1. **State the problem:** Determine if the infinite series $$\sum_{n=1}^{\infty} \frac{n^2 - 1}{2n^3 + 1}$$ converges using the $n^{th}$ term test for convergence. 2. **Recall the $n^{th}$ term test:** If $$\lim_{n \to \infty} a_n \neq 0$$, where $a_n$ is the $n^{th}$ term of the series, then the series diverges. If the limit is zero, the test is inconclusive. 3. **Find the limit of the $n^{th}$ term:** $$ \lim_{n \to \infty} \frac{n^2 - 1}{2n^3 + 1} = \lim_{n \to \infty} \frac{n^2(1 - \frac{1}{n^2})}{n^3(2 + \frac{1}{n^3})} = \lim_{n \to \infty} \frac{1 - \frac{1}{n^2}}{n(2 + \frac{1}{n^3})} $$ 4. **Simplify the limit:** $$ \lim_{n \to \infty} \frac{1 - \frac{1}{n^2}}{n(2 + \frac{1}{n^3})} = \lim_{n \to \infty} \frac{1 - 0}{n(2 + 0)} = \lim_{n \to \infty} \frac{1}{2n} = 0 $$ 5. **Interpretation:** Since $$\lim_{n \to \infty} a_n = 0$$, the $n^{th}$ term test is inconclusive. It does not prove convergence. 6. **Conclusion:** The statement "According to the $n^{th}$ term test, the series converges" is **False** because the test only confirms divergence if the limit is not zero, but cannot confirm convergence if the limit is zero. **Final answer:** B. False