Series & Sequences · Convergence Tests

Alternating
Series

Signs that flip, partial sums that zigzag, and a beautifully simple test that pins down convergence, all with an error bound built in.

Leibniz Test AST Estimation Theorem Absolute Convergence Conditional Convergence Rearrangement Theorem
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  1. What Is an Alternating Series?
  2. The Alternating Series Test (Leibniz Test)
  3. The Alternating Series Estimation Theorem
  4. Interactive Partial-Sum Visualiser
  5. Absolute vs Conditional Convergence
  6. The Riemann Rearrangement Theorem
  7. Convergence Test Strategy
  8. Worked Examples
  9. Common Mistakes
  10. 10-Question Quiz
  11. Continue in Series & Sequences

§ 01What Is an Alternating Series?

An alternating series is one whose terms change sign with every step: positive, then negative, then positive, and so on forever. That relentless oscillation is both the challenge and the key.

A series alternates when consecutive terms have opposite signs. The standard form uses (−1)n or (−1)n+1 as a sign-flipping factor:

Standard Forms — Alternating Series
Σ (−1)n bn  = − b1 + b2 − b3 + ··· Σ (−1)n+1 bn = + b1 − b2 + b3 − ··· where   bn > 0   for all n

The (−1)n factor simply flips sign; the sequence {bn} of absolute values is what determines convergence behaviour. The two forms differ only in whether the first term is positive or negative: this affects the sign of the partial sums but not convergence.

Canonical examples

The three most famous alternating series are:

The Alternating Harmonic Series
Σn=1∞ (−1)n+1/n = 1 − 1/2 + 1/3 − 1/4 + ··· = ln 2
Leibniz's Formula for π/4
Σn=0∞ (−1)n/(2n+1) = 1 − 1/3 + 1/5 − 1/7 + ··· = π/4
Alternating p-Series
Σn=1∞ (−1)n+1/np   (converges for p > 0)

The alternating harmonic series is especially important because it illustrates a subtle point: the series converges (to ln 2), yet the corresponding series of absolute values (the ordinary harmonic series Σ 1/n) diverges. This distinction between absolute and conditional convergence is one of the core themes of this page.

Geometric picture Think of the partial sums Sn as a player walking on a number line. They take a long step right (+b1), then a shorter step left (−b2), then a shorter step right (+b3), and so on. If the step sizes decrease to zero, the player is trapped in a narrowing interval and must converge to a limit. If the steps don't shrink, the player keeps oscillating and never settles.

§ 02The Alternating Series Test (Leibniz Test)

The Alternating Series Test (AST), also called the Leibniz Test, gives a clean sufficient condition for convergence. It requires only two things about the sequence {bn}:

Alternating Series Test — Statement
If: (1) bn > 0   for all n       (2) bn+1 ≤ bn   (bn is eventually non-increasing)       (3) limn→∞ bn = 0 Then: Σ (−1)n bn converges.

These three conditions are often compressed into: "the terms are eventually positive, decreasing, and tend to zero." All three are necessary for the test to apply, though failing one of them does not immediately mean the series diverges (except for condition 3, which is the Divergence Test).

Proof sketch — the nesting interval argument

Group partial sums in pairs. The even-indexed partial sums S2n form an increasing sequence, the odd-indexed partial sums S2n+1 form a decreasing sequence, and every even partial sum lies below every odd partial sum. Both sequences are bounded and monotone, so both converge. Since bn → 0, the two limits must be equal, so the full sequence Sn converges to their common value.

A rigorous insight from the proof
S2 ≤ S4 ≤ S6 ≤ ··· ≤ S ≤ ··· ≤ S5 ≤ S3 ≤ S1

The series sum S is sandwiched between the even and odd partial sums at every step. This nesting is precisely what the Estimation Theorem (§03) exploits.

