\( \sum_{n=1}^{\infty}(-1)^{n} \frac{\ln n}{n} \) \( \begin{array}{l}\text { Does the series converge } \\ \text { conditionally, converge absolutely, or } \\ \text { diverge? }\end{array} \)

To analyze the series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{\ln n}{n} \), we can apply the alternating series test. The terms \( a_n = \frac{\ln n}{n} \) must be positive, decreasing, and approach zero as \( n \) approaches infinity. We see that \( \frac{\ln n}{n} \to 0 \) as \( n \to \infty \) and it is a decreasing function for sufficiently large \( n \). Thus, the series converges conditionally. Now, to check for absolute convergence, we consider \( \sum_{n=1}^{\infty} \left| (-1)^{n} \frac{\ln n}{n} \right| = \sum_{n=1}^{\infty} \frac{\ln n}{n} \). This series diverges by the integral test, since \( \int_{2}^{\infty} \frac{\ln x}{x} \, dx \) diverges. Therefore, the original series converges conditionally.