We consider the limits of sequences defined by recurrence relations. We learn a technique for finding the limiting value by an equation: “If the sequence converges, what is its limit?”
Understanding Limits of Recurrence Relations
When a_{n+1} = f(a_n) converges to α, taking the limit n→∞ on both sides gives
α = f(α)
and the limiting value can be found (assuming convergence).
Example ①
Find the limiting value of a_1=1、a_{n+1}=(1/2)a_n+3.
Solution:α = (1/2)α + 3 → α = 6
Find the limiting value of a_1=1、a_{n+1}=(1/2)a_n+3.
Solution:α = (1/2)α + 3 → α = 6
General Terms and Convergence of Linear Recurrence Relations
For a_{n+1} = r·a_n + c, let α = c/(1−r). Then
a_{n+1} − α = r(a_n − α)
a_n = α + r^{n−1}(a_1 − α)
When |r|<1, a_n → α (convergence).
Example ②
Find the general term and limit of a_1=5、a_{n+1}=(1/3)a_n+2.
Solution:α = 2/(2/3) = 3. a_n = 3 + 2·(1/3)^{n−1}. lim = 3
Find the general term and limit of a_1=5、a_{n+1}=(1/3)a_n+2.
Solution:α = 2/(2/3) = 3. a_n = 3 + 2·(1/3)^{n−1}. lim = 3
Squeeze Theorem
If b_n ≦ a_n ≦ c_n and lim b_n = lim c_n = α, then lim a_n = α.
Example: lim (sin n)/n. Since −1/n ≦ (sin n)/n ≦ 1/n and both endpoints → 0, lim = 0.
Key Points
- Limit of a recurrence relation: solve α = f(α) (assuming convergence)
- Linear recurrence relation: α = c/(1−r), converges when |r|<1
- General term: a_n = α + r^{n−1}(a_1 − α)
- Use the squeeze theorem to bound a sequence above and below by the same limit
Practice Problems
- Find the limit of a_1=0、a_{n+1}=(1/4)a_n+3.
- Find lim_{n→∞} (cos n²)/n using the squeeze theorem.
- Find the general term and limit of a_1=8、a_{n+1}=(3/4)a_n+1.
Answers and Explanations
- Answer:4
Explanation: α = 3/(3/4) = 4. - Answer:0
Explanation: −1/n ≦ (cos n²)/n ≦ 1/n, and both endpoints → 0. - Answer:a_n = 4 + 4·(3/4)^{n−1}, lim = 4
Explanation: α = 1/(1/4) = 4. a_n − 4 = (3/4)^{n−1}·(8−4) = 4·(3/4)^{n−1}.