Final answer:
The series is convergent as it behaves like a p-series with p = 3/2, which is greater than 1, for large values of n.
Step-by-step explanation:
To determine whether the series ∑n=1∞ &frac;√(3 n2+5);n2 √n+4 n; is convergent or divergent, we can use the comparison test. We see that √(3n2+5) is approximately √(3n2) = √3 * n when n is very large, and n2√n + 4n is approximately n2√n because the 4n term becomes insignificant as n becomes very large.
Therefore, our original series behaves like the series ∑n=1∞ &frac;√3 * n;n2√n; = ∑n=1∞ &frac;√3;n3/2; for large values of n. This new series is a p-series with p = 3/2, which is greater than 1. Therefore, the p-series is convergent, and by comparison, so is our original series.