 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/11f/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

u_{n} = 2^{- n} n^{k}

\frac{u _{n + 1}}{u _{n}} = \frac{1}{2} (1 + \frac{1}{n})^{k} ⟶ \frac{1}{2} .

Thus for all sufficiently large

n

the ratio is, for example, at most

3/4

. Comparison with a decaying geometric sequence, equivalently the ratio test, yields

n \to \infty lim 2^{- n} n^{k} = 0 .

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