sum_(n=2)^oo 1/(sqrt(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=2)^oo1/(sqrt(n)-1)
Let b_n=1/(sqrt(n)-1) and a_n=1/sqrt(n)=1/n^(1/2)
1/(sqrt(n)-1)>1/sqrt(n)>0 for n>=2
sum_(n=2)^oo1/n^(1/2) is a p-series
The p-series sum_(n=1)^oo1/n^p , is convergent if p>1 and divergent if 0For the series sum_(n=2)^oo1/n^(1/2) p=1/2<1 so it diverges as per the p-series test.
Since the series sum_(n=2)^oo1/sqrt(n) diverges, so the series sum_(n=2)^oo1/(sqrt(n)-1) diverges as well by the direct comparison test.