Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 1 / 11F / c

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- Solution c

Solution

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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