\( \sum _ { n = 1 } ^ { \infty } \frac { n ^ { k - 1 } } { n ^ { k } + 7 } , k > 2 \)

To analyze the series \[ S = \sum_{n=1}^{\infty} \frac{n^{k-1}}{n^k + 7} \] for \( k > 2 \), we start by simplifying the term inside the summation. We can rewrite the term as follows: \[ \frac{n^{k-1}}{n^k + 7} = \frac{n^{k-1}}{n^k(1 + \frac{7}{n^k})} = \frac{1}{n(1 + \frac{7}{n^k})} \] This allows us to express the series as: \[ S = \sum_{n=1}^{\infty} \frac{1}{n(1 + \frac{7}{n^k})} \] Next, we analyze the behavior of the term \( \frac{1}{n(1 + \frac{7}{n^k})} \) as \( n \) becomes large. For large \( n \), \( \frac{7}{n^k} \) becomes very small, and we can approximate: \[ 1 + \frac{7}{n^k} \approx 1 \] Thus, for large \( n \): \[ \frac{1}{n(1 + \frac{7}{n^k})} \approx \frac{1}{n} \] This suggests that the series behaves similarly to the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \), which diverges. However, we need to be more precise about the convergence of our original series. To analyze the convergence more rigorously, we can use the limit comparison test. We compare our series with the harmonic series: \[ \frac{n^{k-1}}{n^k + 7} \sim \frac{1}{n} \quad \text{as } n \to \infty \] We can compute the limit: \[ \lim_{n \to \infty} \frac{\frac{n^{k-1}}{n^k + 7}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^n}{n^k + 7} = \lim_{n \to \infty} \frac{n^{k}}{n^k + 7} = 1 \] Since this limit is a positive finite number, by the limit comparison test, the convergence of \( S \) is the same as that of \( \sum_{n=1}^{\infty} \frac{1}{n} \). Since the harmonic series diverges, we conclude that: \[ \sum_{n=1}^{\infty} \frac{n^{k-1}}{n^k + 7} \text{ diverges for } k > 2. \] Thus, the final result is: \[ \text{The series diverges for } k > 2. \]