AP Calculus: Tests for Convergence (nth Term, Integral, p?Series, Comparison, Limit Comparison)

Tests for Convergence (nth Term, Integral, p?Series, Comparison, Limit Comparison)

Concept Summary

• nth-Term Test: If (\lim_{n \to \infty} a_n \neq 0), the series (\sum a_n) diverges; if the limit is 0, the test is inconclusive.
• Integral Test: If (f(n) = a_n) is continuous, positive, and decreasing, (\sum a_n) converges iff (\int_1^\infty f(x) \, dx) converges.
• p-Series: (\sum \frac{1}{n^p}) converges iff (p > 1).
• Comparison Test: If (0 \leq a_n \leq b_n) and (\sum b_n) converges, then (\sum a_n) converges; if (\sum a_n) diverges, then (\sum b_n) diverges.
• Limit Comparison Test: If (\lim_{n \to \infty} \frac{a_n}{b_n} = L) (finite and positive), then (\sum a_n) and (\sum b_n) either both converge or both diverge.

Core Questions

WHAT (definitional)

Q: What is the nth-Term Test? A: A test to determine divergence: if (\lim_{n \to \infty} a_n \neq 0), the series (\sum a_n) diverges. Trap/Clarification: A limit of 0 does not imply convergence (e.g., harmonic series).

Q: What is a p-series? A: A series of the form (\sum \frac{1}{n^p}), which converges only if (p > 1). Trap/Clarification: (p = 1) (harmonic series) is a classic divergent case.

WHY (causal/explanatory)

Q: Why is the Integral Test useful? A: It connects series convergence to improper integrals, leveraging calculus techniques for non-algebraic terms (e.g., (\frac{1}{n \ln n})). Trap/Clarification: The function (f(x)) must be decreasing—violating this invalidates the test.

Q: Why does the Limit Comparison Test work? A: It compares the "growth rate" of two series: if (\frac{a_n}{b_n}) approaches a finite positive limit, their convergence behaviors match. Trap/Clarification: (L = 0) or (\infty) is inconclusive (use Direct Comparison instead).

HOW (process/application)

Q: How do you apply the Comparison Test? A: Find a benchmark series (\sum b_n) where (0 \leq a_n \leq b_n) (or (a_n \geq b_n \geq 0)) and compare convergence/divergence. Trap/Clarification: The inequality must hold for all (n) (or eventually); one counterexample breaks the test.

Q: How is the Limit Comparison Test calculated? A: Compute (\lim_{n \to \infty} \frac{a_n}{b_n} = L). If (0 < L < \infty), both series share the same fate. Trap/Clarification: Simplify (\frac{a_n}{b_n}) before taking the limit (e.g., divide numerator/denominator by highest-degree term).

CAN (conditions/possibilities)

Q: Can the Integral Test be used for non-positive series? A: No; the function (f(x)) must be positive (and continuous/decreasing) for the test to apply. Trap/Clarification: For alternating series, use the Alternating Series Test instead.

Q: Under what conditions does the nth-Term Test guarantee convergence? A: Never; it only guarantees divergence if (\lim a_n \neq 0). Trap/Clarification: Convergence requires additional tests (e.g., p-Series, Comparison).

Quick Facts & Traps

• Fact: The harmonic series ((\sum \frac{1}{n})) diverges despite (\lim \frac{1}{n} = 0).
• Trap: Assuming (\sum a_n) converges if (a_n \to 0)-Reality: The nth-Term Test is one-way (divergence only).
• Fact: For the Integral Test, the lower bound of the integral can be any (a \geq 1) (not just 1).
• Trap: Using (\sum \frac{1}{n^2}) as a comparison for (\sum \frac{1}{n^2 + 1}) without verifying (a_n \leq b_n)-Reality: The inequality must hold for all (n).
• Fact: The Limit Comparison Test is ideal when (a_n) and (b_n) are "similar" (e.g., rational functions with same leading terms).
• Trap: Forgetting to check if (L) is finite/positive-Reality: (L = 0) or (\infty) requires a different approach.

Rapid-Fire True/False

• Statement: If (\lim_{n \to \infty} a_n = 0), then (\sum a_n) converges. Answer: FALSE Why the common mistake happens: Overgeneralizing the nth-Term Test’s inconclusive case.

• Statement: The series (\sum \frac{1}{n^{0.9}}) converges because (0.9 < 1). Answer: FALSE Why the common mistake happens: Misapplying the p-Series rule (converges only if (p > 1)).

• Statement: The Integral Test can be used to prove (\sum \frac{\sin n}{n^2}) converges. Answer: FALSE Why the common mistake happens: (\sin n) is not always positive (Integral Test requires positivity).