\( \sum_{k=2}^{\infty} \frac{1}{\sqrt{k^{3}-3 k+6}} \) By the an appropriate test to determine whether the series converges. \( \square \)

To determine whether the series \[ \sum_{k=2}^{\infty} \frac{1}{\sqrt{k^{3}-3k+6}} \] converges, we can use the **Limit Comparison Test**. First, we need to analyze the behavior of the term \( \sqrt{k^3 - 3k + 6} \) as \( k \) becomes large. For large \( k \), the term \( k^3 - 3k + 6 \) behaves like \( k^3 \). Therefore, we can approximate: \[ \sqrt{k^3 - 3k + 6} \sim \sqrt{k^3} = k^{3/2}. \] This leads us to consider the series \[ \sum_{k=2}^{\infty} \frac{1}{\sqrt{k^3}} = \sum_{k=2}^{\infty} \frac{1}{k^{3/2}}. \] The series \( \sum_{k=2}^{\infty} \frac{1}{k^{3/2}} \) is a \( p \)-series with \( p = \frac{3}{2} > 1 \), which is known to converge. Now, we apply the Limit Comparison Test. We compute the limit: \[ \lim_{k \to \infty} \frac{\frac{1}{\sqrt{k^3 - 3k + 6}}}{\frac{1}{\sqrt{k^3}}} = \lim_{k \to \infty} \frac{\sqrt{k^3}}{\sqrt{k^3 - 3k + 6}}. \] This simplifies to: \[ \lim_{k \to \infty} \sqrt{\frac{k^3}{k^3 - 3k + 6}} = \lim_{k \to \infty} \sqrt{\frac{1}{1 - \frac{3}{k^2} + \frac{6}{k^3}}}. \] As \( k \to \infty \), the terms \( \frac{3}{k^2} \) and \( \frac{6}{k^3} \) approach 0, so we have: \[ \lim_{k \to \infty} \sqrt{\frac{1}{1 - 0 + 0}} = \sqrt{1} = 1. \] Since the limit is a positive finite number, by the Limit Comparison Test, the convergence of \( \sum_{k=2}^{\infty} \frac{1}{\sqrt{k^3 - 3k + 6}} \) is the same as the convergence of \( \sum_{k=2}^{\infty} \frac{1}{k^{3/2}} \). Since \( \sum_{k=2}^{\infty} \frac{1}{k^{3/2}} \) converges, we conclude that: \[ \sum_{k=2}^{\infty} \frac{1}{\sqrt{k^3 - 3k + 6}} \text{ converges.} \] Thus, the series converges. \(\boxed{\text{converges}}\)