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

Limit below one freezes the running max

Let (an)n≥1(a_{n})_{n}\ge_{1} be a sequence of positive real numbers. Let bn=an/max⁡b_{n} = a_{n} / \max{a1,…,ana_{1}, \dots, a_{n}}, n ≥ 1. Which of the following statements are necessarily true?

  1. A.If limn→∞bnlim_{n}\to\infty b_{n} exists in R\mathbb{R}, then {an:n≥1a_{n} : n \ge 1} is bounded
  2. B.If limn→∞bn=1lim_{n}\to\infty b_{n} = 1, then limn→∞anlim_{n}\to\infty a_{n} exists in R\mathbb{R}
  3. C.If limn→∞bn=1/2lim_{n}\to\infty b_{n} = 1/2, then limn→∞anlim_{n}\to\infty a_{n} exists in R\mathbb{R}✓ Correct
  4. D.If limn→∞bn=0lim_{n}\to\infty b_{n} = 0, then limn→∞an=0lim_{n}\to\infty a_{n} = 0✓ Correct

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, and why this one tests limit assumed to exist.

See pricing

50 are analysed free — try those first.

The trap it tests

Limit assumed to exist

Reasoning with a limit before establishing there is one. limsup is not lim.

Drill statements like this

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

More on this topic

The chapter behind this: Sequences in ℝ — the implication map — free to read

From The Real Line › Sequences: convergence, monotone, Bolzano–Weierstrass, Cauchy

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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