 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/2f/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

S = \sum_{n = 1}^{\infty} \frac{m _{n}}{n !} .

Assume for a contradiction that

S = a / b

is rational. For every

N \geq b

, both

N! S

and

N! \sum_{n = 1}^{N} \frac{m _{n}}{n !}

are integers, because the

m_{n}

are integers. Their positive difference

R_{N} = N! \sum_{n = N + 1}^{\infty} \frac{m _{n}}{n !} = \sum_{k = 1}^{\infty} \frac{m _{N + k}}{( N + 1 ) \dots ( N + k )}

must therefore be a positive integer.

On the other hand, concavity of

x^{α}

for

0 \leq α < 1

gives

(N + k)^{α} \leq N^{α} + k^{α} \leq N^{α} + k

. Since

(N + 1) \dots (N + k) \geq (N + 1)^{k}

0 < R_{N} \leq A \sum_{k = 1}^{\infty} \frac{N ^{α} + k}{( N + 1 ) ^{k}} = A (N^{α - 1} + \frac{N + 1}{N ^{2}}) ⟶ 0.

Eventually

0 < R_{N} < 1

, contradicting its integrality. Hence

n = 1 \sum \infty \frac{m _{n}}{n !} is irrational .

The integrality assumption is essential. For example, take

α = 0

A = 2

, and

m_{n} = \frac{2}{e - 1} (n \geq 1) .

Then

1 < m_{n} < 2

\sum_{n = 1}^{\infty} \frac{m _{n}}{n !} = \frac{2}{e - 1} \sum_{n = 1}^{\infty} \frac{1}{n !} = 2.

Thus the result does not remain true for arbitrary real $m_{n}$ .

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