 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ia/paper-2/4f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

e^{- 1} / j!

E [e^{tX}] = e^{- 1} j = 0 \sum \infty \frac{e ^{t j}}{j !} = exp (e^{t} - 1) .

k \geq 1

, multiply the exponential probability bound by

e

to obtain

\sum_{j = k}^{\infty} 1/ j! \leq exp (e^{t} - k t)

. The exponent is minimized at

e^{t} = k

, hence

t = lo g k \geq 0

, including

t = 0

when

k = 1

. Consequently

j = k \sum \infty \frac{1}{j !} \leq e^{k - k l o g k} = (\frac{e}{k})^{k} .

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