I assume the series in question is
[tex]\displaystyle\sum_{n=2}^\infty\frac{\ln n}{n^p}[/tex]
Compare the series to the larger integral,
[tex]\displaystyle\sum_{n=2}^\infty\frac{\ln n}{n^p}\le\int_2^\infty\frac{\ln x}{x^p}\,\mathrm dx[/tex]
Integrate by parts taking,
[tex]u=\ln x\implies\mathrm du=\dfrac{\mathrm dx}x[/tex]
[tex]\mathrm dv=\dfrac{\mathrm dx}{x^p}=x^{-p}\,\mathrm dx\implies v=\dfrac{x^{1-p}}{1-p}[/tex]
So the integral is equivalent to
[tex]\displaystyle\dfrac{x^{1-p}\ln x}{1-p}\bigg|_{x=2}^{x\to\infty}-\frac1{1-p}\int_2^\infty x^{-p}\,\mathrm dx[/tex]
[tex]=\displaystyle\frac1{(1-p)^2}\lim_{x\to\infty}((1-p)\lnx-1)x^{1-p}-\frac1{(1-p)^2}((1-p)\ln2-1)2^{1-p}[/tex]
The constant term explodes when [tex]p=1[/tex], so we require [tex]p\neq1[/tex]. The remaining limit term can be written as
[tex]\displaystyle\lim_{x\to\infty}((1-p)\ln x-1)x^{p-1}=\lim_{x\to\infty}x^{1-p}[/tex]
where the last equality is obtained by applying L'Hopital's rule once.
Now, the limit will only exist if [tex]1-p<0\implies p>1[/tex], so this is the condition required for the integral to converge, and hence for the series to converge.
