QuestionJune 5, 2025

7: sum _(n=1)(-1)^n(lnn)/(n)

7: sum _(n=1)(-1)^n(lnn)/(n)
7: sum _(n=1)(-1)^n(lnn)/(n)

Solution
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Answer

The series converges. Explanation 1. Determine convergence The series \sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{n} is an alternating series. Use the Alternating Series Test. 2. Check conditions of Alternating Series Test For convergence, b_n = \frac{\ln n}{n} must be positive, decreasing, and \lim_{n \to \infty} b_n = 0. 3. Verify positivity \ln n > 0 for n \geq 2, so b_n > 0 for n \geq 2. 4. Verify decreasing nature Check if b_{n+1} < b_n: \frac{\ln(n+1)}{n+1} < \frac{\ln n}{n} holds for large n. 5. Calculate limit \lim_{n \to \infty} \frac{\ln n}{n} = 0 using L'Hôpital's Rule.

Explanation

1. Determine convergence<br /> The series $\sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{n}$ is an alternating series. Use the Alternating Series Test.<br />2. Check conditions of Alternating Series Test<br /> For convergence, $b_n = \frac{\ln n}{n}$ must be positive, decreasing, and $\lim_{n \to \infty} b_n = 0$.<br />3. Verify positivity<br /> $\ln n > 0$ for $n \geq 2$, so $b_n > 0$ for $n \geq 2$.<br />4. Verify decreasing nature<br /> Check if $b_{n+1} < b_n$: $\frac{\ln(n+1)}{n+1} < \frac{\ln n}{n}$ holds for large $n$.<br />5. Calculate limit<br /> $\lim_{n \to \infty} \frac{\ln n}{n} = 0$ using L'Hôpital's Rule.
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