10.6 Comparison Tests for Convergence

Direct Comparison Test

If 0 ≤ aₙ ≤ bₙ for all n (eventually), then: if Σbₙ converges, so does Σaₙ; if Σaₙ diverges, so does Σbₙ. Think of it as trapping your unknown series between a known one. You must find a benchmark series (often geometric or p-series) that's always bigger or always smaller for large n.

Limit Comparison Test

For positive series aₙ and bₙ, compute L = lim(n→∞) aₙ/bₙ. If 0 < L < ∞, both series converge or both diverge together. This is useful when direct inequality is messy — just pick bₙ that matches the dominant behavior of aₙ as n→∞ (e.g., drop lower-order terms).

Choosing benchmark series

Common comparisons: p-series Σ1/n^p (converges if p>1, diverges if p≤1) and geometric series Σr^n (converges if |r|<1). For rational/algebraic expressions, compare to the highest-power term; for series with exponentials, compare to a geometric series.

Key terms

Direct Comparison Test

If 0 ≤ aₙ ≤ bₙ, Σbₙ converges ⟹ Σaₙ converges; Σaₙ diverges ⟹ Σbₙ diverges.

Limit Comparison Test

If lim aₙ/bₙ = L with 0 < L < ∞, then Σaₙ and Σbₙ share the same convergence behavior.

p-series convergence rule

Σ1/n^p converges if p > 1, diverges if p ≤ 1.

Geometric series convergence rule

Σr^n converges if |r| < 1, diverges otherwise.

Choosing a comparison series

Match dominant behavior for large n — drop lower-order terms and compare to a p-series or geometric series.

Try an AP-style question

Why does Σ 1/(n² + 1) converge?

1. It is a geometric series
2. 1/(n² + 1) > 1/n², and Σ 1/n² converges
3. Its terms approach 0
4. 0 < 1/(n² + 1) < 1/n², and Σ 1/n² converges