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Part CCSIR NET June 2024splitting-a-conditionally-convergent-series

Splitting a conditionally convergent series

Let ∑n\sum_{n}₌1∞an_{1}^\infty a_{n} be a convergent series of real numbers. For n ≥ 1 define An=anA_{n} = a_{n} if an>0a_{n} > 0 and 0 otherwise; Bn=anB_{n} = a_{n} if an<0a_{n} < 0 and 0 otherwise. Which of the following statements are necessarily true?

  1. A.An→0A_{n} \to 0 and Bn→0B_{n} \to 0 as n→∞n \to \infty✓ Correct
  2. B.If ∑n\sum_{n}₌1∞an_{1}^\infty a_{n} is absolutely convergent, then both ∑n\sum_{n}₌1∞An_{1}^\infty A_{n} and ∑n\sum_{n}₌1∞Bn_{1}^\infty B_{n} are absolutely convergent✓ Correct
  3. C.Both ∑n\sum_{n}₌1∞An_{1}^\infty A_{n} and ∑n\sum_{n}₌1∞Bn_{1}^\infty B_{n} are convergent
  4. D.If ∑n\sum_{n}₌1∞an_{1}^\infty a_{n} is not absolutely convergent, then both ∑n\sum_{n}₌1∞An_{1}^\infty A_{n} and ∑n\sum_{n}₌1∞Bn_{1}^\infty B_{n} are divergent✓ Correct

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: Σaₙ convergent ⇒ Σaₙ² convergent

More on this topic

The chapter behind this: Series tests — decision tree and the trap list — free to read

From The Real Line › Series: comparison, ratio, root, Raabe, condensation, alternating, rearrangements

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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