Mathematics

Mathematics

Problem

Consider the series ∑n=2∞1n(ln⁡n)2\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n(\ln n)^2}. Which statement best justifies that an improper integral of the related function may be used to study the series? a) For x≥2x\ge 2, f(x)=1x(ln⁡x)2f(x)=\dfrac{1}{x(\ln x)^2} is positive and continuous, therefore the integral test can be applied. b) Because lim⁡n→∞1n(ln⁡n)2=0\displaystyle\lim_{n\to\infty}\dfrac{1}{n(\ln n)^2}=0, the integral test is valid. c) For x≥2x\ge 2, f(x)=1x(ln⁡x)2f(x)=\dfrac{1}{x(\ln x)^2} is positive, continuous, and decreasing, so the integral test may be used. d) Since f′(x)=1−2ln⁡xx2(ln⁡x)3f'(x)=\dfrac{1-2\ln x}{x^2(\ln x)^3} is positive for x≥2x\ge 2, ff is increasing, and monotonicity alone guarantees the integral test applies.

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