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Calculus IIChapter 4: Series and SequencesSection 4-8

Alternating Series Test — Question 4

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Question 4

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

  1. Verify the AST hypotheses and classify absolute versus conditional convergence.

  2. Find the smallest NN for which the next-term bound bN+1b_{N+1} is strictly below 0.010.01, thereby guaranteeing |S−sN|<0.01|S-s_N|<0.01.

  3. Explain the role of the strict inequality.

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Question 4 – Solution

Step 1: Verify convergence.

With bn=1/nb_n=1/\sqrt n, positivity is clear, bnb_n decreases, and bn→0b_n\to0. The AST proves convergence. The absolute series ∑1/n\sum1/\sqrt n diverges (p=1/2p=1/2), so convergence is conditional.

Step 2: Apply the error estimate.

|S−sN|≤bN+1=1N+1.|S-s_N|\le b_{N+1}=\frac1{\sqrt{N+1}}. To guarantee an error strictly below 0.010.01 it is sufficient that 1N+1<1100⇔N+1>10000.\frac1{\sqrt{N+1}}<\frac1{100} \iff N+1>10000. Thus the smallest integer furnished by this bound is N=10000\boxed{N=10000}.

Step 3: Verify strictness.

At N=9999N=9999, the bound equals 0.010.01, not less than it. At N=10000N=10000, it is 1/10001<0.011/\sqrt{10001}<0.01.

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