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Calculus IIChapter 4: Series and SequencesSection 4-6

Integral Test — Question 1

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Question 1

Consider the series ∑n=1∞1n3/2.\sum_{n=1}^{\infty}\frac1{n^{3/2}}.

  1. Define a continuous function ff with f(n)=1/n3/2f(n)=1/n^{3/2} and verify that it is positive, continuous, and decreasing on the required interval.

  2. Use the Integral Test to determine whether the series converges or diverges.

  3. If sNs_N is the NNth partial sum and RN=S−sNR_N=S-s_N, derive both an upper and a lower bound for RNR_N.

Show the improper-integral limit and all antiderivative work.

Original worksheet page 1: question and worked solution for 4-6-001
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Question 1 – Solution

Step 1: Choose the continuous function.

Let f(x)=x−3/2f(x)=x^{-3/2} for x≥1x\ge1. It is continuous and positive, and f′(x)=−32x−5/2<0,f'(x)=-\frac32x^{-5/2}<0, so it is decreasing. Also f(n)=1/n3/2f(n)=1/n^{3/2}.

Step 2: Evaluate the improper integral.

∫1∞x−3/2dx=limb→∞[−2x−1/2]1b=2.\int_1^{\infty}x^{-3/2}\,dx =\lim_{b\to\infty}\left[-2x^{-1/2}\right]_1^b=2. The integral converges, so the series converges by the Integral Test.

Step 3: Bound the remainder.

For a positive decreasing ff, ∫N+1∞f(x)dx≤RN≤∫N∞f(x)dx.\int_{N+1}^{\infty}f(x)\,dx\le R_N\le\int_N^{\infty}f(x)\,dx. Therefore 2N+1≤RN≤2N.\boxed{\frac2{\sqrt{N+1}}\le R_N\le\frac2{\sqrt N}}. The test proves convergence; these inequalities additionally measure the truncation error.

Original worksheet page 2: question and worked solution for 4-6-001

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