Absolute vs conditional convergence: Worked examples

This page is the drill. The parent note gave you the theory; here we walk through every kind of series that the topic can throw at you, one worked example per case. Each example makes you guess first, then reasons step by step.

Before starting, recall the single question that runs through everything:

The engine behind cells A, D, F, G, H is one named result — let us state it plainly so it is never invoked as a mystery:

The other engine — used whenever the twin fails — is the Alternating Series Test. Since we lean on it in cells B, C and I, let us state it in full so it is never a black box:

The scenario matrix

Every problem in this topic lands in one of these classes. The examples below are labelled with the class they hit, so by the end you will have seen them all.

#	Case class	What makes it tricky	Example
A	Alternating, twin is a $p$ -series with $p > 1$	signs irrelevant → absolute	Ex 1
B	Alternating, twin is the harmonic series ( $p = 1$ )	classic conditional trap	Ex 2
C	Alternating, twin is $p$ -series with $0 < p < 1$	twin diverges, but AST still saves it	Ex 3
D	Geometric with a sign, ratio test	quick absolute verdict	Ex 4
E	Terms don't go to $0$	fails at the very first gate → divergent	Ex 5
F	Zero / degenerate terms mixed in	do zeros break anything?	Ex 6
G	Non-obvious signs ( $sin n$ , not pure $(- 1)^{n}$ )	absolute test via Comparison Test	Ex 7
H	Word problem (oscillating physical sum)	translate story → series	Ex 8
I	Exam twist: value $x$ inside a [[Power series & radius of convergence	power series]], behaviour at the endpoint	absolute vs conditional at the boundary

We take them in order.

Case A — alternating, twin converges (absolute)

Worked example Ex 1 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{3}}$ Forecast: guess now — absolutely convergent, conditionally convergent, or divergent?

Step 1. Form the all-positive twin: $∣ a_{n} ∣ = \frac{( - 1 ) ^{n}}{n ^{3}} = \frac{1}{n ^{3}}$ . Why this step? Absolute convergence is defined by the twin $\sum ∣ a_{n} ∣$ , so we build it first.

Step 2. Recognise $\sum \frac{1}{n ^{3}}$ as a p-series with $p = 3$ . Why this step? $p$ -series is the fastest classifier we have: it converges exactly when $p > 1$ .

Step 3. Since $p = 3 > 1$ , the twin converges. By the Absolute Convergence Theorem (stated above) the original converges too. Why this step? Once the twin converges, that theorem hands us convergence of $\sum a_{n}$ for free — no Alternating Series Test needed.

Verdict: absolutely convergent (cell A).

Verify: partial sum $\sum_{n = 1}^{6} \frac{( - 1 ) ^{n}}{n ^{3}} \approx - 0.8971$ , and the true value is $- \frac{3}{4} ζ (3) \approx - 0.9015$ — the partial sums are already hugging the limit, exactly what robust convergence looks like. (Here $ζ (3) = \sum_{n \geq 1} \frac{1}{n ^{3}}$ is just shorthand for "the value of that $p = 3$ sum"; you don't need any theory of the zeta function, only that it names a finite number $\approx 1.202$ .)

Case B — alternating, twin is harmonic (conditional trap)

Worked example Ex 2 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n}$ Forecast: absolute, conditional, or divergent? (This is the model conditional series — commit to a guess.)

Step 1. Twin: $∣ a_{n} ∣ = \frac{1}{n}$ , so $\sum ∣ a_{n} ∣ = \sum \frac{1}{n}$ — the Harmonic series. Why this step? Always test the twin first.

Step 2. The harmonic series is a $p$ -series with $p = 1$ , which diverges. So the series is not absolutely convergent. Why this step? $p = 1$ is the exact borderline where $p$ -series stops converging.

Step 3. Now test the original with the Alternating Series Test. Writing $b_{n} = \frac{1}{n}$ : (i) positive, (ii) decreasing ( $\frac{1}{n + 1} < \frac{1}{n}$ ), (iii) $\to 0$ . All three AST conditions hold, so $\sum a_{n}$ converges. Why this step? The twin diverged, so we fall to the AST branch of the recipe.

Verdict: conditionally convergent (cell B). It equals $ln 2$ .

Verify: $\sum_{n = 1}^{8} \frac{( - 1 ) ^{n + 1}}{n} \approx 0.6345$ vs $ln 2 \approx 0.6931$ — converging, but slowly (the partial sums crawl), the fingerprint of fragile convergence.

