Difficulty: Intermediate to Advanced | Prerequisites: Sequence limits, integration techniques (for the Integral Test), familiarity with p-series and geometric series.
This is the heart of the sequences-and-series unit: deciding whether an infinite sum converges or diverges, and when possible, finding its exact value. The exam gives you a menu of tests (Divergence, Integral, Comparison, Limit Comparison, Ratio, Root, Alternating Series) and asks you to pick the right one. Choosing well is more than half the battle. If you can look at a series and instinctively reach for the right test, you are in strong shape.
TL;DR
Geometric series converge when |r| < 1 (sum = a/(1 − r)). Telescoping series collapse to a closed form via partial sums. For everything else, pick the convergence test that best matches the series structure: factorials suggest the Ratio Test, nth powers suggest the Root Test, polynomial-type terms suggest Comparison or Limit Comparison with a p-series, and alternating signs suggest the Alternating Series Test.
Geometric series
A series of the form Σ(n=0 to ∞) arⁿ. Converges to a/(1 − r) when |r| < 1. Diverges when |r| ≥ 1. Think of it as the simplest infinite sum you can actually compute.
Telescoping series
A series whose partial sums collapse because most terms cancel in pairs. In simple terms, you write out several partial sums, watch the interior terms vanish, and read off what remains.
p-series
Σ(n=1 to ∞) 1/nᵖ. Converges when p > 1, diverges when p ≤ 1. The case p = 1 is the harmonic series.
Harmonic series
Σ(n=1 to ∞) 1/n. Diverges, even though the terms approach zero. This is the classic example showing that aₙ → 0 is necessary but not sufficient for convergence.
Divergence Test (nth-Term Test)
If lim(n→∞) aₙ ≠ 0, the series Σaₙ diverges. If the limit is 0, the test is inconclusive. This test can only prove divergence, never convergence.
Integral Test
If f(x) is continuous, positive, and decreasing for x ≥ 1, and aₙ = f(n), then Σaₙ and ∫[1 to ∞] f(x) dx either both converge or both diverge.
Comparison Test (Direct)
If 0 ≤ aₙ ≤ bₙ for all n, and Σbₙ converges, then Σaₙ converges. If Σaₙ diverges, then Σbₙ diverges. In simple terms, a series smaller than a convergent one also converges.
Limit Comparison Test (LCT)
If lim(n→∞) aₙ/bₙ = L where 0 < L < ∞, then Σaₙ and Σbₙ either both converge or both diverge. Useful when direct comparison is hard to set up algebraically.
Ratio Test
Compute L = lim(n→∞) |aₙ₊₁/aₙ|. If L < 1, the series converges absolutely. If L > 1 (or L = ∞), it diverges. If L = 1, the test is inconclusive. Best for series with factorials or products.
Root Test
Compute L = lim(n→∞) |aₙ|^(1/n). Same conclusions as the Ratio Test: L < 1 converges, L > 1 diverges, L = 1 inconclusive. Best for series where the entire term is raised to the nth power.
Alternating Series Test (AST / Leibniz Test)
For Σ(−1)ⁿbₙ with bₙ > 0: if {bₙ} is decreasing and lim(n→∞) bₙ = 0, the series converges. This only proves convergence, not absolute convergence.
Absolute convergence
Σaₙ converges absolutely if Σ|aₙ| converges. Absolute convergence implies convergence.
Conditional convergence
Σaₙ converges conditionally if Σaₙ converges but Σ|aₙ| diverges. The alternating harmonic series is the classic example.
A geometric series Σ(n=0 to ∞) arⁿ converges to a/(1 − r) when |r| < 1.
Worked example (Exam Problem 7): Find the sum of Σ(n=0 to ∞) 9/7ⁿ.
Rewrite as Σ 9·(1/7)ⁿ. Here a = 9, r = 1/7.
Sum = 9 / (1 − 1/7) = 9 / (6/7) = 63/6 = 21/2.
Worked example (Exam Problem 8): Express 0.676767... as a ratio of two integers.
Write 0.6767... = 0.67 + 0.0067 + 0.000067 + ... = Σ(n=0 to ∞) (67/100)·(1/100)ⁿ.
This is geometric with a = 67/100, r = 1/100.
Sum = (67/100) / (1 − 1/100) = (67/100) / (99/100) = 67/99.
A telescoping series is one where writing out the partial sums reveals cancellations, leaving only a few surviving terms.
Worked example (Exam Problem 9): Determine if Σ(k=1 to ∞) (1/√(k+1) − 1/√(k+2)) converges or diverges.
