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$\{a_n\}$ 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.

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marked as duplicate by YuiTo Cheng, José Carlos Santos calculus Jul 11 at 10:22

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    $\begingroup$ 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., $\lim_{n \to \infty} a_{n}$. $\endgroup$ – layman Mar 12 '15 at 23:51
  • $\begingroup$ Do you know the Monotone Convergence Theorem? $\endgroup$ – Ishfaaq Mar 12 '15 at 23:52
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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 $L$.

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  • $\begingroup$ That's established that $L=\sqrt{L+6}$ But can you explain how can you conclude that the $L=\sqrt{3+6}=\sqrt{9}=3$? $\endgroup$ – JaVaPG Mar 13 '15 at 0:28
  • $\begingroup$ @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. $\endgroup$ – Kim Jong Un Mar 13 '15 at 2:17
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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: $\quad a_{n}\leq 3$, $\quad n=1,2,3,...$.
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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 $L \ge 0$

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