There is no single universal convergence test; instead there is a toolkit, and part of the skill this week teaches is recognising which tool fits which series. Every test below answers the same question — does ∑a_n converge? — using a different piece of earlier machinery.
15.1 The Integral Test
If f(x) is continuous, positive, and decreasing for x ≥ 1, and an = f(n), then ∑n=1∞ an and ∫1∞ f(x) dx either both converge or both diverge. The area under the continuous curve and the sum of rectangle areas built from it (exactly Week 7's Riemann-sum picture) trap each other within a bounded difference, so one is finite exactly when the other is.
15.2 The Harmonic Series and p-Series
The harmonic series ∑ 1/n has terms → 0 (so the nth-term test is silent), yet it diverges — the Integral Test confirms this immediately, since ∫1∞ (1/x)dx diverges (Week 13, §13.10). More generally, the p-series:
— identical to the improper integral p-test of Eq. 13.3, by the Integral Test itself. The p-series family is the standard benchmark the comparison tests below measure other series against.
15.3 The Direct Comparison Test
If 0 ≤ an ≤ bn for all n beyond some point, then:
- ∑bn converges ⇒ ∑an converges (a smaller series is bounded by a convergent one).
- ∑an diverges ⇒ ∑bn diverges (a larger series cannot be smaller than a divergent one).
Example 15.1 — Direct comparison
Determine whether ∑n=1∞ 1/(n2+3) converges.
Step 1 — Find a larger, known series
1/(n2+3) < 1/n2 for every n, and ∑1/n2 is a p-series with p=2>1, so it converges (Eq. 15.1).
Step 2 — Apply the comparison
Since the given series is smaller termwise than a convergent series, it converges too.
15.4 The Limit Comparison Test
If an, bn > 0 and limn→∞ an/bn = c for some finite c > 0, then ∑an and ∑bn either both converge or both diverge. This avoids the sometimes-awkward inequality of §15.3, at the cost of computing a limit instead (Week 3 §3.9's degree-comparison shortcut is often the fastest route).
Example 15.2 — Limit comparison
Determine whether ∑n=1∞ (2n+1)/(n3+n) converges.
Step 1 — Choose a comparison series based on dominant terms
For large n, the terms behave like 2n/n3 = 2/n2, so compare with ∑1/n2 (a convergent p-series).
Step 2 — Compute the limit ratio
limn→∞ (2n+1)/(n3+n)1/n2 = limn→∞ 2n3+n2n3+n = 2 (Week 3, §3.9)
Step 3 — Apply the test
The limit is a finite positive number (2), and ∑1/n2 converges, so the given series converges too.
15.5 Absolute Convergence
A series ∑an converges absolutely if ∑|an| converges. Absolute convergence implies convergence (the series itself must converge too), but not conversely — a series that converges only because of cancellation between positive and negative terms is called conditionally convergent (§15.9).
15.6 The Ratio Test
Let ρ = limn→∞ |an+1/an|. Then:
- ρ < 1: ∑an converges absolutely.
- ρ > 1 (or infinite): ∑an diverges.
- ρ = 1: inconclusive — try another test.
The Ratio Test is the standard choice whenever an involves factorials or nth powers, since consecutive-term ratios cause massive cancellation there.
Example 15.3 — The Ratio Test with a factorial
Determine whether ∑n=1∞ n!/nn converges.
Step 1 — Set up the ratio
an+1an = (n+1)!/(n+1)n+1n!/nn = (n+1)·nn(n+1)n+1 = nn(n+1)n = 1(1+1/n)n
Step 2 — Take the limit (recognise Example 12.3's limit, generalised)
limn→∞ (1+1/n)n = e (a standard limit, provable the way Week 12 §12.4 handled indeterminate powers), so ρ = 1/e < 1.
Step 3 — Conclude
Since ρ = 1/e < 1, the series converges (absolutely).
