Hint to solve $\lim_{n\to\infty} n! \ n^{-\ln(n)}$ Can you give me some hint to find where it converges please? 
$$\lim_{n\to\infty} n! \ n^{-\ln(n)}$$
Thank you in advance.
 A: Hint: A look at the cases $n$ even, $n$ odd, will show that $n!>(n/2)^{n/2}.$ Thus the expression is greater than
$$\frac{ (n/2)^{n/2} }{ n^{\ln n} } = \frac{n^{n/2}}{2^{n/2}n^{\ln n}}.$$
Now apply $\ln$ to this.
A: Firstly it's unclear what you mean by "where it converges" since this sequence, $x_{n} = n! n^{-\ln(n)}$ has no parameters; it just depends upon $n$ so that in the limit $n \rightarrow \infty$ it either converges or not.  
I would consider taking logs, defining a new sequence $y_{n} = \ln(x_{n})$. Since log is a monotonic function, the convergence properties of $x_{n}$ are transferred to $y_{n}$. Using the rules of logs you will find yourself having to deal with $\ln(n!)$. For large $n$ you can use Stiring's approximation.
Your task will turn into that of showing that
$$n\ln(n) - n - (\ln(n))^{2}$$
does not converge. For this, recall that $\ln(n) / n$ converges to zero, so that you can make $\ln(n)$ arbitrarily smaller than $n$ by taking $n$ large enough. 
Then if $y_{n}$ diverges, then $x_{n} = e^{y_{n}}$ diverges too. 
