AP Calculus BC 10.8 Ratio Test for Convergence Study Notes - New Syllabus
AP Calculus BC 10.8 Ratio Test for Convergence Study Notes- New syllabus
AP Calculus BC 10.8 Ratio 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:
- Ratio Test for Convergence
Ratio Test for Convergence
Ratio Test for Convergence
The Ratio Test is a powerful tool for determining the convergence or divergence of infinite series, particularly when the series contains factorials, exponential functions, or powers of \( n \).
It works by examining how fast the terms of the series shrink (or grow) as \( n \) becomes large. The idea is to look at the ratio of successive terms \( \dfrac{a_{n+1}}{a_n} \) in absolute value and check if this ratio is consistently less than 1 for large \( n \).
Formal Statement of the Ratio Test:
Let \( \displaystyle \sum_{n=1}^\infty a_n \) be a series, and define
\( L = \lim_{n \to \infty} \left| \dfrac{a_{n+1}}{a_n} \right| \).
- If \( L < 1 \) → The series converges absolutely (and hence converges).
- If \( L > 1 \) or \( L = \infty \) → The series diverges.
- If \( L = 1 \) → The test is inconclusive, and another convergence test must be used.
Why it works:
The Ratio Test compares the given series to a geometric series. If the magnitude of each term is eventually less than a fixed fraction of the previous term (\( L < 1 \)), the series behaves like a geometric series with ratio less than 1, and therefore converges. If the terms do not shrink fast enough (\( L \ge 1 \)), the series either stays the same size or grows, leading to divergence.
When to use:
- Series with factorials \( n! \)
- Series with exponential terms \( a^n \)
- Series with products or quotients of polynomials and exponentials
- When other basic tests like the Geometric Series Test or p-Series Test do not apply
Example
Test the convergence of \( \displaystyle \sum_{n=1}^\infty \dfrac{n!}{n^n} \).
▶️ Answer/Explanation
\( a_n = \dfrac{n!}{n^n} \)
\( \dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{(n+1)^{n+1}} \cdot \dfrac{n^n}{n!} = \dfrac{n+1}{(n+1)^{n+1}} \cdot n^n = \dfrac{n^n}{(n+1)^n} \)
This is \( \left( \dfrac{n}{n+1} \right)^n \). As \( n \to \infty \), \( \left( \dfrac{n}{n+1} \right)^n \to e^{-1} \approx 0.367 < 1 \).
By the Ratio Test, the series converges absolutely.
Final Answer: Converges absolutely.
Example
Test the convergence of \( \displaystyle \sum_{n=1}^\infty \dfrac{3^n}{n!} \).
▶️ Answer/Explanation
\( a_n = \dfrac{3^n}{n!} \)
\( \dfrac{a_{n+1}}{a_n} = \dfrac{3^{n+1}}{(n+1)!} \cdot \dfrac{n!}{3^n} = \dfrac{3}{n+1} \)
Taking \( n \to \infty \), \( L = \lim_{n \to \infty} \dfrac{3}{n+1} = 0 \).
Since \( L < 1 \), the series converges absolutely.
Final Answer: Converges absolutely.
Example
For the series \( \displaystyle \sum_{n=1}^\infty \dfrac{n^5}{5^n} \), the Ratio Test gives:
A. \( L = \dfrac{1}{5} \), converges absolutely.
B. \( L = \dfrac{5}{n} \), diverges.
C. \( L = 1 \), inconclusive.
D. \( L = \infty \), diverges.
▶️ Answer/Explanation
\( a_n = \dfrac{n^5}{5^n} \)
\( \dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^5}{5^{n+1}} \cdot \dfrac{5^n}{n^5} = \dfrac{(n+1)^5}{5 n^5} \)
As \( n \to \infty \), \( \dfrac{(n+1)^5}{n^5} \to 1 \), so \( L = \dfrac{1}{5} \).
Since \( L = \dfrac{1}{5} < 1 \), the series converges absolutely.
Final Answer: A