that the functions tend to have the same value. Rather, it means the functions tend to have the same relative value. For a classic example, consider the equivalent functions $n$ and $n+\sqrt n $.
