Source: MATH 308, Texas A&M University, Sections 11.5–11.6
Tags: alternating series, alternating series test, AST, absolute convergence, conditional convergence, ratio test, convergence tests, series, Leibniz test, harmonic series
The alternating series test (AST) lets you confirm convergence of series whose signs flip back and forth, provided the terms decrease to zero. Once alternating series are covered, the next question is whether a series converges absolutely or only conditionally, which is settled by checking whether the series of absolute values converges. The ratio test is the workhorse for series involving factorials and exponentials, and it sets the stage for power series.
Alternating series
A series whose successive terms alternate in sign. The standard form is ∑(−1)^(n−1) bₙ = b₁ − b₂ + b₃ − b₄ + … where each bₙ > 0.
Alternating series test (AST) / Leibniz test
If ∑(−1)^(n−1) bₙ is alternating with 0 ≤ bₙ₊₁ ≤ bₙ (terms are positive and decreasing) and lim bₙ = 0, then the series converges.
Alternating series estimation theorem (ASET)
For a convergent alternating series, the remainder after n terms satisfies |Rₙ| = |s − sₙ| ≤ bₙ₊₁. In plain terms, the error is no larger than the first omitted term. This only works for alternating series.
Absolute convergence
A series ∑aₙ converges absolutely if the series of positive terms ∑|aₙ| converges. If a series converges absolutely, it converges (full stop).
Conditional convergence
A series ∑aₙ converges conditionally if ∑aₙ converges but ∑|aₙ| does not. It converges, but just barely, so to speak.
Ratio test
For ∑aₙ, compute L = lim |aₙ₊₁ / aₙ|. If L < 1 the series converges absolutely. If L > 1 it diverges. If L = 1 the test is inconclusive.
A series is alternating when signs strictly alternate: +, −, +, −, … or −, +, −, +, …
1 − 1/2 + 1/3 − 1/4 + 1/5 − … is alternating.
1 + 1/2 − 1/3 − 1/4 + 1/5 + 1/6 − … is not alternating (signs do not strictly flip each term).
To apply the AST to ∑(−1)^(n−1) bₙ you check exactly two things:
The bₙ are eventually decreasing: bₙ₊₁ ≤ bₙ for all sufficiently large n.
The terms tend to zero: lim(n→∞) bₙ = 0.
Both conditions met means the series converges.
Classic example: the alternating harmonic series ∑(−1)ⁿ (1/n) = 1 − 1/2 + 1/3 − 1/4 + … converges by the AST. Meanwhile the ordinary harmonic series ∑1/n diverges. This is the textbook illustration of conditional convergence.
Worked examples from the notes:
∑(−1)ⁿ / √(n + 3): bₙ = 1/√(n + 3) is decreasing and → 0, so it converges by AST.
∑(−1)ⁿ · n / (n² + 1): check whether bₙ = n/(n² + 1) → 0. It does (divide top and bottom by n²), and it is eventually decreasing, so AST applies.
∑(−1)ⁿ · n² / (n² + 1): here bₙ = n²/(n² + 1) → 1 ≠ 0. The series diverges by the divergence test (terms do not approach zero).
When an alternating series converges, you can bound the error of a partial-sum approximation:
|Rₙ| = |s − sₙ| ≤ bₙ₊₁
This is specific to alternating series. For a non-alternating series like ∑1/n², the remainder Rₙ is not necessarily smaller than 1/(n + 1)².
Example: For ∑(−1)^(n−1) (1/n⁴), using the partial sum s₂ = 1 − 1/16 = 15/16 gives an error |R₂| ≤ b₃ = 1/81 ≈ 0.012. To get the error below 0.001, find the smallest n such that bₙ₊₁ = 1/(n + 1)⁴ < 0.001, i.e. (n + 1)⁴ > 1000, so n + 1 ≥ 6, meaning n = 5 terms suffice.
Every series ∑aₙ falls into exactly one bucket:
Converges absolutely: ∑|aₙ| converges (which forces ∑aₙ to converge too). Example: ∑(−1)ⁿ / n² converges absolutely because ∑1/n² converges (p-series, p = 2).
Converges conditionally: ∑aₙ converges but ∑|aₙ| does not. Example: ∑(−1)ⁿ / n converges (AST) but ∑1/n diverges (harmonic series).
Diverges: the series does not converge at all.
You can use the integral test, comparison test, or limit comparison test on ∑|aₙ| to establish absolute convergence.
Worked classification (Example 11.6.4):
∑(−1)ⁿ / √(n + 3): |aₙ| = 1/√(n + 3) behaves like 1/√n (p-series, p = 1/2 < 1), so ∑|aₙ| diverges. But AST gives convergence. Verdict: conditionally convergent.
∑(−1)ⁿ √n / √(n + 3): |aₙ| = √n / √(n + 3) → 1 ≠ 0. Diverges outright.
