Absolute vs conditional convergence: Question bank

Before we start, one anchor for every question. We are always looking at two objects at once:

True or false — justify

Every absolutely convergent series is convergent.

True. This is the Absolute Convergence Theorem:

\sum ∣ a_{n} ∣

converging forces

\sum a_{n}

to converge, because cancellation can only help a sum that already converges with all-positive terms.

Every convergent series is absolutely convergent.

False. The alternating harmonic series

\sum (- 1)^{n + 1} / n

converges (to

ln 2

) yet

\sum 1/ n

diverges, so it converges without converging absolutely.

\sum ∣ a_{n} ∣

diverges, then

\sum a_{n}

must diverge.

False. Divergence of the absolute series only rules out absolute convergence; the signed series can still converge conditionally, exactly as

\sum (- 1)^{n + 1} / n

does.

A conditionally convergent series is one that "barely" or "slowly" converges.

False. It converges fully to a genuine finite real number; "conditional" means the absolute series diverges, not that the signed limit is shaky or partial.

\sum a_{n}

converges and every term is positive, it is absolutely convergent.

True. With all

a_{n} \geq 0

we have

∣ a_{n} ∣ = a_{n}

, so the absolute series is the original series and converges too — a positive-term convergent series is automatically absolute.

\sum a_{n}

converges absolutely, then

\sum a_{n}^{2}

converges.

True. Absolute convergence forces

∣ a_{n} ∣ \to 0

, so eventually

∣ a_{n} ∣ < 1

and

a_{n}^{2} \leq ∣ a_{n} ∣

; comparison with the convergent

\sum ∣ a_{n} ∣

finishes it (see Comparison Test).

For a conditionally convergent series, rearranging the terms never changes the sum.

False. Riemann's Rearrangement Theorem says you can reorder a conditionally convergent series to reach any real number or

\pm \infty

— the value depends on the order (see Riemann Rearrangement Theorem).

For an absolutely convergent series, any rearrangement gives the same sum.

True. Absolute convergence makes the series behave like a finite sum, so order is irrelevant and every rearrangement lands on the identical limit.

\sum (- 1)^{n} / n^{2}

is conditionally convergent because it alternates.

False. Alternating signs do not decide the type; here

\sum 1/ n^{2}

(a p-series with

p = 2 > 1

) converges, so it is absolutely convergent.

If a power series converges at a point strictly inside its interval of convergence, it converges absolutely there.

True. Inside the radius of convergence convergence is always absolute; only at the endpoints can convergence turn merely conditional.

Why questions

Why does absolute convergence imply ordinary convergence, intuitively?

Stripping signs makes every partial sum as large as possible; if even that all-positive sum settles down, the signed sum — where cancellation only pulls terms closer together — must settle too.

Why do we test

\sum ∣ a_{n} ∣

first in the decision recipe?

If it converges we are instantly done with the strongest verdict (absolute) and skip all sign-based tests; only if it diverges do we bother with the more delicate Alternating Series Test.

Why can't the Comparison Test be applied directly to a signed series like

\sum (- 1)^{n} / n

Comparison requires non-negative terms so the inequalities point consistently; that is exactly why the proof of the Absolute Convergence Theorem first builds the non-negative quantity

a_{n} + ∣ a_{n} ∣

Why does the Ratio Test automatically test absolute convergence?

It works with

∣ a_{n + 1} ∣/∣ a_{n} ∣

, ratios of magnitudes, so when

L < 1

it is

\sum ∣ a_{n} ∣

that is shown to converge — hence the verdict is absolute, not merely conditional.

Why do mathematicians prize absolute over conditional convergence?

Only for absolute convergence is "the sum" a well-defined number independent of term order; conditional sums are artifacts of ordering, as Riemann's theorem dramatically shows.

Why must a conditionally convergent series have infinitely many positive and infinitely many negative terms?

If only finitely many terms had one sign, the tail would be all one sign, making

\sum a_{n}

and

\sum ∣ a_{n} ∣

agree in the tail — so convergence of one would force the other, contradicting "conditional".

Edge cases

Is a finite sum (only finitely many nonzero terms) absolutely convergent?

Yes. Finitely many terms always sum to a finite number with signs stripped, so both

\sum a_{n}

and

\sum ∣ a_{n} ∣

trivially converge — it is absolutely convergent.

Can a series with

a_{n} \geq 0

ever be conditionally convergent?

No. With no negative terms

\sum a_{n} = \sum ∣ a_{n} ∣

, so if the series converges it converges absolutely; conditional convergence needs sign cancellation that positive terms can't provide.

\sum (- 1)^{n}

(terms

- 1, 1, - 1, \dots

) conditionally convergent?

No. The terms do not tend to zero, so the signed series diverges outright — it fails the necessary condition and is neither absolutely nor conditionally convergent.

What is the classification of

\sum 0

(every term zero)?

Absolutely convergent (to

0

). The absolute series is also all zeros and converges, so it sits firmly in the absolute box — a trivial but valid case.

At the endpoints of a power series' interval of convergence, which convergence types are possible?

Any of the three: it may converge absolutely, converge only conditionally, or diverge — each endpoint must be tested separately (see Power series & radius of convergence).

Does the

p

-series

\sum 1/ n^{p}

ever converge conditionally?

No. Its terms are already positive, so convergence (which happens exactly when

p > 1

) is always absolute; there are no signs to cancel, so conditional convergence is impossible here.

Recall One-line self-check

If you can state, for any series thrown at you, which of the two series (signed or absolute) each test is actually examining, you will never misclassify. The signed series is tested by ::: the Alternating Series Test (and the definition of convergence itself). The absolute series is tested by ::: the Ratio Test, Comparison Test, and $p$ -series applied to $∣ a_{n} ∣$ .

Connections

Alternating Series Test — the only tool that legitimately certifies the conditional case
Comparison Test — powers most absolute-convergence checks
Ratio Test — tests $∣ a_{n} ∣$ , so its verdict is always about absolute convergence
p-series · Harmonic series — the benchmark examples ( $p > 1$ vs $p = 1$ )
Riemann Rearrangement Theorem — why conditional sums depend on order
Power series & radius of convergence — where endpoint conditional convergence appears

Question bank — Absolute vs conditional convergence