 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/19/a/solution

All

a_{n}

are positive, and

a_{1} = e^{- 1}

. The logarithm turns consecutive ratios into differences, producing the finite telescoping sum

\sum_{k = 1}^{n} u_{k} = \sum_{k = 1}^{n} (ln a_{k + 1} - ln a_{k}) = ln a_{n + 1} + 1.

Thus, already for every finite

n

a_{n + 1} = exp (- 1 + \sum_{k = 1}^{n} u_{k}) .

If the partial sums converge to

S

n \to \infty lim a_{n} = exp (- 1 + n \to \infty lim k = 1 \sum n u_{k}) = e^{- 1 + S} .

The shift from

a_{n + 1}

a_{n}

preserves the limit of a sequence. Convergence of the partial sums is established in part (f), so no convergence assumption is left unsupported.

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