∑(−1)ⁿ / (√(n + 3))³: |aₙ| ~ 1/n^(3/2) (p-series, p = 3/2 > 1), so ∑|aₙ| converges. Verdict: absolutely convergent.
For ∑aₙ, compute:
L = lim(n→∞) |aₙ₊₁ / aₙ|
L < 1 → absolutely convergent
L > 1 → divergent
L = 1 → inconclusive (try another test)
Why it works (rough idea): if the ratio of successive terms approaches L < 1, then eventually aₙ behaves like C · Lⁿ, and the series is bounded by a convergent geometric series.
When to reach for the ratio test: whenever factorials (n!) or exponentials (aⁿ) appear in the general term. The ratio test handles these naturally because consecutive factorials and exponentials simplify cleanly in the ratio aₙ₊₁/aₙ.
When the ratio test fails: for p-series ∑1/nᵖ the ratio always gives L = 1, so it is inconclusive. Use the integral test or comparison instead.
Worked examples:
aₙ = 1/5ⁿ: |aₙ₊₁/aₙ| = 1/5 < 1. Converges absolutely.
aₙ = n²/5ⁿ: |aₙ₊₁/aₙ| = ((n+1)²/n²)(1/5) → 1/5 < 1. Converges absolutely.
aₙ = 1/nᵖ: |aₙ₊₁/aₙ| = (n/(n+1))ᵖ → 1. Inconclusive.
∑ n!/6^(2n+1): ratio involves (n+1)!/n! = n+1, which grows without bound. L = ∞ > 1. Diverges.
∑ (2n+1)! / (n!(n+1)!): simplify the ratio carefully using factorial cancellation. L = 4 > 1. Diverges.
∑ 2^(2n) / (n!)²: |aₙ₊₁/aₙ| = 4/(n+1)² → 0 < 1. Converges absolutely.
The main use of the ratio test going forward is for power series, where it determines the radius of convergence.
Alternating series test: ∑(−1)^(n−1) bₙ converges if bₙ₊₁ ≤ bₙ and lim bₙ = 0.
ASET error bound: |Rₙ| ≤ bₙ₊₁ (alternating series only).
Ratio test: L = lim |aₙ₊₁/aₙ|. L < 1 → converges absolutely. L > 1 → diverges. L = 1 → inconclusive.
Absolute convergence implies convergence: ∑|aₙ| converges ⟹ ∑aₙ converges.
⚠️ The AST requires both conditions: decreasing terms and limit zero. Showing only one is incomplete.
⚠️ The ASET error bound |Rₙ| ≤ bₙ₊₁ applies only to alternating series. Do not use it for a general positive-term series.
⚠️ Absolute convergence is strictly stronger than conditional convergence. "Converges absolutely" automatically means "converges." The converse is false.
⚠️ The ratio test giving L = 1 means nothing. You must try another test. This happens for every p-series.
⚠️ When asked "does this series converge absolutely, converge conditionally, or diverge?" you need to check two things: convergence of ∑|aₙ| and convergence of ∑aₙ. A one-line answer is not enough.
⚠️ In ratio-test problems involving factorials, the key simplification is (n+1)!/n! = n + 1. Practise this until it is automatic.
Q: State the two conditions of the alternating series test.
A: For ∑(−1)^(n−1) bₙ with bₙ > 0, the series converges if (1) bₙ is eventually decreasing, and (2) lim(n→∞) bₙ = 0.
Q: The alternating harmonic series ∑(−1)ⁿ/n converges. Does it converge absolutely or conditionally?
A: Conditionally. ∑|aₙ| = ∑1/n is the harmonic series, which diverges.
Q: Using the ASET, how many terms of ∑(−1)^(n−1)/n⁴ do you need for the error to be below 0.001?
A: You need bₙ₊₁ = 1/(n+1)⁴ < 0.001, so (n+1)⁴ > 1000. Since 5⁴ = 625 < 1000 and 6⁴ = 1296 > 1000, you need n + 1 = 6, i.e. n = 5 terms.
Q: Apply the ratio test to ∑ n!/6^(2n+1).
A: |aₙ₊₁/aₙ| = (n+1)! · 6^(2n+1) / (n! · 6^(2n+3)) = (n+1)/36. As n → ∞ this → ∞ > 1. The series diverges.
Q: Why is the ratio test inconclusive for ∑1/n²?
A: |aₙ₊₁/aₙ| = n²/(n+1)² → 1. L = 1 gives no information, so another test (e.g. p-series test) is needed.
Q: ∑ sin(n)/n² – does this converge absolutely?
A: Yes. |sin(n)/n²| ≤ 1/n², and ∑1/n² converges (p = 2 > 1). By comparison, ∑|aₙ| converges, so the series converges absolutely.
alternating series test, Leibniz test, AST, ASET, alternating series estimation theorem, alternating series error bound, absolute convergence, conditional convergence, ratio test, convergence tests, series convergence, divergence test, harmonic series, p-series, factorial series, geometric series comparison, MATH 308, calculus II, Texas A&M