 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-ii/1/c/i/solution

n = 2 k

, the terms are

1 + 1/ (2 k) \to 1

. At odd integers they are

- 1 - 1/ (2 k - 1) \to - 1

. The largest eventual values therefore come from the subsequence at even integers, and

n \to \infty lim sup (- 1)^{n} (1 + 1/ n) = 1.

Explicitly, every sufficiently late term is below

1 + ε

, while infinitely many even integers give terms greater than

1 - ε

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