Convergence Tests for Series: Comparison, Cauchy, d’Alembert
Convergence tests decide whether a series converges without computing its sum; the main ones are the comparison test, the integral test, the ratio test (d’Alembert), the root test (Cauchy) and the alternating series test. In brief: for Σ an with positive terms, if the limit of an+1/an is L, then the series converges when L < 1, diverges when L > 1 and the test is inconclusive for L = 1. For Σ n/2n, L = 1/2, so the series converges.
Most series have no closed form for the sum, and the first question is whether it exists. Each test has its domain of success: the ratio and root tests work well with factorials and powers, comparison works for rational terms and the integral test connects series with improper integrals. This guide states each test, gives the logic, shows worked examples and a table that helps to choose the test.
What is the divergence test?
If an does not tend to 0, the series Σ an diverges. This should always be the first check. Example: Σ (2n + 1)/(3n − 2) has terms tending to 2/3 ≠ 0, so it diverges. If an → 0, the test says nothing, and another test is needed.
How does the comparison test work?
Let 0 ≤ an ≤ bn for all large n.
- If Σ bn converges, then Σ an converges.
- If Σ an diverges, then Σ bn diverges.
The limit comparison test: if an, bn > 0 and an/bn → c with 0 < c < ∞, then both series converge or both diverge.
Example 1
Σ 1/(n2 + 1): since 1/(n2 + 1) < 1/n2 and Σ 1/n2 converges (p = 2), the series converges.
Example 2
Σ 1/√(n2 − 1) for n ≥ 2: 1/√(n2 − 1) > 1/n and Σ 1/n diverges, so the series diverges.
Example 3: limit comparison
Σ (3n + 1)/(n3 + 2n): compare with 1/n2. The ratio of terms (3n + 1)n2/(n3 + 2n) → 3. Since Σ 1/n2 converges, so does the given series.
What is d’Alembert’s ratio test?
For an > 0, let L = lim an+1/an. If L < 1, the series converges; if L > 1 (including ∞), it diverges; if L = 1, the test is inconclusive. The test compares the series with a geometric series with ratio L. For series with signs, apply the test to |an|, which gives absolute convergence.
Example 4
Σ n/2n: an+1/an = (n + 1)/(2n) → 1/2 < 1, so the series converges. (Its sum is 2.)
Example 5
Σ 2n/n!: an+1/an = 2/(n + 1) → 0 < 1, so the series converges (its sum is e2 − 1 ≈ 6.389 when starting at n = 1).
Example 6
Σ n!/10n: an+1/an = (n + 1)/10 → ∞, so the series diverges.
Example 7: inconclusive case
Σ 1/n and Σ 1/n2 both give L = 1, but the first diverges and the second converges. Use another test.
What is Cauchy’s root test?
Let L = lim n√|an|. If L < 1, the series converges absolutely, if L > 1 it diverges and if L = 1 the test fails. It is natural when the term is an nth power.
Example 8
Σ ((2n + 1)/(3n + 4))n: n√an = (2n + 1)/(3n + 4) → 2/3 < 1. The series converges.
Example 9
Σ (1 + 1/n)n2: the nth root is (1 + 1/n)n → e > 1, so the series diverges (the terms even tend to infinity).
What is the integral test?
If f is positive, continuous and decreasing on [1, ∞) with f(n) = an, then Σ an and ∫1∞ f(x) dx both converge or both diverge. The idea is to compare the sum with the area under the graph. This gives the p-series result: ∫1∞ x−p dx converges for p > 1 and diverges for p ≤ 1.
Example 10
Σ 1/(n ln n) for n ≥ 2: ∫ dx/(x ln x) = ln(ln x), which tends to infinity. So the series diverges (very slowly). And Σ 1/(n (ln n)2) converges, since ∫ dx/(x (ln x)2) = −1/ln x converges.
What is the alternating series test?
If bn > 0, bn is decreasing (at least eventually) and bn → 0, then Σ (−1)nbn converges, and the error after N terms is at most bN+1.
Example 11
Σ (−1)n+1/√n: bn = 1/√n decreases to 0, so the series converges (conditionally, since Σ 1/√n diverges). Σ (−1)n n/(n + 1) does not converge: the terms do not tend to 0.
Which test should you choose?
| Form of the terms | Good test | Example | Result |
|---|---|---|---|
| terms do not tend to 0 | divergence test | Σ n/(n + 1) | diverges |
| rational function of n | (limit) comparison with p-series | Σ 1/(n2 + 1) | converges |
| factorials, exponentials | ratio test | Σ 2n/n! | converges |
| nth powers | root test | Σ ((2n + 1)/(3n + 4))n | converges |
| f(n) with f easy to integrate | integral test | Σ 1/(n ln n) | diverges |
| alternating signs | Leibniz test | Σ (−1)n+1/√n | converges |
Where do the tests come from?
Jean le Rond d’Alembert published the ratio test in 1768 as part of his work on the binomial series. Augustin-Louis Cauchy gave the root test and a rigorous approach to convergence in his 1821 Cours d’analyse. The alternating test is credited to Leibniz in 1682, and the integral test was formulated by Cauchy and Maclaurin. These tests turned the manipulation of infinite series from a collection of tricks into a theory.
Where are they used today?
The ratio test determines the radius of convergence of power series, which define functions such as ex, sin x and Bessel functions. For Σ xn/n!, the ratio is |x|/(n + 1) → 0 for every x, so the series converges for all x. In numerical analysis, the tests tell whether an iterative method converges, and in physics, they validate series solutions of equations.
Which mistakes are common?
Concluding from L = 1
In the ratio and root tests, L = 1 gives no information. Try another test.
Applying comparison in the wrong direction
A series smaller than a divergent one proves nothing. Compare with a larger convergent series or a smaller divergent one.
Using the Leibniz test without checking monotonicity
The terms bn must decrease to 0. For bn with an irregular pattern, the test may fail.
Forgetting that an → 0 is not sufficient
The harmonic series has terms tending to 0 but diverges.