# $\{a_n\}$ 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.

## marked as duplicate by YuiTo Cheng, José Carlos Santos calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 11 at 10:22

• I just want to make a comment on your notation. I think by $\lim_{x \to \infty} \{ a_{n} \}$ you mean $\lim_{n \to \infty} a_{n}$. You really shouldn't keep the curly braces around the $a_{n}$'s when writing it with the limit because $\lim_{n \to \infty} \{ a_{n} \}$ to me depicts the limit of the sets $\{a_{n} \}$, while you really are asking for the limit of the sequence, i.e., $\lim_{n \to \infty} a_{n}$. – layman Mar 12 '15 at 23:51
• Do you know the Monotone Convergence Theorem? – Ishfaaq Mar 12 '15 at 23:52
• – YuiTo Cheng Jul 11 at 8:14

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 $L$.

• That's established that $L=\sqrt{L+6}$ But can you explain how can you conclude that the $L=\sqrt{3+6}=\sqrt{9}=3$? – JaVaPG Mar 13 '15 at 0:28
• @JaVaPG From $L=\sqrt{L+6}$, you get $0=L^2-L-6=(L-3)(L+2)$ so $L=3$. As for how I guessed the upperbound $3$, I just simply used $L$, but if you were uncomfortable with $3$, you could use a more conservative bound, say, $4$: $\sqrt{6+a_n}\leq\sqrt{6+4}<\sqrt{16}=4$. All that mattered was that $a_n$ is increasing and bounded from above. – Kim Jong Un Mar 13 '15 at 2:17

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: $\quad a_{n}\leq 3$, $\quad n=1,2,3,...$.

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 $L \ge 0$