Case C — alternating, twin is $p < 1$ (still conditional)

Worked example Ex 3 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{n}$ Forecast: the twin diverges harder than the harmonic series here — does the original still converge?

Step 1. Twin: $∣ a_{n} ∣ = \frac{1}{n} = \frac{1}{n ^{1/2}}$ , a p-series with $p = \frac{1}{2}$ . Why this step? Build the twin, classify by $p$ .

Step 2. $p = \frac{1}{2} < 1$ , so the twin diverges. Not absolutely convergent. Why this step? Any $p \leq 1$ makes a $p$ -series diverge.

Step 3. Apply the Alternating Series Test to the original with $b_{n} = \frac{1}{n}$ : positive ✓, decreasing ✓ (bigger $n$ → bigger $n$ → smaller reciprocal), $\to 0$ ✓. So $\sum a_{n}$ converges. Why this step? Twin failed; AST is the fallback and all three conditions pass.

Verdict: conditionally convergent (cell C).

Verify: $\sum_{n = 1}^{10} \frac{( - 1 ) ^{n + 1}}{n} \approx 0.6068$ ; the true value is $\approx 0.6049$ . The AST error bound (from the theorem above) says the error is $\leq$ the first omitted term $\frac{1}{11} \approx 0.3015$ — and indeed $∣0.6068 - 0.6049∣ \approx 0.0019$ is well within it.

Case D — geometric, ratio test (absolute)

Worked example Ex 4 —

$n = 1 \sum \infty \frac{( - 3 ) ^{n}}{4 ^{n}}$ Forecast: ratio $\frac{- 3}{4} = \frac{3}{4} < 1$ — what does that tell you?

Step 1. Twin: $∣ a_{n} ∣ = \frac{3 ^{n}}{4 ^{n}} = (\frac{3}{4})^{n}$ . Why this step? We want absolute convergence, so strip the sign first.

Step 2. Apply the Ratio Test to the twin: $L = lim_{n \to \infty} \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = lim_{n \to \infty} \frac{( 3/4 ) ^{n + 1}}{( 3/4 ) ^{n}} = \frac{3}{4} .$ Why this step? The Ratio Test on $∣ a_{n} ∣$ directly tests absolute convergence — and it's the natural tool when terms are powers.

Step 3. Since $L = \frac{3}{4} < 1$ , the twin converges → the series is absolutely convergent. Why this step? Ratio test rule: $L < 1 \Rightarrow$ convergence of $\sum ∣ a_{n} ∣$ .

Verdict: absolutely convergent (cell D). As a geometric series with first term $- \frac{3}{4}$ and ratio $- \frac{3}{4}$ , the sum is $\frac{- 3/4}{1 - ( - 3/4 )} = - \frac{3}{7}$ .

Verify: $\frac{- 3/4}{1 + 3/4} = \frac{- 3/4}{7/4} = - \frac{3}{7} \approx - 0.4286$ , and $\sum_{n = 1}^{12} \approx - 0.4286$ already. ✓

Case E — terms don't reach zero (divergent)

Worked example Ex 5 —

$n = 1 \sum \infty (- 1)^{n} \frac{n}{n + 1}$ Forecast: the signs alternate nicely — tempting to say "conditional". Is it a trap?

Step 1. Look at the magnitude: $∣ a_{n} ∣ = \frac{n}{n + 1}$ . As $n \to \infty$ , this $\to 1$ , not $0$ . Why this step? Before any convergence test, the terms of any convergent series must go to $0$ — the term test / divergence test. Check this gate first.

Step 2. Because $a_{n}$ oscillates between values near $+ 1$ and $- 1$ and does not approach $0$ , the series diverges. Why this step? If $a_{n} \neq \to 0$ , the partial sums cannot settle. This kills the series before AST or the twin are even relevant.

Step 3. Do not attempt AST here — AST requires magnitudes $\to 0$ (condition 3), which fails, so AST does not apply. Why this step? Reaching for AST when its hypothesis fails is a classic error; the divergence test already gave the verdict.

Verdict: divergent (cell E) — the third outcome in the recipe: neither absolutely nor conditionally convergent, it simply has no finite sum.

