AP Calculus BC

Convergent but Not Absolutely Convergent

Absolute convergence implies convergence, and the implication runs one way only. The alternating harmonic series converges while the series of its absolute values diverges, which is exactly what the phrase conditionally convergent records.

Absolute Convergence Theorem

∑∣an∣ converges  ⟹  ∑an converges\sum |a_n| \text{ converges} \implies \sum a_n \text{ converges}

If the series of absolute values converges, then the original series converges too. The converse is false.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

  1. 1

    The series of absolute values converges

    ∑n=1∞∣an∣<∞\sum_{n=1}^{\infty} |a_n| < \infty

    This is the only hypothesis, and the theorem is a one-way implication. The useful failure is the converse: knowing the original series converges tells you nothing at all about the absolute one, and half of series questions turn on keeping the two apart.

    Drop it and the theorem fails

    ∑n=1∞(−1)n+1n=ln⁡2,∑n=1∞1n=∞\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = \ln 2, \qquad \sum_{n=1}^{\infty} \frac{1}{n} = \infty

    The alternating harmonic series, whose absolute version is harmonic

    The alternating harmonic series converges, to ln⁡2\ln 2, by the Alternating Series Test. Strip the signs and you have the harmonic series, and the check confirms it is a p-series with p=1p = 1, which diverges. So convergence of the original series did not carry back to the absolute one. That gap is what conditional convergence means.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

  1. Write an=(an+∣an∣)−∣an∣a_n = \left(a_n + |a_n|\right) - |a_n|, a difference of two non-negative pieces.
  2. The first piece satisfies 0≤an+∣an∣≤2∣an∣0 \le a_n + |a_n| \le 2|a_n|, so it converges by direct comparison with twice the absolute series.
  3. The absolute series converges by hypothesis.
  4. A difference of two convergent series converges, so ∑an\sum a_n converges. The argument leans entirely on the absolute series being available as a bound, which is why nothing runs in the other direction.

What it does not say

Convergence implies absolute convergence.

The converse is false, and the alternating harmonic series is the standard witness. This is the point of the whole vocabulary of conditional convergence.

A conditionally convergent series has a sum you can rearrange freely.

It does not. Riemann's rearrangement theorem says a conditionally convergent series can be reordered to sum to any real number you like, or to diverge. Absolutely convergent series are the ones that behave under rearrangement.

The ratio and root tests prove ordinary convergence.

They prove absolute convergence, which is stronger, and then this theorem supplies ordinary convergence as a consequence. Worth knowing, because it means a ratio test verdict is telling you more than it appears to.

Frequently asked questions

How do I test for absolute convergence?

Apply any positive-term test to the series of absolute values: comparison, limit comparison, integral, ratio or root. If it converges, this theorem hands you convergence of the original.

What does conditionally convergent mean exactly?

The series converges but the series of absolute values does not. In practice it means the alternating structure is doing the work, and the convergence would collapse the moment the signs were removed.