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

To find the convergence of the series sum_(n=2)^oo (ln(n))/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=k)^oo a_n converges if and only if the improper integral int_k^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 (ln(n))/n^p , we have:
a_n =(ln(n))/n^p
Then, f(x) =(ln(x))/x^p
The f(x) satisfies the conditions for integral test when pgt0 . We set-up the improper integral as:
int_2^oo (ln(x))/x^pdx
Apply integration by parts: int u dv = uv - int v du.
Let: u=ln(x) then du = 1/xdx
       dv = 1/x^p dx
Then , v = int dv
              =int 1/x^p dx
              = int x^(-p) dx
             = x^(-p+1)/(-p+1)
The indefinite integral will be:
int (ln(x))/x^pdx = ln(x)x^(-p+1)/(-p+1)- intx^(-p+1)/(-p+1) *1/x dx
                    = ln(x)x^(-p+1)/(-p+1)-1/(-p+1) int (x^(-p)x)/x dx
                   = ln(x)x^(-p+1)/(-p+1)-1/(-p+1) intx^(-p) dx        
                  = ln(x)x^(-p+1)/(-p+1)-1/(-p+1) *x^(-p+1)/(-p+1)
                   =(ln(x)x^(-p+1))/(-p+1)-x^(-p+1)/(-p+1)^2
                  =(ln(x)x^(-p+1))/(-p+1)*(-p+1)/(-p+1)-x^(-p+1)/(-p+1)^2
                  =(ln(x)x^(-p+1)(-p+1))/(-p+1)^2-x^(-p+1)/(-p+1)^2
                 =(ln(x)x^(-p+1)(-p+1)-x^(-p+1))/(-p+1)^2|_2^oo
The definite integral will only be finite if 1-p<0 or pgt1 .
Thus, the series  sum_(n=2)^oo(ln(n))/n^p converges when pgt1 .