 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-ii/2/b/ii/solution

The implication is false. Set

a_{n} = \frac{1}{n l o g ( n + 1 )} .

Then

n a_{n} = 1/ lo g (n + 1) \to 0

. For

n \geq 2

lo g (n + 1) \leq lo g (2 n) \leq 2 lo g n

, so

a_{n} \geq 1/ (2 n lo g n)

\int_{2}^{R} \frac{d x}{x l o g x} = lo g lo g R - lo g lo g 2 ⟶ \infty.

Thus the comparison test shows that

\sum a_{n}

diverges.

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