15.7 The Root Test
Let ρ = limn→∞ ⁿ√|an|. The same three cases as the Ratio Test apply. The Root Test is the natural choice when an already involves an nth power throughout.
Example 15.4 — The Root Test
Determine whether ∑n=1∞ (2n/(3n+1))n converges.
Step 1 — Take the nth root, which cancels the outer power directly
ⁿ√|an| = 2n/(3n+1)
Step 2 — Take the limit (Week 3, §3.9)
limn→∞ 2n/(3n+1) = 2/3
Step 3 — Conclude
ρ=2/3<1, so the series converges.
15.8 Alternating Series and the Alternating Series Test
An alternating series has terms that strictly alternate sign, ∑(−1)nbn with bn > 0. The Alternating Series Test: if bn is eventually decreasing and bn → 0, the series converges.
Example 15.5 — The Alternating Series Test
Show that the alternating harmonic series ∑n=1∞ (−1)n+1/n converges.
Step 1 — Check b_n is decreasing
bn = 1/n is decreasing for every n ≥ 1.
Step 2 — Check b_n → 0
limn→∞ 1/n = 0.
Step 3 — Conclude
Both conditions hold, so the alternating harmonic series converges (to ln 2, though the test does not reveal the sum's value).
15.9 Conditional Convergence
The alternating harmonic series converges (§15.8), yet the series of its absolute values, ∑1/n, is the harmonic series, which diverges (§15.2). It is therefore conditionally convergent: convergent as written, but not absolutely convergent. This is why §15.5's Absolute Convergence and §15.8's Alternating Series Test must be checked separately and in that order — a series can pass one and fail the other.
15.10 Worked Capstone Example
Example 15.6 — Choosing the right test
Classify ∑n=1∞ (−1)nn/2n as absolutely convergent, conditionally convergent, or divergent.
Step 1 — Test for absolute convergence first (§15.5) using the Ratio Test (§15.6)
|an| = n/2n. |an+1||an| = (n+1)/2n+1n/2n = n+12n
Step 2 — Take the limit
ρ = limn→∞ (n+1)/(2n) = 1/2 < 1.
Step 3 — Conclude directly, without needing the Alternating Series Test at all
By the Ratio Test, ∑|an| converges, so the original series converges absolutely. Since absolute convergence already implies convergence, there is no need to separately verify the alternating series conditions of §15.8.
15.11 Nomenclature
| Symbol | Meaning |
|---|---|
| p-series | ∑1/n^p; converges iff p>1 |
| ρ | The limiting ratio or root used in the Ratio/Root Tests |
| absolutely convergent | ∑|a_n| converges |
| conditionally convergent | ∑a_n converges but ∑|a_n| diverges |
| alternating series | A series whose terms strictly alternate in sign |
15.12 Problems
- Integral Test. Use the Integral Test to determine whether ∑ ne−n2 converges.
- p-series. State, without further work, whether ∑1/n1.5 and ∑1/√n converge.
- Direct comparison. Determine whether ∑1/(n3+1) converges.
- Limit comparison. Determine whether ∑(n+1)/(n2√n) converges.
- Ratio Test. Determine whether ∑3n/n! converges.
- Root Test. Determine whether ∑(n/(2n+1))n converges.
- Alternating Series Test. Determine whether ∑(−1)n/√n converges.
- Absolute vs. conditional. Classify ∑(−1)n/n2 as absolutely convergent, conditionally convergent, or divergent.
- Choosing a test. Determine whether ∑(2n−1)/(n3+2n) converges, explaining why you chose the test you used.
- Combined. Classify ∑(−1)nn2/n! as absolutely convergent, conditionally convergent, or divergent.
15.13 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §9.3 (the integral test), §9.4 (comparison tests), §9.5 (absolute convergence; the ratio and root tests), §9.6 (alternating series and conditional convergence).
- Pearson MyLab Math — the Week 15 homework set is assigned directly from the sections above; see the Calendar for the due date.