 Solution

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

(n 2 n) \leq \sum_{k = 0}^{2 n} (k 2 n) = 4^{n}

. Therefore

(n 2 n) 3^{- 2 n} sin n \leq (4/9)^{n} .

4/9

converges. By the comparison test, the given series is absolutely convergent.

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