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Part BCSIR NET December 2025bound-the-sequence-first-then-the-recursion-forces-convergence-to-1-not-just-boundedness

Bound the sequence first then the recursion forces convergence to 1 not just boundedness

For a non-negative real number a, let a\sqrt{a} denote its non-negative square-root. Consider the sequence {xnx_{n}}n≥1_{n}\ge_{1} defined by x1=1x_{1} = 1 and xnx_{n}₊1=1+xn2/n_{1} = \sqrt{1 + x_{n}^{2}/n} for n ≥ 1. Which of the following statements is true?

  1. A.lim sup⁡xn=1\limsup x_{n} = 1✓
  2. B.lim sup⁡xn=2\limsup x_{n} = \sqrt{2}
  3. C.lim sup⁡xn=2\limsup x_{n} = 2
  4. D.lim sup⁡xn=1+2\limsup x_{n} = 1 + \sqrt{2}

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: limsup(aₙ + bₙ) = limsup aₙ + limsup bₙ

More on this topic

The chapter behind this: limsup and liminf — three equivalent definitions and the algebra — free to read

From The Real Line › limsup, liminf and subsequential limits

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