Convergence Test: Does \sum(-1)^n-1(lnp(n)/n) {p>0} converge or diverge?

[Dell]

i need to tell if the following sequence converges or diverges

\sum(-1)^n-1ln^p(n)/n {p>0}
(n from 1 to infinity)

what i think i need to do is take the positive version of this series

\sumln^p(n)/n {p>0}
(n from 1 to infinity)
if the positive series converges than the original must also, if not i can use leibnitz to tell me its behaviour

the problem it the "p", surely {p>0} is not enough information, do i not need to know if p>1 or p<1?

i can integrate \int(ln^p(x)/x)dx ===> t=lnx dt=dx/x

\intt^pdt
=t^p+1/(p+1) =...

...ln^p+1(n)|^{infinity}_{1}

if P>-1 \sumln^p(n)/n diverges
if P<-1 \sumln^p(n)/n converges
if p=-1 \sumln^p(n)/n diverges

but i am told that {p>0} therefor i am only left with one option
if P>-1 \sumln^p(n)/n diverges

so now i need to check if:
- lim An = 0
- An+1< An

 

[Dell]

lim An= ln^p(n)/n = 0

so now how do i prove that An > A(n+1)