$\{a_n\}$ sequence $a_1=\sqrt{6}$ for $n \geq 1$ and $a_{n+1}=\sqrt{6+a_n}$ show it that convergence and also find $\lim_{x \to \infty} \{a_n\}$ [duplicate]

$\{a_n\}$ sequence $a_1=\sqrt{6}$ for $n \geq 1$ and $a_{n+1}=\sqrt{6+a_n}$ show that it convergence and as well find $\lim \limits_{n \to \infty} a_n$

In order to show that that sequence convergence I need to show that :

$$\lim_{n \to \infty} a_n= L$$

While $L$ is finite.

Using the calculator. I assume that L=3 because :

$$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6........}}}}=2.999 \cong 3$$

I really don't think that this method is good enough to established that $\lim \limits_{n \to \infty} a_n= 3$ since it based on intuition.

I'll be glad to hear any ideas for an established method to show this.?

Any help will be appreciated.

Solution 1:

Hint.

You may prove

• by induction, that your sequence is increasing: $\quad a_{n}\leq a_{n+1}$, $\quad n=1,2,3,...$.
• by induction, that your sequence is bounded: $\quad a_{n}\leq 3$, $\quad n=1,2,3,...$.

Solution 2:

You can prove by induction that $a_n$ is increasing and is bounded from above by $3$. The induction steps are: $$ a_{n+2}=\sqrt{6+a_{n+1}}\geq\sqrt{6+a_n}=a_{n+1} $$ and $$ a_{n+1}=\sqrt{6+a_n}\leq\sqrt{6+3}=3. $$ Together, these properties say that there is some $L$ such that $a_n\to L$. Then, you can do the usual trick that $L=\sqrt{6+L}$ to solve for $L$.

Solution 3:

Hints:

$ 0 \le a_1 \le 3$ and for $n \in \Bbb N\;\;$ $ 0 \le a_n \le 3 \implies 0 \le a_n + 6 \le 9 \implies a_{n + 1} = \sqrt{a_n + 6} \le 3$

Hence by induction the sequence is bounded above.

Furthermore, $L$ must satisfy $ L = \sqrt{L + 6 } $ and $L \ge 0$