1. The problem is to determine whether the infinite series $$\sum_{n=1}^{\infty} \frac{5n^2 + 7}{n^3 + 9}$$ converges or diverges. 2. To analyze convergence, we use the Comparison Test or Limit Comparison Test. We compare the given series to a simpler series whose convergence is known. 3. For large $n$, the dominant terms in numerator and denominator are $5n^2$ and $n^3$ respectively, so the general term behaves like $$\frac{5n^2}{n^3} = \frac{5}{n}.$$ 4. The series $$\sum_{n=1}^{\infty} \frac{1}{n}$$ is the harmonic series, which is known to diverge. 5. Using the Limit Comparison Test, let $$a_n = \frac{5n^2 + 7}{n^3 + 9}$$ and $$b_n = \frac{1}{n}.$$ Compute $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{5n^2 + 7}{n^3 + 9}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{(5n^2 + 7) n}{n^3 + 9} = \lim_{n \to \infty} \frac{5n^3 + 7n}{n^3 + 9}.$$ 6. Dividing numerator and denominator by $n^3$, we get $$\lim_{n \to \infty} \frac{5 + \frac{7}{n^2}}{1 + \frac{9}{n^3}} = \frac{5 + 0}{1 + 0} = 5.$$ 7. Since the limit is a finite positive number and $$\sum b_n = \sum \frac{1}{n}$$ diverges, by the Limit Comparison Test, the original series $$\sum a_n$$ also diverges. Final answer: The series diverges.