AP Calculus BC 10.6 Comparison Tests for Convergence Study Notes - New Syllabus
AP Calculus BC 10.6 Comparison Tests for Convergence Study Notes- New syllabus
AP Calculus BC 10.6 Comparison Tests for Convergence Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Applying limits may allow us to determine the finite sum of infinitely many terms.
Key Concepts:
- Comparison Tests for Convergence
Comparison Tests for Convergence
Comparison Tests for Convergence
Comparison tests are used to determine whether an infinite series converges or diverges by comparing it to another series whose behavior (convergent or divergent) is already known.
There are two main types: the Direct Comparison Test (DCT) and the Limit Comparison Test (LCT).
1. Direct Comparison Test (DCT):
- Suppose \( 0 \leq a_n \leq b_n \) for all \( n \) large enough.
- If \( \sum b_n \) converges, then \( \sum a_n \) also converges.
- If \( \sum a_n \) diverges, then \( \sum b_n \) also diverges.
- This works because if a “bigger” convergent series exists, the smaller one must also converge; if a “smaller” divergent series exists, the bigger one must also diverge.
2. Limit Comparison Test (LCT):
- Given positive terms \( a_n, b_n > 0 \), compute \( L = \displaystyle \lim_{n \to \infty} \dfrac{a_n}{b_n} \).
- If \( 0 < L < \infty \), both series either converge or both diverge.
- If \( L = 0 \) and \( \sum b_n \) converges, then \( \sum a_n \) converges.
- If \( L = \infty \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges.
- LCT is especially useful when terms are complicated but have the same asymptotic behavior as a known \( p \)-series or geometric series.
Example
Test the convergence of \( \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2 + 5} \).
▶️ Answer/Explanation
We compare with the series \( \displaystyle \sum \dfrac{1}{n^2} \), which is a convergent \( p \)-series with \( p = 2 > 1 \).
For all \( n \geq 1 \), we have \( 0 < \dfrac{1}{n^2 + 5} \leq \dfrac{1}{n^2} \).
Since \( \sum \dfrac{1}{n^2} \) converges and our given series is smaller term-by-term, the given series also converges by the Direct Comparison Test.
Final Answer: Convergent.
Example
Determine the convergence of \( \displaystyle \sum_{n=1}^\infty \dfrac{n^2 + 3}{n^3 + 4n} \).
▶️ Answer/Explanation
For large \( n \), the leading term in the numerator is \( n^2 \) and in the denominator is \( n^3 \), so the series behaves like \( \dfrac{1}{n} \).
Let \( b_n = \dfrac{1}{n} \), the harmonic series, which diverges.
We compute:
\( \displaystyle L = \lim_{n \to \infty} \dfrac{\dfrac{n^2 + 3}{n^3 + 4n}}{\dfrac{1}{n}} = \lim_{n \to \infty} \dfrac{n^2 + 3}{n^3 + 4n} \cdot n = \lim_{n \to \infty} \dfrac{n^3 + 3n}{n^3 + 4n} = 1 \).
Since \( 0 < L < \infty \) and \( \sum b_n \) diverges, the given series also diverges.
Final Answer: Divergent.
Example
Which of the following series converge?
\( \text{(A) } \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2 + n} \)
\( \text{(B) } \displaystyle \sum_{n=1}^\infty \dfrac{n}{n^3 + 5} \)
\( \text{(C) } \displaystyle \sum_{n=1}^\infty \dfrac{2n^2 + 1}{n^4 + n^2} \)
\( \text{(D) } \displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}} \)
▶️ Answer/Explanation
Option (A):
\( \dfrac{1}{n^2 + n} \leq \dfrac{1}{n^2} \) for \( n \geq 1 \). Since \( \sum \dfrac{1}{n^2} \) converges (\( p = 2 > 1 \)), (A) converges by the Direct Comparison Test.
Option (B):
For large \( n \), \( \dfrac{n}{n^3 + 5} \approx \dfrac{1}{n^2} \), a convergent \( p \)-series (\( p = 2 \)). LCT with \( b_n = \dfrac{1}{n^2} \) gives \( L = 1 \), so (B) converges.
Option (C):
For large \( n \), \( \dfrac{2n^2 + 1}{n^4 + n^2} \approx \dfrac{2}{n^2} \), which is a convergent \( p \)-series (\( p = 2 \)). Thus (C) converges.
Option (D):
\( \dfrac{1}{\sqrt{n}} = \dfrac{1}{n^{1/2}} \) is a \( p \)-series with \( p = 1/2 \leq 1 \), which diverges. So (D) diverges.
Final Answer: (A), (B), and (C) converge; (D) diverges.