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Part BCSIR NET December 2023

Series: comparison, ratio, root, Raabe, condensation, alternating, rearrangements

Consider the following infinite series: (a)∑n≥1(a) \sum_{n\ge1} sin⁡(nπ/2)/n,(b)∑n≥1\sin(n\pi/2)/\sqrt{n}, (b) \sum_{n\ge1} log⁡(1+1/n2)\log(1 + 1/n^{2}). Which one of the following statements is true?

  1. A.(a) is convergent, but (b) is not convergent.
  2. B.(a) is not convergent, but (b) is convergent.
  3. C.Both (a) and (b) are convergent.✓ Correct
  4. D.Neither (a) nor (b) is convergent.

The idea, in plain words

One of these series has terms that alternate in sign, and that changes everything: alternating terms that shrink to zero always add up to something finite, even when the same terms without the signs would blow up. The other shrinks fast enough to converge outright.

Solution

(a): the non-zero terms are (−1)k/2k+1(-1)^{k}/\sqrt{2k+1}, an alternating series with terms decreasing to 0 (Leibniz).

(b):log⁡(1+1/n2)(b): \log(1 + 1/n^{2}) ~ 1/n21/n^{2}, comparison with a convergent p-series.

Why each option is right or wrong

Checked against 3 symbolic computations
  • A.WrongExecution slip

    Half right.

    (a) does converge

    but so does (b):log⁡(1+1/n2)(b): \log(1 + 1/n^{2}) ~ 1/n21/n^{2}, and SymPy evaluates the sum numerically to ≈ 1.3018 — finite.

    Dismissing log(1 + u) as 'too slow to decay' overlooks that u=1/n2u = 1/n^{2} is already summable, and log(1 + u) ≤ u.

  • B.WrongHypothesis dropped

    False, and this is the engineered trap.

    ∑1/n\sum1/\sqrt{n} diverges (p = 1/2 ≤ 1)

    so comparing term-by-term suggests (a) diverges too.

    But sin⁡(nπ/2)\sin(n\pi/2) runs 1, 0, −1, 0 repeating

    so the non-zero terms alternate: the series is ∑(−1)k/2k+1\sum(-1)^k/\sqrt{2k+1}

    which converges by Leibniz — the terms decrease to 0, and SymPy confirms that limit is 0.

    The mistake: the comparison and p-tests require non-negative terms.

    That hypothesis was dropped.

  • C.Correct

    True.

    (a) converges conditionally by the Leibniz test; (b) converges absolutely by comparison with ∑1/n2\sum1/n^{2}.

    Note the asymmetry worth remembering: (a) converges but not absolutely, (b) converges absolutely.

  • D.Wrong

    False — it makes both errors above at once.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

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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