For what real number $r$ does the series $\sum_{n=1}^{\infty}\frac{\ln(n)}{n^r}$ converge? TL;DR Is there any way to solve this by comparison?
Context: This question is posed in my text book after the integral test, the comparison test, and the limit-comparison test, but before any of the other tests.
I'm pretty sure the answer is $r>1$, and I can arrive at that with the integral test. But that requires some painful integration, and I can't help but think that I should be able to compare the sum with $\sum\frac{1}{n^r}$, which I know converges when $r>1$.
Since $\ln(n)<n$, I tried comparing with $\sum\frac{n}{n^r}=\sum\frac{1}{n^{r-1}}$ to get $r>2$. And while it certainly does converge when $r>2$, that's not entirely correct.
Can this be done by comparison?

By Comparison, our series diverges if $r\le 1$. We show it converges if $r\gt 1$.
Let $r=1+\delta$. Then $n^r=n^{\delta/2}n^{1+\delta/2}$.
Note that $\displaystyle\lim_{n\to\infty}\dfrac{\ln n}{n^{\delta/2}}=0$. So by Limit Comparison with $\displaystyle\sum_{1}^\infty \dfrac{1}{n^{1+\delta/2}}$, our series converges. 
Remark: We sliced off a small piece $\delta/2$ of $r$. The term $n^{\delta/2}$ in the long run grows fast enough to crush $\ln n$. But there enough of $r$ left to force convergence. 