Write out the partial sums:
S₁ = 1/√2 − 1/√3
S₂ = (1/√2 − 1/√3) + (1/√3 − 1/√4) = 1/√2 − 1/√4
S₃ = 1/√2 − 1/√5
Sₙ = 1/√2 − 1/√(n+2)
As n → ∞, 1/√(n+2) → 0, so S = 1/√2. The series converges.
This is the skill the exam is testing. Here is how each test was applied on the exam.
Integral Test (Exam Problems 10, 16):
Use when the terms have a form you can integrate as a continuous function.
Problem 10: Σ n³·e^(−3n⁴). The function f(x) = x³·e^(−3x⁴) is positive, continuous, and decreasing. Use substitution u = −3x⁴: the improper integral converges (evaluates to 1/(12e³)), so the series converges.
Problem 16: Σ 1/(6n − 1). Here f(x) = 1/(6x − 1) is positive, continuous, decreasing. The integral ∫[1 to ∞] dx/(6x − 1) = (1/6)·lim(b→∞) [ln|6b − 1| − ln 5] = ∞. Diverges.
Ratio Test (Exam Problem 11):
Best when the series involves factorials or products.
Problem 11: Σ 7ⁿ/n!. Compute |aₙ₊₁/aₙ| = (7^(n+1)/(n+1)!) · (n!/7ⁿ) = 7/(n+1). As n → ∞, this → 0 < 1. Series converges absolutely.
p-Series Test (Exam Problem 12):
Problem 12: Σ 9/∛n = 9·Σ 1/n^(1/3). This is a p-series with p = 1/3 < 1. Diverges.
Limit Comparison Test (Exam Problems 13, 20):
Compare with a known series when the terms behave like a simpler expression for large n.
Problem 13: Σ (k⁵ − 9)/(k⁶ + 4). For large k, this behaves like k⁵/k⁶ = 1/k. Compare with Σ1/k (harmonic, diverges). LCT: lim(k→∞) [(k⁵ − 9)/(k⁶ + 4)] / (1/k) = lim k⁶/(k⁶) = 1 > 0. Both diverge.
Problem 20: Σ (−1)ᵏ·k⁶/√(k¹⁴ + 4). First check absolute convergence: Σ k⁶/√(k¹⁴ + 4). For large k, this behaves like k⁶/k⁷ = 1/k. LCT with Σ1/k gives limit 1, so the absolute series diverges. Then check conditional convergence via AST: bₖ = k⁶/√(k¹⁴ + 4) is eventually decreasing and → 0. So the series converges conditionally.
Divergence Test (Exam Problems 14, 18):
The quickest check: if lim aₙ ≠ 0, done.
Problem 14: Σ 5k/(ln 4k). As k → ∞, 5k/(ln 4k) → ∞ ≠ 0. Diverges.
Problem 18: Σ (−1)^(k+1) · k/(15k + 1). lim(k→∞) k/(15k + 1) = 1/15 ≠ 0. The terms do not approach zero (they oscillate between +1/15 and −1/15). Diverges.
Alternating Series Test (Exam Problem 15):
Problem 15: Σ (−1)ⁿ/n^(4/3). Let bₙ = 1/n^(4/3). Check: (i) bₙ is decreasing since n < n+1 implies n^(4/3) < (n+1)^(4/3) implies 1/n^(4/3) > 1/(n+1)^(4/3). (ii) lim(n→∞) 1/n^(4/3) = 0. Both conditions met, so the series converges.
Root Test (Exam Problem 17):
Best when the entire term is an nth power.
Problem 17: Σ (1/(2k))ᵏ. Compute |aₖ|^(1/k) = 1/(2k). As k → ∞, this → 0 < 1. Converges absolutely.
Comparison Test (Exam Problem 19):
Problem 19: Σ (2 + 7cos n)/n³. Since −1 ≤ cos n ≤ 1, we have 2 + 7cos n ≤ 9. So |(2 + 7cos n)/n³| ≤ 9/n³. Since Σ9/n³ is a convergent p-series (p = 3 > 1), the original series converges absolutely by comparison.
Geometric series sum:
Σ(n=0 to ∞) arⁿ = a/(1 − r), valid when |r| < 1
p-series:
Σ(n=1 to ∞) 1/nᵖ converges iff p > 1
Ratio Test:
L = lim(n→∞) |aₙ₊₁/aₙ|. Converges if L < 1, diverges if L > 1.
Root Test:
L = lim(n→∞) |aₙ|^(1/n). Converges if L < 1, diverges if L > 1.
