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Calculus IIChapter 4: Series and SequencesSection 4-6

Integral Test — Question 8

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Question 8

Consider ∑n=2∞ln⁡nn2.\sum_{n=2}^{\infty}\frac{\ln n}{n^2}.

  1. Verify the Integral Test hypotheses for f(x)=ln⁡x/x2f(x)=\ln x/x^2 on [2,∞)[2,\infty).

  2. Evaluate the improper integral using integration by parts.

  3. Determine convergence and derive two-sided bounds for RNR_N.

  4. Explain how the result illustrates polynomial growth dominating logarithmic growth.

Original worksheet page 1: question and worked solution for 4-6-008
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Question 8 – Solution

Step 1: Verify the hypotheses on the tail.

Let f(x)=ln⁡x/x2f(x)=\ln x/x^2 for x≥2x\ge2. It is continuous and positive. Also f′(x)=1−2ln⁡xx3<0(x≥2),f'(x)=\frac{1-2\ln x}{x^3}<0\qquad(x\ge2), so it decreases.

Step 2: Integrate by parts.

Take u=ln⁡xu=\ln x and dv=x−2dxdv=x^{-2}dx. Then du=dx/xdu=dx/x and v=−1/xv=-1/x: ∫ln⁡xx2dx=−ln⁡xx−1x+C.\int\frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}-\frac1x+C. Hence ∫2∞ln⁡xx2dx=ln⁡2+12<∞,\int_2^{\infty}\frac{\ln x}{x^2}\,dx=\frac{\ln2+1}{2}<\infty, and the series converges.

Step 3: Estimate the remainder.

Since the tail integral from AA equals (ln⁡A+1)/A(\ln A+1)/A, ln⁡(N+1)+1N+1≤RN≤ln⁡N+1N.\boxed{\frac{\ln(N+1)+1}{N+1}\le R_N\le\frac{\ln N+1}{N}}. Quadratic denominator growth dominates the logarithmic numerator.

Original worksheet page 2: question and worked solution for 4-6-008

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