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

We wish to analyze the series   S = ∑ₙ₌₁∞ (-1)ⁿ (ln n)⁄n. Step 1. Check for Absolute Convergence Consider the absolute series:   ∑ₙ₌₁∞ |(-1)ⁿ (ln n)⁄n| = ∑ₙ₌₁∞ (|ln n|)⁄n = ∑ₙ₌₂∞ (ln n)⁄n   (since ln 1 = 0). For large n, the term ln n ⁄ n behaves similarly to the integrand in   ∫₂∞ (ln x)⁄x dx. Make a substitution: let u = ln x, then du = dx/x. Hence,   ∫ (ln x)⁄x dx = ∫ u du = ½ u² + C = ½ (ln x)² + C. Evaluating from 2 to some large T, we find that   ∫₂ᵀ (ln x)⁄x dx = ½ (ln T)² − ½ (ln 2)². Since (ln T)² → ∞ as T → ∞, the integral diverges. By the Integral Test, the series ∑ (ln n)/n diverges. Therefore, the original series does not converge absolutely. Step 2. Check for Conditional Convergence (Alternating Series Test) The Alternating Series Test states that the series ∑ (-1)ⁿ aₙ converges if: 1. aₙ ≥ 0 for all n, 2. aₙ is eventually nonincreasing, 3. limₙ→∞ aₙ = 0. Here, let aₙ = (ln n)/n for n ≥ 2. Note that aₙ > 0, and   limₙ→∞ (ln n)/n = 0. To check that aₙ is eventually nonincreasing, consider the function f(x) = (ln x)/x for x > 1. Its derivative is   f′(x) = [1 − ln x] / x². For x large enough (specifically, x > e), we have ln x > 1 so that f′(x) < 0, meaning f(x) is decreasing for x > e. Thus, for all n sufficiently large (n > e), the terms aₙ are decreasing. Since the conditions are fulfilled, by the Alternating Series Test, the series S converges. Conclusion The series ∑ₙ₌₁∞ (-1)ⁿ (ln n)/n does not converge absolutely, but it converges conditionally by the Alternating Series Test.