Consider the function:
#f(x) = (lnx)^p/x > 0# for #x in [2,oo)#
For #p < 0# we have:
#lim_(x->oo) (lnx)^p/x = lim_(x->oo) 1/(x(lnx)^absp) = 0#
While for #p > 0# the limit:
#lim_(x->oo) (lnx)^p/x#
is in the indeterminate form #oo/oo# so we can solve it using l'Hospital's rule:
#(1) lim_(x->oo) (lnx)^p/x = lim_(x->oo) (d/dx(lnx)^p)/(d/dx x) = lim_(x->oo) (p(lnx)^(p-1))/x#
For #p in NN# we can then prove by induction that:
#lim_(x->oo) (lnx)^p/x = 0#
since this is true for #p=0#, and #(1)# shows that if the limit is zero for #(p-1)# then it is zero also for #p#.
For any other #p > 0# we can see that if # m < p < n# with #m,n in NN^0#, we have either that:
# (lnx)^m/x < (lnx)^p/x < (lnx)^n/x#
or that:
# (lnx)^n/x < (lnx)^p/x < (lnx)^m/x#
In both cases by the squeeze theorem:
#lim_(x->oo) (lnx)^p/x = 0#
which then is true for any #p in RR#
Now consider the derivative:
#(df)/dx = (p(lnx)^(p-1)-(lnx)^p)/x^2 = (lnx)^(p-1)/x^2(p-lnx)#
we have that for #x > e^p#
#(df)/dx < 0#
so that the function is decreasing over the interval #(e^p,oo)#
All the hypotheses of the integral test are then satisfied and the convergence of the series:
#sum_(n=2)^oo (lnn)^p/n#
is equivalent to the convergence of the integral:
#int_2^oo (lnx)^p/x dx = int_2^oo (lnx)^p d(lnx) = [(lnx)^(p+1)/(p+1)]_2^oo#
Clearly the integral is convergent only if:
#lim_(x->oo) (lnx)^(p+1) < oo#
that is for #p <= -1#.