Verify: the odd partial sums sit near $- 0.5$ and even ones near $0$ ; $a_{100} = \frac{100}{101} \approx 0.990 \neq = 0$ , confirming terms don't vanish.

Case F — degenerate / zero terms (does convergence survive?)

Worked example Ex 6 —

$n = 1 \sum \infty a_{n}, a_{n} = ⎩ ⎨ ⎧ \frac{( - 1 ) ^{n + 1}}{n ^{2}} 0 n odd n even$ Forecast: half the terms are literally $0$ . Does inserting zeros change convergence or the type?

Step 1. Zeros contribute nothing to any partial sum. So $\sum a_{n}$ has the same partial sums (up to constant padding) as $\sum_{k odd} \frac{1}{k ^{2}}$ , a sub-sum of positive terms. Why this step? Adding a term equal to $0$ never changes a running total — degenerate terms are invisible.

Step 2. Twin: $\sum ∣ a_{n} ∣ = \sum_{k odd} \frac{1}{k ^{2}} \leq \sum_{n = 1}^{\infty} \frac{1}{n ^{2}}$ , which converges ( $p = 2$ p-series). By the Comparison Test, the twin converges. Why this step? Bounding a positive series by a known convergent one proves the twin converges.

Step 3. Twin converges ⇒ absolutely convergent (Absolute Convergence Theorem). The zeros neither help nor hurt. Why this step? Absolute convergence is decided entirely by the twin, and $0$ 's twin is $0$ .

Verdict: absolutely convergent (cell F).

Verify: $\sum_{k odd, k \leq 9} \frac{1}{k ^{2}} = 1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81} \approx 1.1838$ ; the full odd-square sum is $\frac{π ^{2}}{8} \approx 1.2337$ — bounded and converging. ✓

Case G — non-obvious signs (comparison for absolute)

Worked example Ex 7 —

$n = 1 \sum \infty \frac{sin n}{n ^{2}}$ Forecast: the sign of $sin n$ jumps around unpredictably — not a clean $(- 1)^{n}$ . Can we still decide?

Step 1. Twin: $∣ a_{n} ∣ = \frac{∣ sin n ∣}{n ^{2}}$ . Use the bound $∣ sin n ∣ \leq 1$ for all $n$ . Why this step? $sin$ is trapped in $[- 1, 1]$ , so its absolute value never exceeds $1$ — this gives a clean over-estimate.

Step 2. Therefore $\frac{∣ sin n ∣}{n ^{2}} \leq \frac{1}{n ^{2}}$ . Since $\sum \frac{1}{n ^{2}}$ converges ( $p = 2$ ), the Comparison Test forces $\sum ∣ a_{n} ∣$ to converge. Why this step? When the signs are messy, the AST is useless — but the twin is bounded by a nice convergent series, so we go straight for absolute convergence.

Step 3. Twin converges ⇒ the original converges absolutely. Why this step? Absolute Convergence Theorem applies once the twin is settled.

Verdict: absolutely convergent (cell G).

Verify: $\sum_{n = 1}^{20} \frac{s i n n}{n ^{2}} \approx 1.0121$ ; every partial sum stays within the bound $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6} \approx 1.6449$ . ✓

Case H — word problem (oscillating physical sum)

A robot on a number line starts at $0$ . On step $n$ it moves a distance $\frac{1}{2 ^{n}}$ metres, but flips direction each step: right, left, right, left, … Where does it end up, and is that final position "robust"?

Figure — Absolute vs conditional convergence

The figure above is the derivation drawn out: read it left-to-right with the steps below. Each coloured arrow is one term $\frac{( - 1 ) ^{n + 1}}{2 ^{n}}$ — the magenta arrow is step 1 (right, length $\frac{1}{2}$ ), the violet arrow step 2 (left, length $\frac{1}{4}$ ), the orange arrow step 3 (right, length $\frac{1}{8}$ ), and so on. Notice two things the picture makes obvious: (a) each arrow is exactly half the length of the one before, and (b) the tips march closer and closer to the dashed navy line at $x = \frac{1}{3}$ (the red dot) without ever crossing far past it. That "closing in" is convergence made visual.

Worked example Ex 8 — total displacement

$n = 1 \sum \infty \frac{( - 1 ) ^{n + 1}}{2 ^{n}}$ Forecast: the robot zig-zags with ever-smaller steps. Absolute, conditional, or divergent displacement?