Applying the test — a step-by-step method

  • Step 1. Identify bn: the absolute value of the general term.
  • Step 2. Check limn→∞ bn. If the limit is not 0, the series diverges by the Divergence Test (stop here).
  • Step 3. Check that {bn} is eventually non-increasing. You can do this by showing bn+1/bn ≤ 1 or bn+1 ≤ bn algebraically, or by showing that the function f(x) corresponding to bn has f′(x) ≤ 0 for large x.
  • Step 4. If both conditions are satisfied, conclude convergence by the AST.
The AST does not test for absolute convergence Passing the AST tells you the series converges, but it says nothing about whether the convergence is absolute or conditional. You must separately test Σ |an| to determine which type.

§ 03The Alternating Series Estimation Theorem

One of the most practical results in all of series theory is the error bound that comes "for free" with the AST. Because the partial sums alternate above and below the true sum, the error from stopping after n terms is controlled by the very next term:

Alternating Series Estimation Theorem (ASET)
If Σ (−1)n bn converges by the AST, and Sn is the nth partial sum, then: |S − Sn| ≤ bn+1

The error in using Sn to approximate the true sum S is no larger than the first omitted term bn+1. Furthermore, the true sum S lies between any two consecutive partial sums:

True Sum is Bracketed
Sn+1 ≤ S ≤ Sn   (or   Sn ≤ S ≤ Sn+1 depending on parity of n)

This error bound is extraordinarily useful: to estimate S to within a given tolerance ε, you simply find the first n such that bn+1 < ε, then compute Sn. No other convergence test offers this kind of automatic error control.

How many terms are needed?

A common exam question is: "how many terms of Σ (−1)n+1/n are needed to approximate ln 2 to within 0.01?" The answer: find n such that bn+1 = 1/(n+1) < 0.01, i.e., n+1 > 100, so n ≥ 100. At least 100 terms are required, a reminder that conditionally convergent series can converge very slowly.

The error has a sign too The ASET also tells you the sign of the error. Since Sn and Sn+1 straddle S, if Sn > Sn+1 (the n-th term was positive), then Sn > S and Sn+1 < S. So partial sums with the last term positive overestimate, and partial sums with the last term negative underestimate.

§ 04Interactive Partial-Sum Visualiser

The chart below plots the partial sums Sn of the series Σ (−1)n+1 / np for the first N terms. Watch the zigzag pattern and see how the partial sums close in on the true sum from alternating sides. The gold dashed line marks the true sum (for p=1 this is ln 2 ≈ 0.6931).

Interactive · Partial Sum Convergence

Σ (−1)n+1 / np — partial sums Sn

SN (last partial sum)
—
|SN − SN−1| = bN
—
Error bound bN+1
—
Known sum (p=1)
ln 2

Observe: for larger p the series converges faster (fewer terms needed to reach a given accuracy), because the terms bn = 1/np shrink more quickly. The factorial series converges so fast it reaches its limit almost immediately.

§ 05Absolute vs Conditional Convergence

There are two distinct ways a series can converge. Understanding the difference is essential for working with power series, Taylor series, and rearrangements.

Definition — Absolute Convergence
Σ an converges absolutely if Σ |an| converges.
Definition — Conditional Convergence
Σ an converges conditionally if Σ an converges but Σ |an| diverges.

The key theorem: absolute implies convergent

If a series converges absolutely, it also converges in the ordinary sense. The converse is false, a series can converge without converging absolutely (the alternating harmonic series is the canonical counterexample).

Theorem
Absolute convergence  ⟹  (ordinary) convergence Convergence  ⟹̸  absolute convergence

How to classify a series

Given an alternating series Σ an, the classification procedure is:

  • Test Σ |an| using any convergence test (ratio test, p-series test, comparison, etc.).
  • If Σ |an| converges → the original series converges absolutely.
  • If Σ |an| diverges → apply the AST to Σ an. If it converges → conditional convergence. If it diverges → diverges altogether.
Σ |aₙ|Σ aₙ (via AST)Conclusion
Converges — Absolutely convergent (and therefore convergent)
Diverges Converges (AST) Conditionally convergent
Diverges Diverges Divergent

Why does the distinction matter?

Absolute convergence is a much stronger property. An absolutely convergent series behaves like a finite sum in almost every way: you can rearrange its terms freely, multiply it by another series term-by-term, and the result is always the same. A conditionally convergent series loses these guarantees: which brings us to one of the most surprising theorems in mathematics.

