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Consider the sequence $x_1 = 0 $, $ x_n = \frac{1}{4} x_{n-1}^2 + 1$. Prove the limit exists by showing monotone and bounded, then find the limit. I am having difficulty proving that is monotone (increasing) and bounded.

Proving this sequence is monotone by induction:

Base case: $x_0 = 0, x_1 = 1$

In general: $$x_n = \frac{1}{4} x_{n-1}^2 + 1$$ $$ x_{n+1} = \frac{1}{4}x_n ^2 + 1 = \frac{1}{16}x_{n-1}^2 + \frac{5}{4} $$ How do I go on to prove that the sequence is increasing? Would it be easier to prove that it has an upper bound of 2 and use that?

To find the limit: $$ L = lim_{n \to \infty} \frac{1}{4}x_n ^2 +1 = \frac{1}{4} (lim_{n \to \infty} x_n)^2 + 1 = \frac{1}{4}L^2 +1 $$ Then solving $ L = \frac{1}{4}L^2 +1 $, I get L = 2

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Note that $x_n-x_{n-1}=(x_{n-1}/2-1)^2\ge0$.

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you need to show $$x_{n+1}-x_{n}\geq 0$$so $$x_n = \frac{1}{4} x_{n-1}^2 + 1\\ x_{n+1} = \frac{1}{4} x_{n}^2 + 1\\ x_{n+1} -x_n= \frac{1}{4} x_{n}^2-x_n + 1\\ x_{n+1} -x_n= \frac{1}{4}( x_{n}^2-4x_n + 4)=\frac{1}{4}( x_{n}-2)^2\geq 0$$

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  • $\begingroup$ Thanks, how do I prove boundedness? $\endgroup$ – kt046172 Dec 8 '19 at 21:17

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