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Part BCSIR NET December 2024a-ratio-of-growing-terms-can-be-bounded

A ratio of growing terms can be bounded

Consider the sequences (an)n≥1(a_{n})_{n}\ge_{1} and (bn)n≥1(b_{n})_{n}\ge_{1} defined by an=(en+ea_{n} = (e^{n} + e⁻n)/2^{n})/2 and bn=anb_{n} = a_{n}₊1/an_{1}/a_{n}. Which of the following statements is true?

  1. A.For every x∈Rx \in \mathbb{R} there exists some n such that an>xa_{n} > x✓ Correct
  2. B.For every x∈Rx \in \mathbb{R} there exists some n such that an<xa_{n} < x
  3. C.For every x∈Rx \in \mathbb{R} there exists some n such that bn>xb_{n} > x
  4. D.For every x∈Rx \in \mathbb{R} there exists some n such that bn<xb_{n} < x

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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