Alternating series test — Leibniz test, proof: Worked examples

This page is the "throw everything at it" companion to the parent Leibniz test note. Before we compute anything, we lay out every kind of case an alternating-series problem can hand you. Then each worked example is tagged with the cell it covers, so by the end you've seen the whole board.

Reminder of the objects we use everywhere below:

The picture below turns those two knobs into a "hopping toward a flag" walk — keep it in your head for every example.

Figure — Alternating series test — Leibniz test, proof

The scenario matrix

Cell	Decreasing?	$lim b_{n}$	Verdict by Leibniz	Covered by
C1 Clean pass	Yes	$0$	Converges	Ex 1
C2 Fails "to zero"	Yes	$\neq = 0$	Diverges (nth-term test)	Ex 2
C3 Fails "decreasing"	No (wiggles)	$0$	Test does not apply	Ex 3
C4 Decreasing only eventually	Yes for large $n$	$0$	Converges	Ex 4
C5 Degenerate / zero terms	trivially	$0$	Converges (finite sum)	Ex 5
C6 Accuracy / error-bound problem	Yes	$0$	Converges + count terms	Ex 6
C7 Real-world word problem	Yes	$0$	Converges + interpret	Ex 7
C8 Exam twist (endpoint of power series)	Yes	$0$	Converges conditionally	Ex 8

Two knobs → four logical corners (C1–C4). C5 is the degenerate corner, C6–C8 are the "so what" applications. Let's fill every cell.

Worked examples

Ex 1 — Cell C1: the clean pass

Worked example Statement

Does $n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{2 n - 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ converge?

Forecast: guess before reading — decreasing? goes to zero? What's your gut verdict?

Identify $b_{n}$ . Strip the sign: $b_{n} = \frac{1}{2 n - 1}$ . Why this step? Leibniz only ever talks about the positive sizes $b_{n}$ , never the signs. We must isolate them first.
Check decreasing. $2 (n + 1) - 1 = 2 n + 1 > 2 n - 1$ , so a bigger denominator gives a smaller fraction: $b_{n + 1} < b_{n}$ . ✓ Why this step? Knob 1. Bigger bottom, smaller value — that's the whole reason it shrinks.
Check limit. As $n \to \infty$ , $2 n - 1 \to \infty$ so $\frac{1}{2 n - 1} \to 0$ . ✓ Why this step? Knob 2. Sizes must vanish for even/odd partial sums to merge.
Conclude. Converges. Why? Both knobs passed, so the "hopping toward a flag" picture applies literally: forward $1$ , back $\frac{1}{3}$ , forward $\frac{1}{5}$ … each hop smaller, so the walk is trapped in an ever-shrinking window and must settle on one point. That settling point is the sum.

Verify: This famous series equals $π /4$ (Leibniz's own formula). Numerically $\sum_{n = 1}^{5000}$ sits very close to $π /4 \approx 0.7854$ . Sanity: it's between $s_{1} = 1$ and $s_{2} = \frac{2}{3}$ , exactly as the proof predicts.

Ex 2 — Cell C2: sizes don't vanish

Worked example Statement

Does $n = 1 \sum \infty (- 1)^{n - 1} \frac{3 n}{2 n + 5}$ converge?

Forecast: the terms keep flipping sign — does flipping alone save it?

Identify $b_{n} = \frac{3 n}{2 n + 5}$ . Why? Isolate the sizes before testing.
Compute the limit. Divide top and bottom by $n$ : $\frac{3}{2 + 5/ n} \to \frac{3}{2} = 1.5 \neq = 0$ . Why this step? Knob 2 is the first thing to check when you suspect trouble — if the size doesn't die, nothing else matters.
Invoke the nth-Term Test for Divergence. Since the term size doesn't go to $0$ , the actual terms $(- 1)^{n - 1} b_{n}$ don't go to $0$ either (they oscillate around $\pm 1.5$ ). A series whose terms don't vanish cannot converge. Why this step? Leibniz is a convergence checker; when its condition 2 fails, we hand the case to the nth-term test, which gives a hard diverges.
Conclude. Diverges. Why? In the hopping picture the hops never shrink — they keep jumping by roughly $1.5$ forever, so the walker never gets trapped in a small window and can never settle. No settling point means no sum. Leibniz never even applies.

