solving this recurrence

Question

I have this homework and it states,

Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of reals defined recursively: $$a_{n+1} = \sqrt{6+a_n},\quad a_1 = 0.$$ Prove that $\lim_{n\to\infty} a_n = 3$

Hint: prove that $|a_{n+1} -3| ≤ \frac13|a_n-3|$ for all $n \in \mathbb N$.

Answer 1

Let $f(x)=\sqrt{6+x}$ and $\alpha_{n+1}=f(\alpha_n), \forall n\in\mathbb{N}^*$,
we know that $f(3)=3$
and we know that $f$ is continuous in $\mathbb{R}$ $\Rightarrow$ $f$ is continuous in $\big[\inf(\alpha_n, 3), \sup(\alpha_n, 3)\big]$ (We don't know yet if $\alpha_n < 3$ or not), and it's differentiable in $\big]\inf(\alpha_n, 3), \sup(\alpha_n, 3)\big[$
Then, according to the Mean Value Theorem: $$\exists c\in\big]\inf(\alpha_n, 3), \sup(\alpha_n, 3)\big[\big/\big|f'(c)\big|\cdot\big|\alpha_n-3\big|=\big|f(\alpha_n)-f(3)\big|$$ $$\Rightarrow \big|f'(c)\big|\cdot\big|\alpha_n-3\big|=\big|\alpha_{n+1}-3\big|$$

We have $\forall x\in\big]\inf(\alpha_n, 3), \sup(\alpha_n, 3)\big[,\ f'(x)=\frac{1}{2\sqrt{6+x}}$
And since $3>0$ and $\alpha_n$ is positive ($\forall n\in\mathbb{N}^*$) then $x$ is positive $\Rightarrow x>-\frac{15}{4}$ $\Rightarrow \sqrt{6+x}\geq\frac{3}{2}\ \Rightarrow \big|f'(x)\big|\leq\frac{1}{3},\ \forall x\in\big]\inf(\alpha_n, 3), \sup(\alpha_n, 3)\big[$
$\Rightarrow \big|f'(c)\big|\leq\frac{1}{3} \Rightarrow \big|f'(c)\big|\cdot\big|\alpha_n-3\big|=\boxed{\big|\alpha_{n+1}-3\big|\leq\frac{1}{3}\big|\alpha_n-3\big|}$

Now we will try to prove that $\big|\alpha_n-3\big|\leq\big(\frac{1}{3}\big)^n, \ \forall n\in\mathbb{N}^*$ using Recursion
For $n=1$, $\alpha_1-3=-3\lt\frac{1}{3}$
We can see that the property is verified in this case.

Let $p\in\mathbb{N}^*\big/\big|\alpha_p-3\big|\leq\big(\frac{1}{3}\big)^p$
We know that $\forall p\in\mathbb{N}^*, \ \big|\alpha_{p+1}-3\big|\leq\frac{1}{3}\big|\alpha_p-3\big|$
$\Rightarrow \frac{1}{3}\big|\alpha_p-3\big|\leq\big(\frac{1}{3}\big)^{p+1}$
$\Rightarrow \big|\alpha_{p+1}-3\big|\leq\big(\frac{1}{3}\big)^{p+1}$
$\Rightarrow \boxed{\forall p\in\mathbb{N}^*, \ p\in\mathbb{N}^*\big/\big|\alpha_p-3\big|\leq\big(\frac{1}{3}\big)^p}$

$$\Rightarrow -\big(\frac{1}{3}\big)^n\leq\alpha_n-3\leq\big(\frac{1}{3}\big)^n$$

$$\lim_{n\to\infty} {-\big(\frac{1}{3}\big)^n}=\lim_{n\to\infty} {\big(\frac{1}{3}\big)^n}=0$$ $$\Rightarrow \lim_{n\to\infty}{\alpha_n-3}=0$$ $$\Rightarrow \boxed{\lim_{n\to\infty}{\alpha_n}=3}$$

Answer 2

Following the hint: It is easily seen that the $a_n$ stay below $3$. Therefore we introduce the quantities $$b_n:=3-a_n>0\qquad(n\geq1)\ .$$ Then $b_1=3$ and $$ b_{n+1}=3-a_{n+1}=3-\sqrt{6+(3-b_n)}=3-\sqrt{9-b_n}\ .$$ Therefore we have $$b_1=3,\qquad b_{n+1}={b_n\over3+\sqrt{9-b_n}}<{b_n\over3}\qquad(n\geq1)\ .$$

Answer 3

Define a sequence of $t_i$'s such that: $$t_i=a_{i+1}-a_i$$

Where $i∈ℕ^+$

Set $i=1$ we have: $$t_1=a_{2}-a_1=\sqrt{6+a_{1}}-a_{1}=\sqrt{6+0}-0=\sqrt{6}>0$$

Now assume $t_i>0$ for $2\le i<n+1$, setting $i↦n+1$ follows:

$$t_{n+1}=a_{n+2}-a_{n+1}=\sqrt{6+a_{n+1}}-\sqrt{6+a_{n}}=\frac{a_{n+1}-a_{n}}{\sqrt{6+a_{n+1}}+\sqrt{6+a_{n}}}$$ By our previous assumption it's clear that $t_{n+1}>0$$\tag{1}$

Hence the sequence is monotone increasing.

Now notice that $a_{2}=\sqrt{6+a_{1}}=\sqrt{6}<3$

Assume for an arbitrary $2\le i<n+1$ also we have the inequality $a_{i}<3$ holds and consider the case $i=n+1$: $$a_{n+1}=\sqrt{6+a_{n}}<\sqrt{6+3}=3$$$\tag{2}$

So using $(1)$ and $(2)$ implies the sequence is monotone increasing and bounded, hence convergent.

Besides denote the limit with $L$, then $$L=\lim_{n\to\infty} a_n = \lim_{n\to\infty}\sqrt{6+a_{n-1}}=\sqrt{\lim_{n\to\infty}6+a_{n-1}}=\sqrt{6+L}$$

Solving $L=\sqrt{6+L}$ there is two cases for $L$, it's either $-2$ or $3$, but since the first term of the sequence is $0$ and the sequence is increasing, implies the limit should be positive,e.g. $$\boxed {L=3}$$