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Part BCSIR NET December 2023periodic-subsequence

Periodic subsequence

Consider the sequence (an)n≥1(a_{n})_{n\ge1}, where an=cos⁡((−1)nnπ/2+nπ/3)a_{n} = \cos((-1)^{n} n\pi/2 + n\pi/3). Which one of the following statements is true?

  1. A.lim sup⁡an=3/2\limsup a_{n} = \sqrt{3}/2.
  2. B.lim sup⁡a2n=1\limsup a_{2n} = 1.✓ Correct
  3. C.lim sup⁡a2n=1/2\limsup a_{2n} = 1/2.
  4. D.lim sup⁡a3n=0\limsup a_{3n} = 0.

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: 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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