I've reached my wit's end with this problem. We have not been taught about Ratio Tests or anything involving Radius of Convergence, but we have a problem asking for it.
For what values of $p$ does the sum converge?
$$\sum_{n=1}^\infty{\ln(n) \over{n^p}}$$
I notice that it is "similar" to
$$\sum_{n=1}^\infty{\frac {1}{n^p}}$$
And we know from the Integral Test of Convergence that this series converges when $\lvert p\rvert > 1$. Does this have any connection to the answer? Because I also notice that (via WolframAlpha) that when $p=1$ the series diverges, but when $p=1.1$ it converges.
Best Answer
Note that $\ln n<n$ so
$$\sum_{n=1}^\infty \frac{\ln n}{n^p}<\sum_{n=1}^\infty\frac{n}{n^p}=\sum_{n=1}^\infty \frac{1}{n^{p-1}}$$
which is the familiar $p$-series. Thus the series converges by the comparison test if $p-1>1$, or if $p>2$.
This is not a full proof since it remains to show whether the series converges for $1< p\le 2$. Note that divergence for $p\le1$ is given by the same comparison test above.