 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-1/10f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Because

3 > e

, each of

x

lo g x

and

lo g lo g x

is positive and increasing for

x \geq 3

. Hence

h (x) = 1/ [x lo g x (lo g lo g x)^{2}]

is positive, continuous and decreasing. With

u = lo g lo g x

d u = d x / (x lo g x)

, so

\int_{3}^{\infty} \frac{d x}{x l o g x ( l o g l o g x ) ^{2}} = \int_{l o g l o g 3}^{\infty} u^{- 2} d u = \frac{1}{l o g l o g 3} < \infty.

The integral test for convergence implies that the series is convergent.

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