sum_(n=2)^oo 1/(n(lnn)^p) Find the positive values of p for which the series converges.

To find the convergence of the series sum_(n=2)^oo 1/(n(ln(n))^p) where pgt0 (positive values of p ), we may apply integral test.
Integral test is applicable if f is positive, continuous, and decreasing function on an interval and let a_n=f(x) . Then the infinite series sum_(n=1)^oo a_n converges if and only if the improper integral int_1^oo f(x) dx converges to a real number. If the integral diverges then the series also diverges.
For the infinte series series sum_(n=2)^oo 1/(n(ln(n))^p) , we have:
a_n =1/(n(ln(n))^p)
Then, f(x) =1/(x(ln(x))^p).
The f(x) satisfies the conditions for integral test based on the following reasons:
- f(x) is continuous since x(ln(x))^p !=0 for any x-value on the interval [2,oo)
- f(x) is positive since 1/(x(ln(x))^p)gt0 for any x-value on the interval [2,oo).
- f(x) is decreasing since f'(x) is negative  for large value of x . It eventually decreases at the tail of the series.
To evaluate the convergence of the series using integral test, we set-up the improper integral as: 
int_2^oo 1/(x(ln(x))^p)dx
Apply u-substitution by letting: u=ln(x) u=ln(x) then du = 1/xdx , a=ln(2) and b=oo .
int_(ln(2))^oo 1/(x(ln(x))^p) dx=int_(ln(2))^oo 1/(ln(x))^p *1/xdx
                           = int_(ln(2))^oo 1/u^pdu
                           =int_(ln(2))^oo u^-p dx
                          = u^(-p+1)/(-p+1)|_(ln(2))^oo  
                          =u^(-(p-1))/(-p+1)|_(ln(2))^oo
                           =1/(u^(p-1)(-p+1))|_(ln(2))^oo
 Apply definite integral formula:  F(x)|a^b = F(b)-F(a) .  
1/(u^(p-1)(-p+1))|_(ln(2))^oo=1/(oo^(p-1)(-p+1))-1/((ln(2))^(p-1)(-p+1))
                               = 1/oo-1/(-p+1)1/(ln(2))^(p-1)
                               =0-1/(-p+1)1/(ln(2))^(p-1)
                               =1/(-p+1)1/(ln(2))^(p-1)
The integral converges to a real number when pgt1 .
Thus, the series sum_(n=2)^oo 1/(n(ln(n))^p) converges whenever pgt1 .