Comparison and Limit Comparison Tests — Question 2

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Question 2

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

  1. Use dominant powers to select a harmonic benchmark.

  2. Compute the Limit Comparison Test ratio.

  3. Classify the series and give an independent direct lower comparison.

Original worksheet page 1: question and worked solution for 4-7-002
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Question 2 – Solution

Step 1: Select a benchmark.

The leading-term quotient is 2n/n2=2/n2n/n^2=2/n, so choose bn=1/nb_n=1/n. All terms are positive.

Step 2: Compute the ratio limit.

L=limn→∞(2n+1)/(n2+1)1/n=limn→∞2n2+nn2+1=2.L=\lim_{n\to\infty}\frac{(2n+1)/(n^2+1)}{1/n} =\lim_{n\to\infty}\frac{2n^2+n}{n^2+1}=2. Because 0<L<∞0<L<\infty and the harmonic series diverges, the given series diverges.

Step 3: Check the direction directly.

For n≥1n\ge1, n2+1≤2n2n^2+1\le2n^2 and 2n+1≥2n2n+1\ge2n, so 2n+1n2+1≥2n2n2=1n.\frac{2n+1}{n^2+1}\ge\frac{2n}{2n^2}=\frac1n. A series larger than the divergent harmonic series must diverge.

Original worksheet page 2: question and worked solution for 4-7-002

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