Alternating Series Test — Question 1

Section 11-8 · Calculus pages 2081–2082

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*Question 1 Consider ∑n=1∞(−1)n−1n\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n.

  1. Verify every Alternating Series Test hypothesis.

  2. Test absolute convergence and classify the series.

  3. State its familiar value and give the error bound after NN terms.

*Question 1 – Solution

Write bn=1/nb_n=1/n. Then bn>0b_n>0, bn+1<bnb_{n+1}<b_n, and bn→0b_n\to0. Hence the series converges by the Alternating Series Test.

∑|(−1)n−1n|=∑1n,\sum\left|\frac{(-1)^{n-1}}n\right|=\sum\frac1n, which diverges. Therefore convergence is conditional.

The power series for ln⁡(1+x)\ln(1+x) at x=1x=1 gives S=ln⁡2S=\ln2. The Alternating Series Estimation Theorem gives |S−sN|≤bN+1=1N+1.|S-s_N|\le b_{N+1}=\frac1{N+1}. Odd and even partial sums approach ln⁡2\ln2 from opposite sides.