Part CCSIR NET June 2024limit-below-one-freezes-the-running-max

Limit below one freezes the running max

Let $(a_{n})_{n} \geq_{1}$ be a sequence of positive real numbers. Let $b_{n} = a_{n} / max$ { $a_{1}, \dots, a_{n}$ }, n ≥ 1. Which of the following statements are necessarily true?

A.If $l i m_{n} \to \infty b_{n}$ exists in $R$ , then { $a_{n} : n \geq 1$ } is bounded
B.If $l i m_{n} \to \infty b_{n} = 1$ , then $l i m_{n} \to \infty a_{n}$ exists in $R$
C.If $l i m_{n} \to \infty b_{n} = 1/2$ , then $l i m_{n} \to \infty a_{n}$ exists in $R$ ✓
D.If $l i m_{n} \to \infty b_{n} = 0$ , then $l i m_{n} \to \infty a_{n} = 0$ ✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: aₙ^{1/n} → L ⇒ aₙ₊₁/aₙ → L

Limit below one freezes the running max

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