Verify: $b_{100} = \frac{300}{205} \approx 1.463$ , $b_{1000} = \frac{3000}{2005} \approx 1.496$ — climbing to $1.5$ , nowhere near $0$ . ✓

Ex 3 — Cell C3: it wiggles

Worked example Statement

Consider the alternating series $\sum (- 1)^{n - 1} b_{n}$ whose sizes are the sequence $b_{n} : 1, \frac{1}{3}, \frac{1}{2}, \frac{1}{5}, \frac{1}{4}, \frac{1}{7}, \frac{1}{6}, \dots$ (the parent note's "wiggle": the odd-position and even-position values are interleaved so the list rises and falls). Does Leibniz apply?

Forecast: the sizes clearly head to zero — is that enough for Leibniz?

Check the limit. Every value is one of $\frac{1}{k}$ for growing $k$ , so $b_{n} \to 0$ . ✓ Why? Knob 2 passes — tempting to declare victory.
Check decreasing. Read the list in order: $b_{1} = 1 > b_{2} = \frac{1}{3}$ (down), but $b_{2} = \frac{1}{3} < b_{3} = \frac{1}{2}$ (up). So it goes down then up — not monotone. Why this step? Knob 1 fails. The proof grouped terms into brackets $(b_{2 k - 1} - b_{2 k})$ and needed each $\geq 0$ . If sizes rise sometimes, a bracket can go negative and the "monotone increasing even sums" argument collapses.
Conclude. Leibniz gives no conclusion. Why? The whole "trapped in a shrinking window" guarantee relied on each hop being smaller than the last. Here a hop can be bigger than the previous one, so the window can widen again — the squeeze that forces settling is broken, and the test stays silent. (It might still converge by other tools, but this test can't say.)

Verify: For this same sequence, the rise at step 3 is $b_{3} - b_{2} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} > 0$ — a genuine increase, so monotonicity is broken exactly as claimed. ✓

Ex 4 — Cell C4: decreasing only eventually

Worked example Statement

Does $n = 1 \sum \infty (- 1)^{n - 1} \frac{ln n}{n}$ converge? (Note $b_{1} = \frac{l n 1}{1} = 0$ , $b_{2} = \frac{l n 2}{2} \approx 0.347$ , $b_{3} \approx 0.366$ — it rises at first!)

Forecast: it increases for the first couple of terms — does that kill it?

Identify $b_{n} = \frac{ln n}{n}$ , limit. By growth rates, $ln n$ crawls while $n$ races, so $\frac{ln n}{n} \to 0$ . ✓ Why? Knob 2 first.
Test decreasing with a derivative. Let $f (x) = \frac{ln x}{x}$ . Why a derivative? Because "is this shrinking?" is exactly the question "is the slope negative?" — the derivative is the tool that measures whether a smooth function goes up or down. Quotient rule: $f^{'} (x) = \frac{1 - l n x}{x ^{2}} .$ This is negative exactly when $ln x > 1$ , i.e. $x > e \approx 2.718$ . So $b_{n}$ decreases for all $n \geq 3$ . The figure below shows this: the curve climbs, peaks near $x = e$ , then falls forever. Why this step? Knob 1 needs only eventual decrease — the Leibniz proof says "for all large $n$ ." A finite pile of early terms is just a fixed number added on; it can't change convergence.
Conclude. Converges (conditionally). Why? The first two rising terms just shift the starting point by a fixed amount; from $n = 3$ onward the hops shrink monotonically to $0$ , which is exactly the shrinking-window setup that forces settling. A fixed head-start added to a convergent tail is still convergent. (Conditionally, since $\sum \frac{l n n}{n}$ diverges.)

Verify: $f^{'} (4) = \frac{1 - ln 4}{16} = \frac{1 - 1.386}{16} \approx - 0.0242 < 0$ , confirming decrease past $e$ . ✓

Ex 5 — Cell C5: degenerate / zero terms

Worked example Statement

What does the Leibniz machinery say about $n = 1 \sum \infty (- 1)^{n - 1} b_{n}$ with $b_{n} = 0$ for all $n \geq 4$ and $b_{1} = 5, b_{2} = 3, b_{3} = 1$ ?

