 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-1/9d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Suppose

a_{n} \to L

. Given

ε > 0

, choose

N

such that

∣ a_{k} - L ∣ < ε /2

for

k > N

. Then

\frac{1}{n} \sum_{k = 1}^{n} a_{k} - L \leq \frac{1}{n} \sum_{k = 1}^{N} ∣ a_{k} - L ∣ + \frac{1}{n} \sum_{k = N + 1}^{n} ∣ a_{k} - L ∣.

The first term tends to zero because its numerator is fixed, and the second is at most

ε /2

. Thus the Cesaro mean tends to

L

The converse is false. For

a_{n} = (- 1)^{n}

, the Cesaro means tend to

0

, while the original sequence alternates between

- 1

and

1

and does not converge.

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