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Calculus IIChapter 4: Series and SequencesSection 4-7

Comparison and Limit Comparison Tests — Question 3

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Question 3

Consider ∑n=1∞sin⁡2nn2\displaystyle\sum_{n=1}^{\infty}\frac{\sin^2 n}{n^2} (radians).

  1. Establish a pointwise trigonometric bound.

  2. Compare with a standard pp-series.

  3. Classify the series and explain why no averaging argument is needed.

Original worksheet page 1: question and worked solution for 4-7-003
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Question 3 – Solution

Step 1: Bound the numerator.

For every real number tt, 0≤sin⁡2t≤10\le\sin^2t\le1. Hence for each positive integer nn, 0≤sin⁡2nn2≤1n2.0\le\frac{\sin^2n}{n^2}\le\frac1{n^2}.

Step 2: Classify the benchmark.

The series ∑1/n2\sum1/n^2 converges because it is a pp-series with p=2>1p=2>1.

Step 3: Apply Direct Comparison.

The given nonnegative series is bounded term by term above by a convergent series, so it converges. No information about the average value or periodic distribution of sin⁡2n\sin^2n is required.

Original worksheet page 2: question and worked solution for 4-7-003

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