For what $z$ is the series $\sum\;z^{\ln n}\; $absolutely convergent?
$$\sum z^{\ln n} = \sum e^{(\ln n)(\ln z)} = \sum n^{\ln z}$$
According to $p$ series test, the series converges when $p>1$, so $$\ln z < -1\\z<e^{-1}\\z<1/e$$
Is this the correct way to solve it and how to show $\vert z\vert<1/e$?
If $z$ is suppose to be a complex number then $\ln z <-1$ does not make sense.
$z^{\ln n}$ is not single valued but this doesn't matter as long as absolute convegrence is concerned.
$|z^{\ln n}|=|z|^{\ln n}=n^{\ln |z|}$ and $\sum n^{r}$ is convergent iff $r <-1$. Hence the given series is absolutely convergent iff $\ln |z| <-1$ or $|z|<\frac 1 e$.