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

A ratio of growing terms can be bounded

Consider the sequences $(a_{n})_{n} \geq_{1}$ and $(b_{n})_{n} \geq_{1}$ defined by $a_{n} = (e^{n} + e$ ⁻ $^{n}) /2$ and $b_{n} = a_{n}$ ₊ $_{1} / a_{n}$ . Which of the following statements is true?

A.For every $x \in R$ there exists some n such that $a_{n} > x$ ✓
B.For every $x \in R$ there exists some n such that $a_{n} < x$
C.For every $x \in R$ there exists some n such that $b_{n} > x$
D.For every $x \in R$ there exists some n such that $b_{n} < x$

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.

See pricing

50 are analysed free — try those first.

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

A ratio of growing terms can be bounded

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