Question

Let \( \{a_n\} \) be the sequence of real numbers such that \[ a_1 = 1 \quad \text{and} \quad a_{n+1} = a_n + a_n^2 \quad \text{for all } n \geq 1. \] Then

• A. \( a_4 = a_1(1 + a_1)(1 + a_2)(1 + a_3) \)
• B. \( \lim_{n \to \infty} \frac{1}{a_n} = 0 \)
• C. \( \lim_{n \to \infty} \frac{1}{a_n} = 1 \)
• D. \( \lim_{n \to \infty} a_n = 0 \)

Hint:

When dealing with recurrence relations where terms grow rapidly, the sequence often tends to infinity, making the reciprocal approach zero.

Answer 1

The Correct Option is A, B

Step 1: Analyze the recurrence relation.
The recurrence \( a_{n+1} = a_n + a_n^2 \) is an increasing function. Since \( a_1 = 1 \), we compute a few terms: \[ a_2 = 1 + 1^2 = 2, \quad a_3 = 2 + 2^2 = 6, \quad a_4 = 6 + 6^2 = 42, \dots \] The sequence grows rapidly, and we observe that \( a_n \to \infty \) as \( n \to \infty \).
Step 2: Compute the limit.
Since \( a_n \to \infty \), we conclude that \( \frac{1}{a_n} \to 0 \) as \( n \to \infty \).
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(B)} \).