AP Calculus BC 10.7 Alternating Series Test for Convergence Study Notes - New Syllabus
AP Calculus BC 10.7 Alternating Series Test for Convergence Study Notes- New syllabus
AP Calculus BC 10.7 Alternating Series Test 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:
- Alternating Series Test for Convergence
Alternating Series Test for Convergence
Alternating Series Test for Convergence
An alternating series is a series whose terms alternate in sign, typically in the form:
\( \displaystyle \sum_{n=1}^\infty (-1)^{n-1} b_n \quad \text{or} \quad \sum_{n=1}^\infty (-1)^n b_n \), where \( b_n > 0 \).
Alternating Series Test (Leibniz’s Test):
The series \( \displaystyle \sum_{n=1}^\infty (-1)^{n-1} b_n \) converges if:
- The terms decrease: \( b_{n+1} \leq b_n \) for all \( n \) large enough.
- The terms approach zero: \( \lim_{n \to \infty} b_n = 0 \).
Absolute vs Conditional Convergence:
- Absolute Convergence: If \( \sum |a_n| \) converges, then \( \sum a_n \) converges absolutely.
- Conditional Convergence: If \( \sum a_n \) converges but \( \sum |a_n| \) diverges, the series is conditionally convergent.
Example
Determine whether \( \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n} \) converges.
▶️ Answer/Explanation
Here, \( b_n = \dfrac{1}{n} \).
Decreasing check: \( \dfrac{1}{n+1} < \dfrac{1}{n} \) for all \( n \), so the sequence is decreasing.
Limit check: \( \lim_{n \to \infty} \dfrac{1}{n} = 0 \), so the second condition holds.
By the Alternating Series Test, the series converges.
However, \( \sum \dfrac{1}{n} \) diverges, so convergence is conditional.
Final Answer: Conditionally convergent.
Example
Test \( \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^n}{n^2} \) for convergence.
▶️ Answer/Explanation
Here, \( b_n = \dfrac{1}{n^2} \).
Decreasing check: \( \dfrac{1}{(n+1)^2} < \dfrac{1}{n^2} \) for all \( n \), so the sequence decreases.
Limit check: \( \lim_{n \to \infty} \dfrac{1}{n^2} = 0 \), so the second condition holds.
By the Alternating Series Test, the series converges.
Also, \( \sum \dfrac{1}{n^2} \) converges (\( p = 2 > 1 \)), so the convergence is absolute.
Final Answer: Absolutely convergent.
Example
Which of the following series converge absolutely, converge conditionally, or diverge?
\( \text{(A) } \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{\sqrt{n}} \)
\( \text{(B) } \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^n}{n^{3/2}} \)
\( \text{(C) } \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^n}{n} \)
▶️ Answer/Explanation
Option (A): \( b_n = \dfrac{1}{\sqrt{n}} \) decreases and tends to 0, so the series converges by AST. However, \( \sum \dfrac{1}{\sqrt{n}} \) diverges (\( p = 1/2 < 1 \)), so convergence is conditional.
Option (B): \( b_n = \dfrac{1}{n^{3/2}} \) decreases and tends to 0, so converges by AST. Also, \( \sum \dfrac{1}{n^{3/2}} \) converges (\( p = 3/2 > 1 \)), so convergence is absolute.
Option (C): This is the alternating harmonic series, already known to converge conditionally (Example 1).
Final Answer: (A) Conditionally convergent; (B) Absolutely convergent; (C) Conditionally convergent.
Example
Consider the series \( \displaystyle \sum_{n=2}^\infty \dfrac{(-1)^n \ln n}{n} \). Which statement is correct?
A. The series converges absolutely.
B. The series converges conditionally.
C. The series diverges.
D. The series fails the Alternating Series Test.
▶️ Answer/Explanation
Here, \( b_n = \dfrac{\ln n}{n} \).
Decreasing check: For \( n \geq 3 \), the function \( f(x) = \dfrac{\ln x}{x} \) decreases because \( f'(x) = \dfrac{1 – \ln x}{x^2} < 0 \) for \( x > e \). So it eventually decreases.
Limit check: \( \lim_{n \to \infty} \dfrac{\ln n}{n} = 0 \), so the second condition holds.
Thus, the Alternating Series Test implies convergence.
Absolute convergence check: \( \sum \dfrac{\ln n}{n} \) diverges by the Integral Test (\( \int \dfrac{\ln x}{x} dx = \dfrac{(\ln x)^2}{2} \) grows without bound). So the series does not converge absolutely.
Conclusion: The series converges conditionally.
Final Answer: B