Forecast: a "series" that's secretly finite — does the test choke on the zeros?

List the terms. $5 - 3 + 1 - 0 + 0 - 0 + \dots = 3$ . Every term past $n = 3$ contributes nothing. Why this step? A tail of exact zeros means the partial sums stop moving — degenerate but perfectly legal.
Check the two knobs. $b_{n}$ : $5, 3, 1, 0, 0, \dots$ is (weakly) decreasing since $b_{n + 1} \leq b_{n}$ throughout (equalities allowed — the test uses $\leq$ , not $<$ ). ✓ And $b_{n} \to 0$ (it's literally $0$ ). ✓ Why this step? This is the corner case that shows the $\leq$ in the definition is deliberate — flat/zero pieces are fine.
Conclude. Converges, and here we read the sum off exactly: $s = 3$ . Why? When the hops become exactly $0$ , the walker has literally stopped — there is nothing left to settle, it is already sitting on its final spot. A walk that halts is the most extreme case of "settles."

Verify: $s_{3} = 5 - 3 + 1 = 3$ , and $s_{N} = 3$ for all $N \geq 3$ since later terms are $0$ . Error bound $∣ s - s_{3} ∣ \leq b_{4} = 0$ — perfectly tight. ✓

Ex 6 — Cell C6: accuracy / error-bound problem

Worked example Statement

For $n = 1 \sum \infty \frac{( - 1 ) ^{n - 1}}{n ^{3}}$ , how many terms guarantee an error below $1 0^{- 3}$ ?

Forecast: cube in the denominator shrinks fast — guess whether you need dozens or just a handful of terms.

Confirm convergence quickly. $b_{n} = \frac{1}{n ^{3}}$ decreases and $\to 0$ ⟹ converges. ✓ Why? You may only use the error bound once the test's conditions hold.
Write the error bound. $∣ s - s_{N} ∣ \leq b_{N + 1} = \frac{1}{( N + 1 ) ^{3}}$ . Why this step? The proof gave this free — the true sum is trapped between consecutive partial sums, so the miss is at most the next hop.
Solve the inequality. Want $\frac{1}{( N + 1 ) ^{3}} < 1 0^{- 3}$ , i.e. $(N + 1)^{3} > 1000$ , so $N + 1 > 10$ , giving $N \geq 10$ . Why this step? We invert the bound to find the smallest $N$ that squeezes the error under target.
Conclude. Summing the first $10$ terms is enough. Why exactly 10? Because $b_{11}$ is the first unused hop, and we forced it below the tolerance; the trapped true sum can't be farther than that hop, so 10 terms already pin the answer to the required precision. (Contrast Ex 2 of the parent, where $1/ n$ needed 200 terms — cubing shrinks far faster.)

Verify: $b_{11} = \frac{1}{1 1 ^{3}} = \frac{1}{1331} \approx 7.51 \times 1 0^{- 4} < 1 0^{- 3}$ ✓, while $b_{10} = \frac{1}{1000} = 1 0^{- 3}$ is not strictly below — so $N = 10$ is exactly the cutoff.

Ex 7 — Cell C7: real-world word problem

Worked example Statement

A ball is dropped and bounces. It loses energy so each bounce height is smaller. A physicist models the signed displacement contributions of a damped oscillation as $\sum_{n = 1}^{\infty} (- 1)^{n - 1} \frac{1}{n n}$ metres (up = $+$ , down = $-$ ). Does the net displacement settle to a finite value, and within what error after 8 bounces?

Forecast: damped, shrinking, sign-flipping — does the ball settle at a finite spot?