§ 06The Riemann Rearrangement Theorem

Riemann proved in 1853 that a conditionally convergent series can be rearranged to converge to any real number: or even to diverge to ±∞. This is not a paradox; it is a precise theorem about the fragility of conditional convergence.

Riemann Rearrangement Theorem
If Σ aₙ converges conditionally, then for any L ∈ ℝ ∪ {±∞}, there exists a rearrangement Σ aσ(n) that converges to L.

The proof is constructive: to converge to L, take enough positive terms to exceed L, then enough negative terms to go below L, then positive again, and so on. Because both the positive parts and the negative parts individually diverge (as they must for conditional convergence), this process never terminates, and the partial sums oscillate ever more closely around L.

Concrete example The alternating harmonic series sums to ln 2 in its standard order. Rearranging it — taking two positive terms for every one negative term — gives a series that sums to (3/2)ln 2 instead. Rearranging to take one positive and two negatives gives (1/2)ln 2. The terms are the same; only the order changes.

For absolutely convergent series, no such pathology occurs: every rearrangement of an absolutely convergent series converges to exactly the same sum. This is the Dirichlet rearrangement theorem, and it is why absolute convergence is the "safe" form.

§ 07Convergence Test Strategy

When you encounter a series that looks like it might alternate, here is the decision flowchart used by experienced problem-solvers:

Strategy for Alternating-Looking Series
1. Is there a (−1)ⁿ or (−1)ⁿ⁺¹ factor? → If no, it is not alternating. 2. Does lim bₙ = 0? → If no, diverges by the Divergence Test. 3. Test Σ |aₙ| first (ratio test, p-series, comparison). → If Σ|aₙ| converges: absolutely convergent. Done. 4. If Σ|aₙ| diverges: is bₙ eventually decreasing? → Yes: conditionally convergent by the AST. → No: further analysis required (or diverges).

Testing for absolute convergence first is always wise, because the standard tests (ratio, root, comparison) all work on positive series. If absolute convergence is established, you are done. Only when the absolute series diverges do you need the AST.

✦   ✦   ✦

§ 08Worked Examples

Eleven fully worked examples covering the AST, the Estimation Theorem, absolute vs conditional convergence, and trickier cases.

