Source: Integral Calculus, University of Texas at Austin
Tags: series convergence, divergence, ratio test, root test, comparison test, alternating series, absolute convergence, conditional convergence
Difficulty: Intermediate to Advanced | Prerequisites: Sequences and series basics, p-series, geometric series, integral test.
This material sits near the end of a standard Integral Calculus (Calculus II) course, after you have covered improper integrals and basic series. The homework tests your ability to decide whether an infinite series converges or diverges, and if it converges, whether that convergence is absolute or conditional. You need fluency with the Comparison Test, Ratio Test, Root Test, Alternating Series Test, and the nth-term Divergence Test. If geometric series and p-series are not second nature yet, review those first.
You have a toolbox of convergence tests, and the skill is picking the right one. The Ratio Test and Root Test handle series with factorials or exponentials. The Comparison and Limit Comparison Tests work well when the general term looks like a tweaked p-series or geometric series. The Alternating Series Test tells you about conditional convergence, and the Divergence Test is the quickest way to rule out convergence when the terms do not approach zero.
Convergent series
A series whose partial sums approach a finite limit. In simple terms, if you keep adding terms forever, the running total settles on a single number.
Divergent series
A series that does not converge. The partial sums either grow without bound, oscillate, or fail to settle.
Absolute convergence
A series Σaₙ is absolutely convergent if Σ|aₙ| converges. Think of it as: even if you strip away all the sign changes (remove the minus signs), the series still converges. Absolute convergence implies ordinary convergence.
Conditional convergence
A series that converges, but does not converge absolutely. The convergence depends on the cancellation between positive and negative terms. The classic example is the alternating harmonic series.
Ratio Test
Compute L = lim |aₙ₊₁ / aₙ| as n → ∞. If L < 1, the series converges absolutely. If L > 1, it diverges. If L = 1, the test is inconclusive. Best suited for series with factorials or exponential terms.
Root Test
Compute L = lim |aₙ|^(1/n) as n → ∞. Same conclusion thresholds as the Ratio Test: L < 1 converges, L > 1 diverges, L = 1 inconclusive. Works well when the general term is raised to the nth power.
Comparison Test (Direct)
If 0 ≤ aₙ ≤ bₙ for all n and Σbₙ converges, then Σaₙ converges. If Σaₙ diverges, so does Σbₙ. Think of it as: a smaller series inherits convergence from a bigger one that converges.
Limit Comparison Test
If lim (aₙ / bₙ) = c where 0 < c < ∞, then Σaₙ and Σbₙ either both converge or both diverge. Useful when the algebra for a direct comparison is awkward but the terms behave like a known series.
Alternating Series Test (Leibniz Test)
For a series Σ(−1)ⁿbₙ with bₙ > 0: if bₙ is eventually decreasing and lim bₙ = 0, the series converges. This test proves convergence but not absolute convergence.
Divergence Test (nth-term test)
If lim aₙ ≠ 0 (or the limit does not exist), the series Σaₙ diverges. If the limit is zero, the test says nothing, so you need another test.
These are your go-to tests when the general term of a series looks like a familiar benchmark (a p-series 1/nᵖ or a geometric series rⁿ) with some extra baggage.
Direct Comparison. Find a known series that bounds yours from above (to prove convergence) or below (to prove divergence). Example: Σ 2ⁿ/(5ⁿ + n). Since 5ⁿ + n > 5ⁿ, we have 2ⁿ/(5ⁿ + n) < 2ⁿ/5ⁿ = (2/5)ⁿ. The geometric series Σ(2/5)ⁿ converges (ratio 2/5 < 1), so the original series converges by direct comparison.
Limit Comparison. When the algebra for a direct bound is messy, compare the asymptotic behaviour instead. For Σ 3√n / (n√n − 2), the dominant behaviour for large n is 3n^(1/2) / n^(3/2) = 3/n, so compare with Σ 1/n (the harmonic series, which diverges). The limit of the ratio is a positive finite constant, so the original series also diverges.
Choosing the benchmark. Keep only the highest-order terms in numerator and denominator. Whatever simplified fraction you get is your comparison series. If it is 1/nᵖ with p > 1, convergence. If p ≤ 1, divergence.
Tricky example from HW12. Series B in Q4: 3/5 + 4/6 + 5/7 + ... The general term is (n+2)/(n+4), which approaches 1, not 0. By the Divergence Test alone this series diverges, no comparison needed.
The Ratio Test is the workhorse for series involving factorials, products of factorials, or terms where each aₙ₊₁ has a clean relationship to aₙ.