Step 1. Model: rightward is $+$ , leftward is $-$ , step size $\frac{1}{2 ^{n}}$ . Total displacement $= \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{2 ^{n}}$ . This is exactly the arrow chain in the figure — trace the magenta → violet → orange arrows and watch them shrink toward the red dot. Why this step? Translating "zig-zag of halving steps" into signs and magnitudes is the whole modelling job.

Step 2. Twin: $∣ a_{n} ∣ = \frac{1}{2 ^{n}} = (\frac{1}{2})^{n}$ , a geometric series with ratio $\frac{1}{2} < 1$ → converges (the total path length is finite, $\sum \frac{1}{2 ^{n}} = 1$ ). Why this step? If even the total distance travelled (ignoring direction) is finite, the position is robustly pinned down.

Step 3. Twin converges ⇒ absolutely convergent (Absolute Convergence Theorem). Sum $= \frac{1/2}{1 - ( - 1/2 )} = \frac{1/2}{3/2} = \frac{1}{3}$ . Why this step? Geometric-series formula with first term $\frac{1}{2}$ , ratio $- \frac{1}{2}$ .

Verdict: absolutely convergent (cell H). The robot settles at $x = \frac{1}{3}$ m — precisely the red dot in the figure — and this position is order-independent: you could take the steps in any order and land at $\frac{1}{3}$ .

Verify: $\sum_{n = 1}^{10} \frac{( - 1 ) ^{n + 1}}{2 ^{n}} \approx 0.3330 \to \frac{1}{3}$ , and total distance $\sum_{n = 1}^{10} \frac{1}{2 ^{n}} \approx 0.999 \to 1$ . ✓ (Units: metres throughout.)

Case I — exam twist: endpoint of a power series

Worked example Ex 9 — For

$n = 1 \sum \infty \frac{x ^{n}}{n}$ , classify convergence at the two endpoints $x = 1$ and $x = - 1$ . Forecast: the power series has radius $1$ . The interesting drama is always at the endpoints — guess the type at each end.

Step 1. At $x = 1$ : the series becomes $\sum \frac{1 ^{n}}{n} = \sum \frac{1}{n}$ , the Harmonic series, which diverges. Why this step? Plug the endpoint value in; here it collapses to a pure $p = 1$ series.

Step 2. So at $x = 1$ the series is divergent — neither absolute nor conditional. (Right endpoint is excluded.) Why this step? Divergence at an endpoint means that endpoint is not part of the interval of convergence.

Step 3. At $x = - 1$ : the series becomes $\sum \frac{( - 1 ) ^{n}}{n}$ . Twin $= \sum \frac{1}{n}$ diverges (not absolute), but by the Alternating Series Test (with $b_{n} = \frac{1}{n}$ decreasing $\to 0$ ) the series converges. Why this step? The same endpoint machinery, but now the alternating signs rescue it.

Verdict: at $x = 1$ divergent; at $x = - 1$ conditionally convergent (cell I). The identical magnitude series can be divergent or conditionally convergent depending purely on the sign the endpoint supplies — the essence of this whole topic.

Verify: at $x = - 1$ , $\sum_{n = 1}^{8} \frac{( - 1 ) ^{n}}{n} \approx - 0.6345 \to - ln 2 \approx - 0.6931$ ; at $x = 1$ the partials $\sum_{n = 1}^{8} \frac{1}{n} \approx 2.7179$ keep climbing (no limit). ✓

Recall Which cell was which? (self-quiz)

Twin is $p$ -series with $p > 1$ ::: absolutely convergent (cells A, F, G) Twin is harmonic, original alternates ::: conditionally convergent (cells B, I at $x = - 1$ ) Twin is $p$ -series with $0 < p < 1$ , original alternates ::: conditionally convergent (cell C) Terms fail to reach $0$ ::: divergent — stop immediately (cell E) Geometric with ratio magnitude $< 1$ ::: absolutely convergent (cells D, H) Endpoint $x = 1$ of $\sum x^{n} / n$ ::: divergent (cell I)

Connections

4.3.11 Absolute vs conditional convergence — the parent theory
Alternating Series Test — rescued cells B, C, I
Comparison Test — decided cells F, G
Ratio Test — decided cell D
p-series — the universal classifier
Harmonic series — the conditional villain
Power series & radius of convergence — endpoint drama in cell I
Riemann Rearrangement Theorem — why the absolute/conditional split matters