Choosing a Test, and Absolute versus Conditional Convergence

You’ll be able to

Select an appropriate convergence test from the form of the terms
Distinguish absolute from conditional convergence and test for each in the right order
State a convergence conclusion with the hypotheses the test actually requires

A decision procedure, read off the terms

The tests are not interchangeable, and the exam rewards picking well. Work down this list. Do the terms fail to go to zero? nth-term test — diverges, stop. Is it geometric or telescoping? Use the formula. Are the terms a rational or algebraic expression in n? Comparison or limit comparison against a p-series, comparing the dominant powers. Do factorials or nth powers appear? Ratio test. Is it alternating? Alternating series test. Is the term a function that is easy to integrate and positive and decreasing? Integral test. Most exam series announce their test in their form.

p-series is the yardstick

Σ1/n^p converges if and only if p > 1. This single fact anchors every comparison: Σ1/n² converges (p = 2), the harmonic series Σ1/n diverges (p = 1), and Σ1/√n diverges (p = 1/2). When comparing, look only at the dominant powers — (3n + 5)/(n³ − 2) behaves like 3/n², so it converges by limit comparison with a p-series of p = 2. The lower-order terms never change the verdict.

Absolute first, then conditional

Σaₙ converges absolutely when Σ|aₙ| converges — and absolute convergence implies convergence, so this is the stronger finding and the right thing to test first. If Σ|aₙ| diverges but Σaₙ itself converges (necessarily by the alternating series test), the convergence is conditional. Σ(−1)ⁿ/n is the standard example: conditionally convergent, since the alternating test passes while the harmonic series of absolute values diverges. Σ(−1)ⁿ/n² is absolutely convergent, because Σ1/n² converges on its own.

The tests, with their conditions

p-series: Σ1/n^p converges ⟺ p > 1 · Ratio: lim|aₙ₊₁/aₙ| = L, converges if L < 1, diverges if L > 1, inconclusive if L = 1 · Alternating: terms decreasing in magnitude AND → 0

The ratio test at L = 1 says nothing at all — that is exactly where p-series live, and where a comparison is needed instead.

Worked example

Determine whether Σ (n = 1 to ∞) of (−1)ⁿ/√(n + 3) converges absolutely, converges conditionally, or diverges.

1.Test absolute convergence first: Σ|aₙ| = Σ1/√(n + 3), which behaves like Σ1/√n — a p-series with p = 1/2.
2.Since p = 1/2 ≤ 1, that p-series diverges, so the series does not converge absolutely.
3.Now test the original alternating series. The magnitudes 1/√(n + 3) are positive and decreasing, since √(n + 3) increases with n.
4.And lim(n→∞) 1/√(n + 3) = 0, so both conditions of the alternating series test are met and the series converges.
5.Converging while its absolute series diverges is the definition of conditional convergence.

Answer: The series converges conditionally: the alternating series test passes, but the absolute series is a divergent p-series with p = 1/2.

Watch out

The alternating series test needs both conditions — decreasing in magnitude and limit zero. A series whose terms alternate and go to zero non-monotonically is not covered by the test, and citing it there is a lost point.

Checkpoint

Which test is best suited to Σ n!/(3ⁿ)?

Checkpoint

Σ(−1)ⁿaₙ converges, but Σ|aₙ| diverges. The series is:

On the exam

Free-response convergence questions want the test named and its hypotheses verified. "By the alternating series test, since 1/√(n+3) is decreasing and tends to 0" earns credit; "it converges by AST" alone often does not.

Answer the 2 checkpoints as you read.