Proving Convergence with BMCT

We usually use BMCT on recursively defined functions.

Example

With the sequence $\{a_n\}$ converges where:

• $a_1=\sqrt{6}$
• $a_{n+1}=\sqrt{6+a_n}$ if $n\ge 1$

Proof

Proving Bounded Above

1. WTS that $\exists M\in \mathbb{R}$ s.t $\forall n\in \mathbb{N},a_n\le M$
2. Note that $a_1=\sqrt{6}<\sqrt{9}=3$
3. Similarly, $a_2=\sqrt{6+\sqrt{6}}<\sqrt{6+3}=\sqrt{9}=3$
4. Similarly, $a_3=\sqrt{6+\sqrt{6+\sqrt{6}}}<\sqrt{6+\sqrt{6+3}}=\sqrt{6+3}=\sqrt{9}=3$
5. So, choose $M=3$
6. Let $n\in \mathbb{N}$ be arbitrary
7. Proof By Induction
8. Choose $P(n)$ as $a_n<3$
9. Note that $P(1)=\sqrt{6}<3$ is true
10. Then, suppose $P(n)$
11. Consider $\sqrt{6+a_n}<\sqrt{6+3}=3$
12. Then, $P(n+1)$ holds by definition of the sequence
13. Thus, we know that $\forall n\in \mathbb{N},a_n<3$
14. Thus, $\{a_n\}$ is bounded above

Proving Strictly Increasing

1. Let $n\in \mathbb{N}$ be arbitrary
2. Consider $a_n^2-a_{n+1}^2$
3. $=a_n^2-{\sqrt{6-a_n}}^2$
4. $=a_n^2-6-a_n$
5. $=(a_n-3)(a_n+2)$
6. Note that the $a_n>0$ by sequence definition
7. Note that $a_n<3$ by 8
8. Thus, $(a_n-3)(a_n+2)<0$ by parity
9. This means $a_n^2-a_{n+1}^2<0$
10. $⟹a_n^2<a_{n+1}^2$
11. $⟹a_n<a_{n+1}$ as $\sqrt{\cdot }$ function is increasing, ineq prop
12. Then, we get $\{a_n\}$ is increasing by definition As we have shown $\{a_n\}$ is bounded above and increasing, then we have shown it converges by BMCT
13. Choose $x_1,x_2\in \mathbb{N}$ arbitrarily where $x_1<x_2$
14. We do Proof By Induction
15. With $P(n)$ as $a_n<a_{n+1}$
16. Note that $a_1=\sqrt{6}$
17. $a_2=\sqrt{6+\sqrt{6}}$