Example 01 — Apply the AST directly
Determine whether Σn=1∞ (−1)n+1 / n converges or diverges.
  1. Identify bn = 1/n (positive for all n ≥ 1).
  2. Check condition 3: limn→∞ 1/n = 0. ✓
  3. Check condition 2: bn+1 = 1/(n+1) < 1/n = bn. The sequence is strictly decreasing. ✓
  4. Both AST conditions are satisfied. The series converges.
  5. Note: Σ |an| = Σ 1/n is the divergent harmonic series, so convergence is conditional.
Converges conditionally to ln 2
Example 02 — AST with a rational bn
Test Σn=1∞ (−1)n · n / (n² + 1) for convergence.
  1. bn = n/(n²+1). Clearly bn > 0 for all n ≥ 1.
  2. limn→∞ n/(n²+1) = lim 1/n = 0. ✓
  3. Check monotone decrease via f(x) = x/(x²+1). Differentiate:
    f'(x) = (x²+1 − x·2x)/(x²+1)² = (1 − x²)/(x²+1)²
  4. For x > 1, f'(x) < 0, so f is decreasing for n ≥ 2. Condition 2 holds eventually. ✓
  5. Both AST conditions hold. The series converges.
  6. For absolute convergence: compare bn = n/(n²+1) ~ 1/n → diverges by limit comparison with Σ1/n. So convergence is conditional.
Converges conditionally
Example 03 — Series that diverges despite alternating signs
Test Σn=1∞ (−1)n · n / (n + 1) for convergence.
  1. bn = n/(n+1).
  2. Check limn→∞ bn = lim n/(n+1) = 1 ≠ 0.
  3. Since the general term an = (−1)n·n/(n+1) does not → 0, the series diverges by the Divergence Test.
  4. The AST cannot be applied, condition 3 fails first.
Diverges (Divergence Test — terms do not → 0)
Example 04 — Absolutely convergent alternating series
Classify Σn=1∞ (−1)n / n² as absolutely convergent, conditionally convergent, or divergent.
  1. Test Σ |an| = Σ 1/n².
  2. This is a p-series with p = 2 > 1. It converges (to π²/6). ✓
  3. Since the absolute series converges, the original series converges absolutely.
  4. No need to apply the AST: absolute convergence implies ordinary convergence.
Converges absolutely
Example 05 — Estimation Theorem: find the error bound
Approximate Σn=1∞ (−1)n+1/n³ using the first 4 terms, and give the error bound.
  1. Compute S4:
    S4 = 1/1³ − 1/2³ + 1/3³ − 1/4³ = 1 − 0.125 + 0.03704 − 0.015625 = 0.896415
  2. By the ASET, |S − S4| ≤ b5 = 1/5³ = 1/125 = 0.008.
  3. So S ≈ 0.896 with error at most 0.008.
  4. The true sum is (3/4)ζ(3) ≈ 0.9016: indeed within 0.008 of our estimate.
S ≈ 0.8964, error ≤ 0.008
Example 06 — How many terms for a given accuracy?
How many terms of Σ (−1)n+1/n² are needed to approximate its sum to within 0.005?
  1. By the ASET, the error after n terms is ≤ bn+1 = 1/(n+1)².
  2. We need 1/(n+1)² < 0.005.
  3. (n+1)² > 200  ⟹  n+1 > √200 ≈ 14.14.
  4. So n + 1 ≥ 15, meaning n ≥ 14.
  5. At least 14 terms are required.
14 terms are sufficient
Example 07 — Alternating series with factorial terms
Classify Σn=0∞ (−1)n / n! for convergence.
  1. Test Σ |an| = Σ 1/n! using the Ratio Test:
  2. lim |an+1/an| = lim 1/(n+1) = 0 < 1. The absolute series converges.
  3. Therefore the series converges absolutely.
  4. Note: the sum is e−1 = 1/e, this is the Maclaurin series of ex evaluated at x = −1.
Converges absolutely to 1/e
Example 08 — Non-monotone bn: be careful
Consider Σn=1∞ (−1)n+1 bn where b1=2, b2=1/2, b3=3/2, b4=1/4, b5=5/6, ... with the odd terms b2k−1 = (2k−1)/(2k) and even terms b2k = 1/(2k). Can you apply the AST?
  1. Check lim bn: odd terms b2k−1 = (2k−1)/(2k) → 1 as k → ∞.
  2. Since the terms do not → 0, the Divergence Test applies: the series diverges.
  3. The AST cannot be applied; its third condition fails.
  4. This example warns that non-obvious sequences may fail condition 3 even when even-indexed subsequences → 0.
Diverges — bn does not tend to 0
Example 09 — Alternating series at the boundary of a power series
The power series Σ xn/n has radius of convergence 1. At x = −1 the series becomes Σ (−1)n/n. Does it converge?
  1. At x = −1: the series is Σ (−1)n/n (alternating harmonic, starting from n=1).
  2. bn = 1/n; lim bn = 0 ✓; bn strictly decreasing ✓.
  3. AST applies → converges conditionally to −ln 2.
  4. At x = +1 the series is Σ 1/n = harmonic series → diverges.
  5. This illustrates how endpoints of a power series require individual testing, the radius tells us nothing about the endpoints.
Converges conditionally at x = −1 (to −ln 2); diverges at x = +1
Example 10 — Classify Σ (−1)n √n / (n+1)
Classify Σn=1∞ (−1)n √n / (n+1) as absolutely convergent, conditionally convergent, or divergent.
  1. bn = √n/(n+1). Check absolute convergence: compare |an| = √n/(n+1) with 1/√n.
  2. lim [√n/(n+1)] / (1/√n) = lim n/(n+1) = 1 > 0.
  3. Since Σ 1/√n is a divergent p-series (p = 1/2 ≤ 1), by limit comparison Σ|an| diverges. Not absolutely convergent.
  4. Apply AST: lim √n/(n+1) = 0 ✓. Check decreasing: f(x) = √x/(x+1), f'(x) = (1−x)/(2√x(x+1)²) < 0 for x > 1. ✓
  5. AST applies → conditionally convergent.
Converges conditionally
Example 11 — Estimation Theorem to bound a remainder
Show that |ln 2 − (1 − 1/2 + 1/3 − ··· + 1/99)| < 0.01, where the sum has 99 terms.
  1. The series Σ (−1)n+1/n converges to ln 2 by the AST.
  2. The partial sum S99 uses terms n = 1 to 99.
  3. By the ASET: |ln 2 − S99| ≤ b100 = 1/100 = 0.01.
  4. The inequality asks for strictly less than 0.01, but in fact the ASET gives ≤ 0.01. For strict inequality, note that the next term b100 = 1/100 and the error is strictly less than b100 (the bound is achieved only in the limit), so the strict inequality holds.
|ln 2 − S99| ≤ 1/100 = 0.01 ✓ — confirmed by ASET

