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Calculus IIChapter 4: Series and SequencesSection 4-6

Integral Test — Question 3

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Question 3

Consider ∑n=2∞1n(ln⁡n)2.\sum_{n=2}^{\infty}\frac1{n(\ln n)^2}.

  1. Verify that f(x)=1/[x(ln⁡x)2]f(x)=1/[x(\ln x)^2] is positive, continuous, and decreasing for x≥2x\ge2.

  2. Apply the Integral Test, showing the substitution and evaluation of the improper integral.

  3. Derive two-sided Integral Test bounds for the remainder RNR_N.

  4. Briefly compare the result with ∑1/(nln⁡n)\sum 1/(n\ln n).

Original worksheet page 1: question and worked solution for 4-6-003
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Question 3 – Solution

Step 1: Verify the hypotheses.

Define f(x)=1/[x(ln⁡x)2]f(x)=1/[x(\ln x)^2] for x≥2x\ge2. It is continuous and positive, and f′(x)=−ln⁡x+2x2(ln⁡x)3<0,f'(x)=-\frac{\ln x+2}{x^2(\ln x)^3}<0, so it decreases.

Step 2: Evaluate the integral.

Let u=ln⁡xu=\ln x, du=dx/xdu=dx/x: ∫2∞dxx(ln⁡x)2=∫ln⁡2∞u−2du=1ln⁡2<∞.\int_2^{\infty}\frac{dx}{x(\ln x)^2} =\int_{\ln2}^{\infty}u^{-2}\,du=\frac1{\ln2}<\infty. Hence the series converges. The extra logarithm power changes ∫du/u\int du/u into the convergent pp-integral ∫du/u2\int du/u^2.

Step 3: Estimate the remainder.

1ln⁡(N+1)≤RN≤1ln⁡N.\boxed{\frac1{\ln(N+1)}\le R_N\le\frac1{\ln N}}. These bounds follow by integrating from N+1N+1 and NN, respectively.

Original worksheet page 2: question and worked solution for 4-6-003

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