Series Convergence Tests

You’ll be able to

Apply the nth-term test, geometric series test, and p-series test
Use the ratio test to determine convergence
Distinguish absolute from conditional convergence

A series is a sum of infinitely many terms

An infinite series Σaₙ adds up a sequence’s terms. It converges if its sequence of partial sums approaches a finite limit, and diverges otherwise. The first check is always the nth-term test: if the terms aₙ do not shrink to 0, the series diverges. But the converse fails — aₙ → 0 does not guarantee convergence (the harmonic series Σ1/n is the famous counterexample). So aₙ → 0 is necessary, not sufficient.

Two benchmark series

geometric Σ arⁿ: converges ⟺ |r| < 1, to a/(1 − r) · p-series Σ 1/nᵖ: converges ⟺ p > 1

These two are the reference points most comparison and ratio arguments lean on.

The ratio test

The ratio test examines L = lim |aₙ₊₁/aₙ|. If L < 1 the series converges (absolutely); if L > 1 it diverges; if L = 1 the test is inconclusive and you need another approach. The ratio test is the default weapon for series with factorials or nth powers, because those simplify beautifully in the ratio. It is also the engine behind finding a power series’ radius of convergence.

Absolute vs. conditional convergence

A series Σaₙ converges absolutely if Σ|aₙ| converges — the strongest form, immune to how the signs are arranged. It converges conditionally if Σaₙ converges but Σ|aₙ| diverges; here the cancellation of alternating signs is doing all the work. The alternating harmonic series Σ(−1)ⁿ⁺¹/n is the classic conditional case: it converges (to ln 2), but its absolute version, the harmonic series, diverges.

Worked example

Use the ratio test to determine whether Σ (n=1 to ∞) 2ⁿ/n! converges.

1.Form the ratio |aₙ₊₁/aₙ| = [2ⁿ⁺¹/(n+1)!] · [n!/2ⁿ].
2.Simplify the powers and factorials: 2ⁿ⁺¹/2ⁿ = 2, and n!/(n+1)! = 1/(n+1).
3.So the ratio is 2/(n + 1).
4.Take the limit as n → ∞: 2/(n + 1) → 0, and L = 0 < 1.

Answer: Since L = 0 < 1, the ratio test guarantees the series converges (absolutely). Factorials in the denominator make ratio-test limits go to 0.

Watch out

The nth-term test can only prove divergence. If aₙ → 0, the test is inconclusive — it never proves convergence. Students who "conclude convergence because the terms go to 0" fall straight into the harmonic-series trap.

Checkpoint

For which values of p does the p-series Σ 1/nᵖ converge?

Checkpoint

The geometric series Σ (n=0 to ∞) 3·(1/2)ⁿ converges. What is its sum?

On the exam

Match the test to the terms: factorials or nth powers → ratio test; a clean 1/nᵖ → p-series; a constant ratio between terms → geometric. Recognizing the form quickly is the skill that makes the series free-response manageable.

Answer the 2 checkpoints as you read.