§ 09Common Mistakes

Mistake 1 — Applying the AST when bn does not tend to 0 If lim bn ≠ 0, the series diverges by the Divergence Test. Many students try to "fix" this by applying the AST anyway, or fail to check condition 3 first. Always verify lim bn = 0 before proceeding.
Mistake 2 — Concluding absolute convergence from the AST The AST proves convergence of the alternating series Σ(−1)nbn. It says nothing about Σ bn. Absolute convergence must be tested separately.
Mistake 3 — Using the Estimation Theorem for non-alternating remainders The ASET applies specifically to alternating series that satisfy the AST conditions. It does not apply to general series. Never use |S − Sn| ≤ bn+1 for a positive series — there is no such automatic bound.
Mistake 4 — Thinking convergence implies absolute convergence The alternating harmonic series converges (to ln 2) but does not converge absolutely. Convergence is a weaker statement than absolute convergence. When a problem asks you to "determine the type of convergence," you must test both.
Mistake 5 — Assuming bn must be strictly decreasing from n = 1 The AST only requires bn to be eventually non-increasing — it may increase for small n before settling into a decreasing pattern. A series that is decreasing for all n ≥ N (some finite N) satisfies the AST.

§ 1010-Question Quiz

Select one answer per question and click Check All Answers to see your score and explanations.

Score: 0 / 10
Question 01
Which condition is NOT required by the Alternating Series Test?
Question 02
Σ (−1)n+1/n converges. What type of convergence is it?
Question 03
Σ (−1)n/n² — what type of convergence?
Question 04
For the series Σ (−1)n+1/n, the Alternating Series Estimation Theorem says the error |S − S₅| is at most:
Question 05
Σ (−1)n · n/(n+1) — does it converge?
Question 06
The Riemann Rearrangement Theorem applies to series that are:
Question 07
How many terms of Σ (−1)n+1/n are needed so the partial sum is within 0.05 of ln 2?
Question 08
If Σ aₙ converges absolutely, which statement must be true?
Question 09
The alternating series Σ (−1)n / n! converges to:
Question 10
For a series Σ aₙ that converges conditionally, which of the following is true?
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§ 11Continue in Series & Sequences

Now that you have mastered the Alternating Series Test and the absolute/conditional convergence distinction, these are the natural next steps in building a complete convergence toolkit.

You are here Alternating Series Leibniz Test, ASET, absolute vs conditional ← Start here Sequences Convergence, monotone sequences, squeeze theorem Series Series Convergence Partial sums, divergence test, geometric & telescoping Series Geometric Series The only series with a closed-form sum Series p-Series Test Σ 1/nᵖ converges iff p > 1 Series Comparison Tests Direct and limit comparison for positive series Series Ratio & Root Tests Best tests for factorials and exponentials Next → Power Series Radius & interval of convergence — uses all tests Series Taylor Series ASET lets you bound the Taylor remainder Series Maclaurin Series Taylor series centred at zero — eˣ, sin x, cos x Integration → Improper Integrals Integral Test uses the same convergence ideas Limits Limits at Infinity Foundation for the divergence test — lim aₙ ≠ 0

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