Find the set of $x>0$ such that the series $\sum\limits_n x^{\ln{n}}$ converges
take $x=e^{-y}$
$$\sum_{n=1}^{\infty}x^{\ln n}=\sum_{n=1}^{\infty}e^{-y\ln n}=\sum_{n=1}^{\infty}\frac{1}{n^y}$$
The last series converges iff $y>1$.
Hence $0<x<\frac{1}{e}$, for which the series converges.