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Calculus IIChapter 4: Series and SequencesSection 4-7

Comparison and Limit Comparison Tests — Question 1

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Question 1

Consider ∑n=1∞1n2+3n\displaystyle\sum_{n=1}^{\infty}\frac1{n^2+3n}.

  1. Identify the dominant power and select a standard benchmark.

  2. Prove a direct-comparison inequality with the correct direction.

  3. Classify the series and confirm the result by limit comparison.

State why the benchmark converges or diverges.

Original worksheet page 1: question and worked solution for 4-7-001
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Question 1 – Solution

Step 1: Select the benchmark.

The denominator behaves like n2n^2, so use bn=1/n2b_n=1/n^2, a convergent pp-series with p=2p=2.

Step 2: Prove direct comparison.

Since n2+3n≥n2>0n^2+3n\ge n^2>0, 0<1n2+3n≤1n2.0<\frac1{n^2+3n}\le\frac1{n^2}. The Direct Comparison Test proves convergence.

Step 3: Verify by limit comparison.

limn→∞1/(n2+3n)1/n2=limn→∞11+3/n=1.\lim_{n\to\infty}\frac{1/(n^2+3n)}{1/n^2} =\lim_{n\to\infty}\frac1{1+3/n}=1. The finite positive limit confirms that both series share the same verdict.

Original worksheet page 2: question and worked solution for 4-7-001

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