Series

case p > 0:
f(n)= ln(n)/n^(p+1), f'(n) < 0, then f(n) is decreasing.
by the integral test
Σ(k=2,inf) ln(n)/n^(p+1)
~ integrate(x=2,inf) ln(x)/x^(p+1) dx
= -(p lnx +1)/(p^2 x^p) for x=2, inf
= (p ln2 +1)/(p^2 * 2^p) conv.
case p < =0:
Σ(k=2,inf) ln(n)/n^(p+1)
> Σ(k=2,inf) ln(n)/n
~ integrate ln(x)/x dx = inf. div.

Ans. p=0