Integral Test:
Σ f(n) and ∫ f(x) dx share convergence behaviour (for f positive, continuous, decreasing).
Alternating Series Test:
Σ(−1)ⁿbₙ converges if bₙ is decreasing and bₙ → 0.
Geometric series underpin present-value calculations in finance: a stream of payments discounted at rate r is exactly a geometric sum. Fourier series (used in signal processing, acoustics, and image compression) are infinite sums whose convergence properties determine how well they approximate the original signal. The Ratio Test is the workhorse behind determining the radius of convergence of power series, which in turn underpins Taylor-series approximations used throughout engineering and physics.
Students often think that if aₙ → 0, the series must converge. The harmonic series (Σ1/n) is the permanent counterexample. The terms go to zero, but the sum is infinite.
Students mix up the Ratio Test and the Root Test and use them interchangeably regardless of the series structure. The Ratio Test is for factorials and products; the Root Test is for nth-power expressions. Using the wrong one usually produces the inconclusive result L = 1.
When applying the Divergence Test, students sometimes conclude convergence from lim aₙ = 0. The Divergence Test can only prove divergence (when the limit is nonzero). A limit of zero tells you nothing about convergence.
Students forget that the Alternating Series Test only proves convergence, not absolute convergence. If you need to determine absolute vs conditional convergence, you must separately test Σ|aₙ|.
⚠️ The exam gives you a list of tests and asks you to use each one only once. Practise identifying which test matches which series at a glance. Factorials = Ratio. nth powers = Root. Polynomial fractions = Limit Comparison with p-series. Alternating sign = AST.
⚠️ Know the p-series threshold cold: p > 1 converges, p ≤ 1 diverges. Many comparison and limit-comparison problems hinge on this.
⚠️ Absolute vs conditional convergence is a common multi-part question. The typical pattern: show Σ|aₙ| diverges (often by LCT with harmonic), then show Σaₙ converges (by AST).
⚠️ Geometric series sums and repeating-decimal conversions are quick marks. Do not lose them to arithmetic errors.
True or false: Σ(n=0 to ∞) (3/4)ⁿ converges to 4.
True. Sum = 1/(1 − 3/4) = 4.
Fill in the blank: Σ 1/n² is a p-series with p = ____. It ____.
p = 2. It converges.
True or false: The Ratio Test can prove convergence of the harmonic series.
False. The Ratio Test gives L = 1 for the harmonic series, which is inconclusive.
Fill in the blank: If a series converges but does not converge absolutely, it is said to converge ______.
Conditionally.
True or false: Σ (−1)ⁿ/n converges absolutely.
False. It converges conditionally. Σ1/n (the harmonic series) diverges.
Q: Determine whether Σ(n=1 to ∞) n!/(10ⁿ) converges or diverges.
A: Use the Ratio Test. |aₙ₊₁/aₙ| = (n+1)!/10^(n+1) · 10ⁿ/n! = (n+1)/10. As n → ∞, this → ∞ > 1. The series diverges.
Q: Does Σ(n=1 to ∞) 1/(n·ln n) converge or diverge? (Assume n ≥ 2.)
A: Use the Integral Test. ∫[2 to ∞] dx/(x·ln x). Let u = ln x, du = dx/x. Integral becomes ∫ du/u = ln|u| = ln(ln x), which → ∞. The integral diverges, so the series diverges.
Q: Find the sum of the geometric series 5 + 5/3 + 5/9 + 5/27 + ...
A: First term a = 5, common ratio r = 1/3. Sum = 5/(1 − 1/3) = 5/(2/3) = 15/2.
Q: Determine whether Σ(n=1 to ∞) (−1)ⁿ·n/(n² + 1) converges absolutely, conditionally, or diverges.
A: Check absolute convergence: Σ n/(n² + 1). For large n this behaves like 1/n. LCT with Σ1/n: lim n/(n² + 1) ÷ (1/n) = lim n²/(n² + 1) = 1 > 0. Harmonic diverges, so absolute convergence fails. Check AST: bₙ = n/(n² + 1) is eventually decreasing (check derivative) and → 0. Converges conditionally.
Series convergence is the gateway to power series and Taylor series, which appear in the next unit. The Ratio Test, in particular, is how you find the radius of convergence of a power series.
The Integral Test connects this material back to improper integrals from earlier in the course. If you struggled with improper integrals, review them before tackling Integral Test problems.
Absolute and conditional convergence become important in real analysis and in understanding when you can rearrange the terms of a series without changing its sum (Riemann's rearrangement theorem).
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