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

The 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 the series is sum_(n=1)^oo1/sqrt(n^3+1)
Let a_n=1/sqrt(n^3+1) and b_n=1/sqrt(n^3)=1/n^(3/2)
1/sqrt(n^3)>1/sqrt(n^3+1)>0  for n>=1
The series sum_(n=1)^oo1/n^(3/2) is a p-series with p=3/2
The p-series sum_(n=1)^oo1/n^p converges if p>1 and diverges if 0As per the p-series test the series sum_(n=1)^oo1/sqrt(n^3) converges, so the series sum_(n=1)^oo1/sqrt(n^3+1) as well, converges by the direct comparison test.

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