Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 11 - Infinite Series - 11.4 Absolute and Conditional Convergence - Exercises - Page 563: 29


The series $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{1 / 2}(\ln n)^{2}}$ converges.

Work Step by Step

For the series $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{1 / 2}(\ln n)^{2}}$, the positive term $b_n=\frac{1}{n^{1 / 2}(\ln n)^{2}}$, which is positive and decreasing. Evaluate the limit: $$\lim_{n\to \infty}b_n=\lim_{n\to \infty}\frac{1}{n^{1 / 2}(\ln n)^{2}}=\frac{1}{\infty}=0$$ Hence, using the alternating series test, the series $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n^{1 / 2}(\ln n)^{2}}$ converges.
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