Use the Comparison Test to determine whether the series converges. \( \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}}+2} \) The Comparison Test with \( \sum_{k=1}^{\infty} \square \) shows that the series

To determine whether the series \[ \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}} + 2} \] converges, we can use the Comparison Test. We will compare it to a known convergent series. First, observe that for large \( k \), the term \( k^{\frac{7}{4}} \) dominates the constant \( 2 \). Therefore, we can approximate: \[ k^{\frac{7}{4}} + 2 \approx k^{\frac{7}{4}} \quad \text{for large } k. \] This implies that: \[ \frac{1}{k^{\frac{7}{4}} + 2} \approx \frac{1}{k^{\frac{7}{4}}} \quad \text{for large } k. \] Now, we know that the series \[ \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}}} \] is a \( p \)-series with \( p = \frac{7}{4} > 1 \), which converges. Next, we need to establish a formal comparison. For \( k \geq 1 \), we have: \[ k^{\frac{7}{4}} + 2 > k^{\frac{7}{4}}. \] Thus, we can write: \[ \frac{1}{k^{\frac{7}{4}} + 2} < \frac{1}{k^{\frac{7}{4}}}. \] Since \( \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}}} \) converges, by the Comparison Test, we conclude that: \[ \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}} + 2} \] also converges. In summary, we have shown that: \[ \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}} + 2} \text{ converges by the Comparison Test with } \sum_{k=1}^{\infty} \frac{1}{k^{\frac{7}{4}}}. \]