Strategy for Series — Question 3

Section 11-12 · Calculus pages 2165–2166

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*Question 3 Classify ∑n=1∞(−1)nn\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt n} as absolutely convergent, conditionally convergent, or divergent. Use separate tests for ordinary and absolute convergence.

*Question 3 – Solution

Put bn=1/nb_n=1/\sqrt n. Then bn>0b_n>0, bn+1<bnb_{n+1}<b_n, and limn→∞bn=0.\lim_{n\to\infty}b_n=0. Therefore ∑(−1)nbn\sum(-1)^n b_n converges by the Alternating Series Test.

∑n=1∞|(−1)nn|=∑n=1∞1n1/2.\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{\sqrt n}\right|=\sum_{n=1}^{\infty}\frac1{n^{1/2}}. This is a divergent pp-series because p=1/2≤1p=1/2\le1.

The original series is conditionally convergent. One test establishes convergence; the second shows that convergence depends essentially on cancellation between alternating signs.