Convergence of $\frac{n^{\ln n}}{(\ln n)^{n}}$ Let $$U_n = \frac{n^{\ln n}}{(\ln n)^{n}}$$
Does the series  $\sum Un$ converge ?
I've tried to do the convergence test, by seeing if $\frac{U_{n+1}}{U_n}$ converges to any real number $k$ with $k < 1$
Therefore I've transformed $\frac{U_{n+1}}{U_n}$ into
$$\frac{ \exp (\ln^2(n+1) + n\ln(\ln n)) }{ \exp(\ln^2(n) + (n+1)\ln(\ln (n+1))) }$$
this seems pretty messy and I don't know whether I'm doing it right.
Any help?
 A: HINT
We have that
$$\frac{n^{\ln n}}{(\ln n)^{n}}=e^{\ln^2n-n\ln(\ln n)}$$
with
$$\frac{\ln^2n-n\ln(\ln n)}n\to -\infty$$
A: If suffices to show that
$${n^{\ln n}\over(\ln n)^n}\lt{1\over n^2}$$
for all sufficiently large $n$, i.e., that $n^{\ln n+2}\lt(\ln n)^n$ for all large $n$. But this follows from the observation that
$$(\ln n+2)\ln n\lt2(\ln n)^2\lt n\lt n\ln\ln n$$
for large $n$; the first inequality in the display starts at $n=8$, the second at $n=14$, and the third at $n=16$.
A: Note that we can write
$$
\frac{n^{ln(n)}}{(ln(n))^{n}}=\frac{(e^{ln(n)})^{ln(n)}}{(ln(n))^{n}}=\frac{e^{ln^{2}(n)}}{(ln(n))^{n}}
$$
We apply the root test
$$
\lim_{n \to \infty}\sqrt[n]\frac{n^{ln(n)}}{(ln(n))^{n}}=\lim_{n \to \infty}\frac{e^{\frac{ln^{2}(n)}{n}}}{ln(n)}
$$
For a sufficiently large n is satisfied
$$
0 \leqslant \frac{e^{\frac{ln^{2}(n)}{n}}}{ln(n)} \leqslant \frac{1}{ln(n)}
$$
Finally, by the compression theorem
$$
\lim_{n \to \infty}0 \leqslant \lim_{n \to \infty} \frac{e^{\frac{ln^{2}(n)}{n}}}{ln(n)} \leqslant \lim_{n \to \infty} \frac{1}{ln(n)}
$$
$$
0 \leqslant \lim_{n \to \infty} \frac{e^{\frac{ln^{2}(n)}{n}}}{ln(n)} \leqslant 0
$$
$$
\lim_{n \to \infty} \frac{e^{\frac{ln^{2}(n)}{n}}}{ln(n)} = 0
$$
Therefore the series converges
A: I added it here, because, may be last change not being noted.
$$n\ln\ln n - \ln^2 n-=n\ln\ln n\cdot\left(1-\frac{\ln^2n}{n\ln\ln n}\right)\sim n\ln\ln n$$
as $e^f=(e^{g})^\frac{f}{g}$, then if root test is correct for $e^g$, then it will be correct for $e^f$ also, so we can use equivalence for exponent also.
