Question

Define \( s_1 = \alpha > 0 \) and \( s_{n+1} = \sqrt{\dfrac{1 + s_n^2}{1 + \alpha}}, \, n \geq 1. \) Which of the following is true?

• A. If \( s_n^2 < \dfrac{1}{\alpha} \), then \(\{s_n\}\) is monotonically increasing and \(\displaystyle \lim_{n \to \infty} s_n = \dfrac{1}{\sqrt{\alpha}}\)
• B. If \( s_n^2 < \dfrac{1}{\alpha} \), then \(\{s_n\}\) is monotonically decreasing and \(\displaystyle \lim_{n \to \infty} s_n = \dfrac{1}{\alpha}\)
• C. If \( s_n^2 > \dfrac{1}{\alpha} \), then \(\{s_n\}\) is monotonically increasing and \(\displaystyle \lim_{n \to \infty} s_n = \dfrac{1}{\sqrt{\alpha}}\)
• D. If \( s_n^2 > \dfrac{1}{\alpha} \), then \(\{s_n\}\) is monotonically decreasing and \(\displaystyle \lim_{n \to \infty} s_n = \dfrac{1}{\sqrt{\alpha}}\)

Hint:

To find the limit of a recurrence sequence, assume convergence to \( L \) and substitute into the recurrence. Then use inequalities to determine monotonicity.

Answer 1

The Correct Option is A

Step 1: Write the recurrence relation. 
\[ s_{n+1} = \sqrt{\dfrac{1 + s_n^2}{1 + \alpha}}. \]

Step 2: Assume the sequence converges to a limit \( L \). 
Then, taking limit on both sides, \[ L = \sqrt{\dfrac{1 + L^2}{1 + \alpha}}. \]

Step 3: Solve for \( L \). 
Squaring both sides: \[ L^2 = \dfrac{1 + L^2}{1 + \alpha} \Rightarrow L^2 (1 + \alpha) = 1 + L^2. \] \[ L^2 \alpha = 1 \Rightarrow L = \dfrac{1}{\sqrt{\alpha}}. \]

Step 4: Determine monotonicity. 
Consider the difference \( s_{n+1} - s_n \). If \( s_n^2 < \dfrac{1}{\alpha} \), then from the recurrence relation, \[ s_{n+1} = \sqrt{\dfrac{1 + s_n^2}{1 + \alpha}} > s_n. \] Thus, \(\{s_n\}\) is monotonically increasing. 
 

Step 5: Boundedness. 
The sequence is bounded above by \( \dfrac{1}{\sqrt{\alpha}} \), since as \( s_n \) increases, \( s_{n+1} \to \dfrac{1}{\sqrt{\alpha}}. \)

Step 6: Conclusion. 
The sequence \(\{s_n\}\) is increasing and bounded, hence convergent, with \[ \boxed{\lim_{n \to \infty} s_n = \dfrac{1}{\sqrt{\alpha}}.} \]