 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2014/1/b/i/solution

Subtract adjacent values of the cumulative distribution function. Its value at zero is zero, so the same calculation applies also to

k = 1

p_{k} = (1 - \frac{1}{k + 1}) - (1 - \frac{1}{k}) = \frac{1}{k ( k + 1 )}, k \geq 1.

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