 Solution

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

For

t \geq 0

∣ sin t ∣ \leq t

, so

0 \leq \frac{n s i n ^{2} ( 1/ n )}{( l o g n ) ^{2}} \leq \frac{1}{n ( l o g n ) ^{2}} .

The integral test applies to the decreasing positive comparison function, and

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

The comparison test proves that the series converges; since all terms are nonnegative, this is also absolute convergence.

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