How it works. Compute L = lim (n→∞) |aₙ₊₁ / aₙ|. Converges absolutely if L < 1. Diverges if L > 1. Inconclusive if L = 1.
Factorials are the signal. If you see n! or (n+k)! in the general term, reach for the Ratio Test first. Example: Σ 2ᵏk² / k! From HW12 Q8, the ratio aₖ₊₁/aₖ simplifies so that the factorial dominates, giving L = 0 < 1, so the series converges.
Exponentials competing with other exponentials. For Σ 4ᵏ / (2 + 3ᵏ), the ratio test gives L = 4/3 > 1, so the series diverges. Meanwhile, for Σ (−3)^(m+1) / 4^(3m) from Q3, the ratio is 3/64 < 1, so absolute convergence.
When it fails. The Ratio Test gives L = 1 for any p-series Σ 1/nᵖ, so it cannot distinguish convergent (p > 1) from divergent (p ≤ 1) p-series. This is exactly the point tested in Q10 and Q11: the Ratio Test is not useful for Σ 1/n³ (it gives L = 1), but the Ratio Test works perfectly for Σ 1/n! (it gives L = 0).
Series with n! growing against exponentials. For Σ 3e^(−2n) n!, use the Ratio Test: aₙ₊₁/aₙ = (n+1)/e², which goes to infinity. The series diverges because n! eventually overwhelms any exponential decay.
The Root Test is a close cousin of the Ratio Test. It shines when the entire general term is raised to the nth power, because taking the nth root peels that exponent right off.
How it works. Compute L = lim (n→∞) |aₙ|^(1/n). Same thresholds: L < 1 converges, L > 1 diverges, L = 1 inconclusive.
When to prefer it over the Ratio Test. If your term looks like [f(n)]ⁿ, the Root Test reduces it to lim f(n), which is often simpler than forming the ratio. Example: Σ 1/(ⁿ√5 − 1)ⁿ from Q9. Taking the nth root gives 1/(ⁿ√5 − 1). As n → ∞, ⁿ√5 → 1, so the denominator goes to 0, meaning L = ∞. The series diverges.
Limitations. Like the Ratio Test, the Root Test is inconclusive for p-series (L = 1 every time). Q11 confirms that the Root Test cannot determine whether Σ k/(3 + k²) converges. That series needs the Limit Comparison Test against Σ 1/k instead.
Several homework questions ask you to classify a series as absolutely convergent, conditionally convergent, or divergent. This is a three-step decision.
Step 1: Check absolute convergence. Take the absolute value of the general term and test Σ|aₙ| using any test (Comparison, Ratio, Root, Integral). If Σ|aₙ| converges, the original series is absolutely convergent and you are done.
Step 2: If Σ|aₙ| diverges, check for conditional convergence. Apply the Alternating Series Test to the original series. If bₙ is decreasing and lim bₙ = 0, the series converges conditionally.
Step 3: If neither works, the series diverges.
Worked examples from HW12:
Q3: Σ (−3)^(m+1) / 4^(3m). The absolute value gives Σ (3/64)^m, a geometric series with ratio 3/64 < 1. Absolutely convergent.
Q5: Σ (−1)^(n−1) e^(1/n) / (n² + 7). For absolute convergence, note e^(1/n) → 1, so |aₙ| behaves like 1/n², which is a convergent p-series (p = 2). Absolutely convergent.
Q7: Σ (−1)ᵏ ln(4k) / k². The absolute series Σ ln(4k)/k² converges by comparison with Σ 1/k^(3/2) (since ln(4k) grows slower than any positive power of k). Absolutely convergent.
Q6: Σ (−1)^(k−1) cos(1/k). As k → ∞, cos(1/k) → cos(0) = 1, so the terms do not approach zero. The series diverges by the Divergence Test.
The Divergence Test is the simplest test in the toolbox, and it should always be your first check.
The rule. If lim (n→∞) aₙ ≠ 0, the series Σaₙ diverges. Full stop.
The trap. The converse is false. If lim aₙ = 0, the series might still diverge. The harmonic series Σ 1/n is the standard counterexample: the terms go to zero but the series diverges. This is exactly what Q10 statement C tests: "If lim aₙ = 0, then Σaₙ converges" is false.
Q10 in full. Statement A ("If Σaₙ diverges, then Σ|aₙ| diverges") is true, because absolute convergence implies convergence, so its contrapositive holds. Statement B ("The Ratio Test can determine whether Σ1/n³ converges") is false, as the Ratio Test gives L = 1 for any p-series. Statement C is false as discussed. Only A is true.