Identify $b_{n} = \frac{1}{n n} = n^{- 3/2}$ . Why? Physical "size of the $n$ -th nudge," always positive.
Check knobs. As $n$ grows, $n^{3/2}$ grows, so $b_{n}$ decreases and $\to 0$ . ✓✓ Why? Both required for the "zeroing in" behaviour that makes a settling physical answer meaningful.
Conclude convergence. The displacement settles to a finite $s$ . Why? The shrinking, sign-flipping nudges are exactly the hopping-toward-a-flag walk; each nudge is smaller than the last, so the ball is trapped in an ever-tightening interval and must approach one resting displacement — that is what "settles" means physically.
Error after 8 terms. $∣ s - s_{8} ∣ \leq b_{9} = 9^{- 3/2} = \frac{1}{27} \approx 0.037$ m. Why? The remainder bound turns "how sure am I?" into an actual distance in metres — note the units survive: $b_{9}$ is in metres, so the error is in metres.

Verify: $9^{3/2} = (9)^{3} = 3^{3} = 27$ , so $b_{9} = 1/27 \approx 0.0370$ m. ✓ Units check: metres in, metres out. ✓

Ex 8 — Cell C8: exam twist (power-series endpoint)

Worked example Statement

The power series $n = 1 \sum \infty \frac{x ^{n}}{n}$ has radius of convergence $1$ . At the endpoint $x = - 1$ , does it converge?

Forecast: the ratio test can't decide endpoints — which test rescues us here?

Substitute $x = - 1$ . Each term becomes $\frac{( - 1 ) ^{n}}{n}$ , so the series is $n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n} = - 1 + \frac{1}{2} - \frac{1}{3} + \dots$ . Why this step? Endpoints must be tested by hand — see Power Series — radius of convergence; the radius test is silent exactly on the boundary.
Recognise the alternating shape. Factor out the sign: this is $- \sum (- 1)^{n - 1} \frac{1}{n}$ , an alternating series with $b_{n} = \frac{1}{n}$ . Why? An overall minus sign doesn't affect convergence; we just need the alternating skeleton.
Apply Leibniz. $b_{n} = \frac{1}{n}$ decreases and $\to 0$ ⟹ converges. Why this step? Leibniz is the standard tool for deciding a conditionally convergent endpoint — compare the Harmonic Series at $x = + 1$ , which diverges.
Conclude. Converges at $x = - 1$ (conditionally), diverges at $x = + 1$ , so the interval of convergence is $[- 1, 1)$ . Why the asymmetry? At $x = - 1$ the signs alternate and the shrinking-hop squeeze kicks in; at $x = + 1$ all terms are positive $\frac{1}{n}$ — no alternation, no squeeze, and the harmonic series marches off to infinity.

Verify: At $x = - 1$ the sum is $- ln 2 \approx - 0.6931$ ; partial sums bracket it, e.g. $s_{2} = - 1 + 0.5 = - 0.5$ and $s_{3} \approx - 0.833$ , straddling $- 0.6931$ . ✓ At $x = + 1$ it's the divergent harmonic series. ✓

Recall Quick self-test on the matrix

Which two knobs decide every Leibniz case? ::: Is $b_{n}$ decreasing (eventually), and does $b_{n} \to 0$ . If $b_{n} \to 0$ but wiggles, what's the verdict? ::: Leibniz is silent — try another method. If $b_{n} \neq \to 0$ , what's the verdict? ::: Diverges by the nth-term test; Leibniz never applies. Why does "decreasing only for large $n$ " still work? ::: A finite head of terms is a fixed constant; convergence depends only on the tail. How do you turn an accuracy demand into a term count? ::: Solve $b_{N + 1} < tolerance$ using the remainder bound.

Connections

Parent: Leibniz test & proof
Convergence of Series — overview
Absolute vs Conditional Convergence
Monotone Convergence Theorem
nth-Term Test for Divergence
Harmonic Series
Power Series — radius of convergence

Worked examples — Alternating series test — Leibniz test, proof

The scenario matrix

Worked examples

Ex 1 — Cell C1: the clean pass

Ex 2 — Cell C2: sizes don't vanish

Ex 3 — Cell C3: it wiggles

Ex 4 — Cell C4: decreasing only eventually

Ex 5 — Cell C5: degenerate / zero terms

Ex 6 — Cell C6: accuracy / error-bound problem

Ex 7 — Cell C7: real-world word problem

Ex 8 — Cell C8: exam twist (power-series endpoint)

Connections

Case Map