sum_(n=1)^oo 1/(4root(3)(n)-1) Use the Direct Comparison Test to determine the convergence or divergence of the series.

Direct comparison test is applicable when suma_n and sumb_n are both positive series for all n such that a_n<=b_n
If sumb_n converges , then suma_n converges
If suma_n diverges , then sumb_n diverges.
Given series is sum_(n=1)^oo1/(4root(3)(n)-1)
Let b_n=1/(4root(3)(n)-1) and a_n=1/(4root(3)(n))=1/4(1/n^(1/3))
1/(4root(3)(n)-1)>1/(4root(3)(n))>0  for n>=1
The series sum_(n=1)^oo1/4(1/n^(1/3)) is a p-series with p=1/3<1
The p-series sum_(n=1)^oo1/n^p converges if p>1 and diverges if 0Since the series diverges sum_(n=1)^oo1/(4root(3)(n)) as per the p-series test and so the series sum_(n=1)^oo1/(4root(3)(n)-1) as well , diverges by the direct comparison test.