 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-4/2e/b/solution

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

1

gives

e = n = 0 \sum \infty \frac{1}{n !} .

For every positive integer

q

, multiply by the factorial

q!

and separate the integer part:

q! e = I_{q} \in Z n = 0 \sum q \frac{q !}{n !} + R_{q}, R_{q} = \sum_{j = 1}^{\infty} \frac{1}{( q + 1 ) ( q + 2 ) \dots ( q + j )} .

Every term in

R_{q}

is positive. Each denominator is at least

(q + 1)^{j}

, and the inequality is strict when

j \geq 2

. Comparison with the geometric series therefore gives

0 < R_{q} < \sum_{j = 1}^{\infty} (q + 1)^{- j} = \frac{1}{q} \leq 1.

e = p / q

p

and

q \geq 1

, then

q! e = p (q - 1)!

would be an integer. Since

I_{q}

is also an integer,

R_{q}

would be an integer strictly between zero and one, which is impossible. Hence

e is irrational .

This proves the irrationality of e by using its rapidly decreasing series tail.

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