The behavior of the geometric sequence a_n = r^n with common ratio r changes greatly depending on the value of r. This is an important topic closely related to infinite geometric series and limits of exponential functions.
The Limit of r^n
As n → ∞, r^n is classified as follows.
- When r > 1: r^n → ∞ (diverges to positive infinity)
- When r = 1: r^n = 1 → 1 (converges)
- When |r| < 1 (−1 < r < 1): r^n → 0 (converges)
- When r = −1: r^n = +1, −1, +1, … → oscillation (divergence)
- When r < −1: The absolute value tends to infinity while the sign alternates → oscillation (divergence)
It converges only when −1 < r ≦ 1.
Example ①
Find the following limits.
(1) lim_{n→∞} (1/2)^n (2) lim_{n→∞} 3^n (3) lim_{n→∞} (−1/3)^n
Solution:
(1) |1/2| < 1 → 0
(2) 3 > 1 → ∞
(3) |−1/3| < 1 → 0
Find the following limits.
(1) lim_{n→∞} (1/2)^n (2) lim_{n→∞} 3^n (3) lim_{n→∞} (−1/3)^n
Solution:
(1) |1/2| < 1 → 0
(2) 3 > 1 → ∞
(3) |−1/3| < 1 → 0
Limits Containing Exponents in the Numerator and Denominator
For expressions such as (a^n + b^n)/(c^n + d^n), the standard method is to divide the numerator and denominator by the term whose base has the greatest absolute value.
Example: lim_{n→∞} (3^n − 2^n)/(3^n + 2^n). Dividing the numerator and denominator by 3^n gives
= lim (1 − (2/3)^n)/(1 + (2/3)^n) = (1 − 0)/(1 + 0) = 1
Example ②
Find lim_{n→∞} (4^n + 1)/(4^n − 2^n).
Solution: Divide the numerator and denominator by 4^n.
= lim (1 + 1/4^n)/(1 − (1/2)^n) = (1 + 0)/(1 − 0) = 1
Find lim_{n→∞} (4^n + 1)/(4^n − 2^n).
Solution: Divide the numerator and denominator by 4^n.
= lim (1 + 1/4^n)/(1 − (1/2)^n) = (1 + 0)/(1 − 0) = 1
Key Points
- r^n converges to 0 when |r| < 1 and to 1 when r = 1. Otherwise, it diverges.
- When r = −1 or r < −1, the sequence oscillates. “Oscillation is also divergence.”
- For limits involving sums or differences of powers, divide by the term with the greatest absolute base.
- Memorize the convergence range −1 < r ≦ 1 accurately.
Practice Problems
- Find lim_{n→∞} (2^n + 3^n)/(3^n + 1).
- State the range of r for which r^n converges.
- Find lim_{n→∞} (3^{n+1} − 2^n)/(3^n + 2^{n+1}).
Answers and Explanations
- Answer:1
Explanation: Divide the numerator and denominator by 3^n. ((2/3)^n + 1)/(1 + 1/3^n) → 1. - Answer:−1 < r ≦ 1
Explanation: It converges to 0 when |r|<1 and to 1 when r=1. - Answer:3
Explanation: 3^{n+1}=3·3^n and 2^{n+1}=2·2^n. Dividing the numerator and denominator by 3^n gives (3−(2/3)^n)/(1+2·(2/3)^n) → 3.