 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/1/a/iii/solution

It is useful to prove the contrapositive. Suppose

k

(a_{n})

and

m

(- a_{n})

. The second condition says

a_{j} \geq a_{m}

for every

j > m

. Hence

j > max {k, m} ⟹ a_{m} \leq a_{j} \leq a_{k} .

Only finitely many earlier terms remain, and their absolute values have a finite maximum. Together these give a bound for every term, so

(a_{n})

is a bounded sequence. Consequently,

(a_{n}) unbounded ⟹ (a_{n}) or (- a_{n}) has no peaks .

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