Strategy for Series — Question 7

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Question 7

Classify ∑n=1∞sin⁡nn\displaystyle\sum_{n=1}^{\infty}\frac{\sin n}{n} as absolutely convergent, conditionally convergent, or divergent. Use bounded partial sums for the sine factor.

Original worksheet page 1: question and worked solution for 4-12-007
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Question 7 – Solution

Step 1: Verify bounded partial sums.

The finite identity ∑n=1Nsin⁡n=sin⁡(N/2)sin⁡((N+1)/2)sin⁡(1/2)\sum_{n=1}^N\sin n=\frac{\sin(N/2)\sin((N+1)/2)}{\sin(1/2)} shows that |∑n=1Nsinn|≤1/|sin⁡(1/2)|\left|\sum_{n=1}^N\sin n\right|\le1/|\sin(1/2)|.

Step 2: Apply Dirichlet’s Test.

The sequence 1/n1/n decreases to 00, and the partial sums of sin⁡n\sin n are bounded. Hence ∑sin⁡n/n\sum \sin n/n converges.

Step 3: Check absolute convergence.

Since |sin⁡n|≥sin⁡2n|\sin n|\ge\sin^2 n, ∑|sin⁡n|n≥∑sin⁡2nn=12∑1n−12∑cos⁡(2n)n.\sum\frac{|\sin n|}{n}\ge\sum\frac{\sin^2 n}{n}=\frac12\sum\frac1n-\frac12\sum\frac{\cos(2n)}n. The cosine series converges by Dirichlet’s Test while the harmonic series diverges, so the absolute-value series diverges.

Conclusion.

The series is conditionally convergent.

Original worksheet page 2: question and worked solution for 4-12-007

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