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Calculus IIChapter 4: Series and SequencesSection 4-8

Alternating Series Test — Question 2

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Question 2

Consider ∑n=1∞(−1)nn2\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}.

  1. Verify that the Alternating Series Test applies.

  2. Determine whether the series converges absolutely or conditionally.

  3. Give an error bound for sNs_N and explain which conclusion is stronger.

Original worksheet page 1: question and worked solution for 4-8-002
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Question 2 – Solution

Step 1: Apply the AST.

For bn=1/n2b_n=1/n^2, we have bn>0b_n>0, bn+1<bnb_{n+1}<b_n, and bn→0b_n\to0. Thus the alternating series converges.

Step 2: Test absolute convergence.

∑|(−1)nn2|=∑1n2,\sum\left|\frac{(-1)^n}{n^2}\right|=\sum\frac1{n^2}, a convergent pp-series with p=2p=2. Hence the original series converges absolutely, a stronger conclusion than the AST alone provides.

Step 3: Bound the alternating error.

|S−sN|≤1(N+1)2.|S-s_N|\le\frac1{(N+1)^2}. In fact S=−π2/12S=-\pi^2/12, though this exact value is not needed for classification.

Original worksheet page 2: question and worked solution for 4-8-002

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