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Part CCSIR NET December 2023unbounded-vs-bounded-sequences

Unbounded vs bounded sequences

Let x be a real number. Which of the following statements are true?

  1. A.There exists an integer n ≥ 1 such that n2sin⁡(1/n)≥xn^{2} \sin(1/n) \ge x.✓ Correct
  2. B.There exists an integer n ≥ 1 such that n cos(1/n) ≥ x.✓ Correct
  3. C.There exists an integer n ≥ 1 such that n e−ne^{-n} ≥ x.
  4. D.There exists an integer n ≥ 2 such that n (log n)−1n)^{-1} ≥ x.✓ Correct

The idea, in plain words

Three of these sequences grow without any ceiling, so each of them eventually passes whatever number you name. The fourth only looks like it grows, because the shrinking exponential beats the growing factor in front of it and the whole thing stays small.

Solution

n2sin⁡(1/n)n^{2} \sin(1/n) ~ n, n cos(1/n) ~ n and n/log n all tend to +∞+\infty

so each eventually exceeds any x. n e−ne^{-n} ≤ 1/e is bounded

so it fails for x > 1/e.

Why each option is right or wrong

Checked against 5 symbolic computations
  • A.Correct

    True.

    For large n,sin⁡(1/n)=1/n−1/(6n3)+…n, \sin(1/n) = 1/n - 1/(6n^{3}) + \dots

    so n2sin⁡(1/n)=n−1/(6n)+…n^{2} \sin(1/n) = n - 1/(6n) + \dots and the sequence grows like n.

    SymPy confirms the limit is ∞\infty

    so the terms eventually pass any real x.

    Trigger: whenever you meet n^k sin(1/n), substitute the first-order expansion sin(u) ≈ u before judging growth — the factor n2n^{2} is not what decides it.

  • B.Correct

    True. cos(1/n) → 1

    so n cos(1/n) grows like n.

    Limit verified as ∞\infty.

  • C.WrongBoundary and endpoint

    False — and the whole question is built around this option. n·e^(−n) is bounded: the function t·e^(−t) peaks at t = 1 with value 1/e ≈ 0.3679 and decays thereafter.

    Sampled across [0.5, 40], the largest value SymPy finds is exactly that peak.

    So for any x > 1/e there is no such n.

    The mistake: reading ′n→∞′'n \to \infty' off the leading factor and assuming the product must grow. e^(−n) decays faster than n grows, and comparing growth rates — not merely spotting an n — is what is being tested.

  • D.Correct

    True. n/log⁡n→∞(n/\log n \to \infty (verified).

    Slower than n

    but still unbounded: log n grows more slowly than any positive power of n.

1 of these 1 wrong options correspond to a named reasoning error — the same errors recur across subjects.

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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