sum_(n=3)^oo 1/(nlnn[ln(lnn)]^p) Find the positive values of p for which the series converges.

To find the positive values of p in which the series sum_(n=3)^oo 1/(nln(n)(ln(ln(n)))^p) , we may apply the integral test.
Integral test is applicable if f is positive, continuous, and decreasing function and a_n=f(x) . The infinite series sum_(n=k)^oo a_n converges if and only of the improper integral int _k^oo f(x)dx converges to a finite number. If the integral diverges then the series also diverges.
For the infinite series sum_(n=3)^oo 1/(nln(n)(ln(ln(x)))^p) , we have:
a_n=1/(nln(n)(ln(ln(n)))^p)
Then, f(x) =1/(xln(x)(ln(ln(x)))^p)
The f(x) satisfies the conditions for integral test for the interval [3,oo)
We set-up the improper integral as:
int_3^oo1/(xln(x)(ln(ln(x)))^p) dx.
Apply u-substitution by letting u = ln(x) then du =1/xdx .
int 1/(xln(x)(ln(ln(x)))^p) dx=int 1/(ln(x)(ln(ln(x)))^p)* 1/xdx
                      =int 1/(u(ln(u))^p) du
Apply another set of substitution: let v =ln(u) and dv = 1/u du .
int 1/(u(ln(u))^p) du=int 1/(ln(u))^p* 1/u du
                       =int 1/v^p* dv
                        =int v^(-p) dv
                        = v^(-p+1)/(-p+1)
Recall u =ln(x) and v = ln(u)  then v =ln(ln(x)) .
v^(-p+1)/(-p+1)=(ln(ln(x)))^(-p+1)/(-p+1)|3^oo
The integral is finite when -p+1lt0 or pgt1 .
Note: When (ln(ln(x))) has positive power on the numerator side then the integral diverges. 
 When (ln(ln(x))) has negative power on the numerator side then the integral converges. 
Thus, the series sum_(n=3)^oo 1/(nln(n)(ln(ln(n)))^p) converges when pgt1 .