 Solution

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/i/solution

g (x) = 1/ (x ln x)

is positive and decreasing for

x \geq 2

, since

g^{'} (x) = - (ln x + 1) / (x^{2} ln^{2} x) < 0

\int_{2}^{R} \frac{d x}{x l n x} = ln (ln R) - ln (ln 2) ⟶ + \infty.

Consequently $n = 2 \sum \infty 1/ (n ln n)$ diverges.

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