 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/5/a/ii/solution

q = 1 - p

G_{N} (s) = E [s^{N}] = n = 0 \sum \infty p (q s)^{n} = \frac{p}{1 - q s}, ∣ s ∣ < q^{- 1} .

The displayed restriction ensures absolute convergence.

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