 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ia/paper-1/9f/iv/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

The PDF has the doubly exponential denominator

e^{e^{n}}

. For its positive summands, the successive-term ratio is

\frac{a _{n + 1}}{a _{n}} = (1 + \frac{1}{n})^{3} exp (e^{n} - e^{n + 1}) = (1 + \frac{1}{n})^{3} e^{- (e - 1) e^{n}} ⟶ 0.

Thus the ratio test proves absolute convergence. Concretely, the tail is eventually dominated by a geometric series of ratio

1/2

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