Quick use in HW12. Q4 series B and Q6 both have terms whose limit is not zero (the terms approach 1 in both cases), so the Divergence Test settles them immediately.
Geometric series: Σ rⁿ converges if and only if |r| < 1. Sum = a/(1 − r) where a is the first term.
p-series: Σ 1/nᵖ converges if p > 1, diverges if p ≤ 1.
Ratio Test threshold: L = lim |aₙ₊₁/aₙ|. L < 1 → converges absolutely. L > 1 → diverges. L = 1 → inconclusive.
Root Test threshold: L = lim |aₙ|^(1/n). Same thresholds as Ratio Test.
Alternating Series conditions: For Σ(−1)ⁿbₙ, convergence requires (i) bₙ₊₁ ≤ bₙ eventually, and (ii) lim bₙ = 0.
Useful limit: ⁿ√n → 1 as n → ∞. Also ⁿ√c → 1 for any constant c > 0.
Factorial dominance: n! grows faster than any exponential cⁿ. So Σ cⁿ/n! always converges (Ratio Test gives L = 0).
Students often believe that if lim aₙ = 0, the series must converge. It does not. The harmonic series is the standard counterexample. The Divergence Test only works in one direction: nonzero limit proves divergence, but a zero limit proves nothing.
Students confuse absolute convergence with conditional convergence. If Σ|aₙ| converges, the series is absolutely convergent, period. You only get conditional convergence when the series converges but Σ|aₙ| does not.
Students try to apply the Ratio Test to every series. It is inconclusive (L = 1) for all p-series and many algebraic-term series. When the general term is a rational function of n with no factorials or exponentials, the Comparison or Limit Comparison Test is almost always the better tool.
Students forget that absolute convergence implies convergence. If you have already shown Σ|aₙ| converges, you do not need a separate argument for Σaₙ.
⚠️ Expect at least one question testing whether you know that lim aₙ = 0 is necessary but not sufficient for convergence.
⚠️ Know which tests are inconclusive for p-series (both Ratio and Root give L = 1). This is a favourite conceptual question.
⚠️ Absolute vs conditional classification questions are common. The three-step workflow (check absolute first, then alternating series test, then diverge) is the template.
⚠️ Be ready for series where the Divergence Test immediately settles the question. If the terms do not go to zero, no further work is needed.
True or false: If lim aₙ = 0, then Σaₙ converges. (False.)
True or false: Absolute convergence implies convergence. (True.)
Fill in the blank: The Ratio Test is inconclusive when L = ___. (1.)
True or false: The Ratio Test can determine whether Σ1/n³ converges. (False, it gives L = 1.)
True or false: If Σaₙ diverges, then Σ|aₙ| must also diverge. (True, by contrapositive of "absolute convergence implies convergence.")
Q: Does Σ 2ⁿ/(5ⁿ + n) converge or diverge?
A: Converges. Compare with (2/5)ⁿ, a convergent geometric series.
Q: Is Σ (−3)^(m+1) / 4^(3m) absolutely convergent, conditionally convergent, or divergent?
A: Absolutely convergent. The absolute series is geometric with ratio 3/64 < 1.
Q: Σ (−1)^(k−1) cos(1/k). Classify its convergence.
A: Divergent. cos(1/k) → 1, so the terms do not approach zero.
Q: Does Σ 2ᵏk²/k! converge or diverge?
A: Converges. The Ratio Test gives L = 0 because k! grows faster than any polynomial times exponential.
Q: Series A is Σ 3e^(−2n) n! and Series B is Σ 1/(ⁿ√5 − 1)ⁿ. Which converge?
A: Neither. Series A diverges because n! dominates e^(2n). Series B diverges because ⁿ√5 → 1, making the base go to infinity.
Q: Which statements are true? (A) Σaₙ diverges implies Σ|aₙ| diverges. (B) Ratio Test works for Σ1/n³. (C) lim aₙ = 0 implies Σaₙ converges.
A: Only A is true. B fails because L = 1 for p-series. C is false by the harmonic series.
These convergence tests are the gateway to power series (covered in the second set of study notes). You cannot determine the radius or interval of convergence of a power series without the Ratio Test or Root Test. The Alternating Series Test comes back when checking endpoint convergence. The Comparison Test connects directly to the Integral Test, which relates series convergence to improper integrals from earlier in the course.
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