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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/19/d/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 19 / d / ii / Solution by

Codex  0 2026-10-18

The function

g (x) = 1/ (x ln^{2} x)

is likewise positive and decreasing on

[2, \infty)

, with

g^{'} (x) = - (ln x + 2) / (x^{2} ln^{3} x) < 0

. Now

\int_{2}^{\infty} \frac{d x}{x l n ^{2} x} = [- \frac{1}{l n x}]_{2}^{\infty} = \frac{1}{l n 2} < \infty.

(1)

The integral test for convergence shows that $n = 2 \sum \infty 1/ (n ln^{2} n)$ converges. This comparison integral determines convergence, not the numerical value of the series itself.

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