Question

Let \( a_1 = 5 \) and define recursively \[ a_{n+1} = \frac{1}{3} \left(a_n\right)^{\frac{3}{4}}, \quad n \ge 1. \] Then, which of the following statements is TRUE?

• A. \(\{a_n\}\) is monotone increasing, and \(\lim_{n \to \infty} a_n = 3\)
• B. \(\{a_n\}\) is monotone decreasing, and \(\lim_{n \to \infty} a_n = 3\)
• C. \(\{a_n\}\) is non-monotone, and \(\lim_{n \to \infty} a_n = 3\)
• D. \(\{a_n\}\) is decreasing, and \(\lim_{n \to \infty} a_n = 0\)

Hint:

For recursive sequences of the form \( a_{n+1} = f(a_n) \), fixed points are found by solving \( f(L) = L \), and stability is checked by comparing \( |f'(L)|<1 \).

Answer 1

The Correct Option is B

Step 1: Determine the fixed point.
Let the limit be \(L\). Then, taking limit on both sides, \[ L = \frac{1}{3} L^{3/4}. \] If \(L>0\), we get \(L^{1/4} = \frac{1}{3} \Rightarrow L = \frac{1}{81}\). However, since \(a_1 = 5\), we need to check the direction of monotonicity.
Step 2: Check monotonicity.
Compute a few terms: \[ a_2 = \frac{1}{3}(5)^{3/4} \approx \frac{1}{3} (3.34) = 1.11, \] \[ a_3 = \frac{1}{3}(1.11)^{3/4} \approx 0.37, \quad a_4 \approx 0.18. \] Hence, the sequence is decreasing.
Step 3: Find the limit behavior.
Since \(a_n>0\) and \(a_{n+1}<a_n\), it is monotone decreasing and bounded below by 0. Therefore, \[ \lim_{n \to \infty} a_n = 0. \] Final Answer: \[ \boxed{\{a_n\} \text{ is decreasing, and } \lim_{n \to \infty} a_n = 0.} \]