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We have a sequence: $$a_1=\sqrt{6}$$ $$a_{n+1} = \sqrt{6+a_n}$$

The problem is to check convergence and then find the limit. We know that sequence converges when it is monotonic and bounded. With a little trying I can guess, that this sequence must be bounded by 3 and it's monotonic. Then I can prove my guess:

Proving upper boundary I guessed $a_n<3$:

  • For $a_1 = \sqrt{6} < 3$
  • $a_{n+1}= \sqrt{6+a_n}$
  • $a_{n+1} = \sqrt{6+3} = 3$

However, I personally don't like this proof. It requires guessing and checking if I was right.

Later I could use this and prove monotonicity, but I'd like to ask you for some better proofs you may know.

stil
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4 Answers4

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You just need to guess that it is indeed bounded. In that case, the limit if it exists will be a bound (since the sequence is monotonic). In this case, it is easy to see that the limit if it exists is 3 (by solving $a^2 - a - 6 = 0$). So $a_n < 3$ does indeed hold, it is just a matter of proving that by induction.

Wonder
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To avoid "guessing" first establish the fixed points of the recurrence, i.e. the solutions of $$a=\sqrt{6+a}$$ or $$a^2-a-6=0$$ This gives $a=3, a=-2$ and since $a$ is positive the likely limit is $3$ (it could be infinite, for example).

Note that $a_1=\sqrt 6\lt 3$ and if $a_n\lt 3$ then $$a_{n+1}^2=a_n+6\lt 9 \dots (1)$$so $a_{n+1}\lt 3$ and the elements of the recurrence are positive and bounded above by $3$.

Now note also that $a_{n+1}^2-a_n^2=a_n-a_{n-1}$ (by subtracting successive instances of the equation which appears in $(1)$ above) which shows that the recurrence is monotonic. It is easy to show that it is increasing.

Mark Bennet
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Define the function $f$ on $\mathbb{R}_+$ by $$\forall x\in\mathbb{R}_+,\ f(x)=\sqrt{6+x}.$$ For all $x\in\mathbb{R}_+$, $$f'(x)=\frac1{2\sqrt{6+x}}$$ hence $$\bigl\lvert f'(x)\bigr\rvert\leq\frac1{2\sqrt6}<1.$$ Hence $f$ is a contraction mapping on $\mathbb{R}_+$ and since the codomain of $f$ is included in the domain of $f$ and $\mathbb{R}_+$ is closed we can conclude, by the fixed point theorem, that $f$ possesses a unique fixed point $c\in\mathbb{R}_+$; moreover, for all $a_0\in\mathbb{R}_+$, the sequence $(a_n)_{n\in\mathbb{N}}$ defined by $$\forall n\in\mathbb{N},\ a_{n+1}=f(a_n)$$ converges to $c$. Moreover, $$\forall n\in\mathbb{N},\ \lvert a_n-c\rvert\leq\frac{k^n}{1-k}\lvert a_0-a_1\rvert$$ where $$k=\frac1{2\sqrt6}.$$

Conclusion: your sequence converges to $c$ and you have a rough idea of how it converges to $c$.

It's quite easy to determine $c$ explicitly by solving $f(c)=c$ (left to the reader—but you already have it: $c=3$).

A note about the variations of $(a_n)$: since $f$ is increasing, $(a_n)$ is monotone, so you only need to compare $a_0$ and $a_1$. In your case, $a_0=\sqrt{6}$ and $a_1=\sqrt{6+\sqrt{6}}>a_0$, hence $(a_n)$ is increasing.

That was a good proof I know.

gniourf_gniourf
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Once you guessed that the limit, if it exists, is $3$ because this is the intercept of the first diagonal $y=x$ and of the graph $y=\sqrt{x+6}$, you can check the convergence, using the identity $$ a_{n+1}-3=\frac{a_n-3}{a_{n+1}+3}. $$ Since $a_{n+1}\geqslant0$, this implies that $$ |a_{n+1}-3|\leqslant\frac13\cdot|a_n-3|.$$

Did
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