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Computing the first few terms $$a_1=1, a_2=\sqrt{3}=1.732....,a_3=1.9318....,a_4=1.9828...$$ I feel that $(a_n)_{n\in \mathbb{N}}$ is bounded above by 2, although I have no logical reasoning for this. Since, $(a_n)_{n\in \mathbb{N}}$ is monotone increasing sequence, it must converge by monotone convergence theorem, and converge to 2.

Can anyone help me to make this more formal? Besides, I would really appreciate if anyone could shed some light on how to find the bound and limit of such sequences (that are not in closed form but in recursion).

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marked as duplicate by Martin Sleziak, user127.0.0.1, Davide Giraudo, user61527, M Turgeon Mar 4 '14 at 20:12

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    $\begingroup$ You can (try to) show that $0 \leqslant x < 2 \Rightarrow x < \sqrt{2+x} < 2$. Then you know the sequence has a limit. This inequality also shows that the limit can't be smaller than $2$. If the sequence is defined by a recursion $a_n = f(a_{n-1})$, when $f$ is continuous, the limit $\lambda$ (if it exists) must be a fixed point of $f$, i.e., it must satisfy $f(\lambda) = \lambda$. $\endgroup$ – Daniel Fischer Sep 30 '13 at 20:35
  • $\begingroup$ sorry Daniel, we haven't learn continuity yet! So, I am not allowed to use it! But that was nice idea. I will use it in future. $\endgroup$ – math Sep 30 '13 at 20:40
  • $\begingroup$ Continuity is not some special thing, it's just a statement about our ability to constrain the value of a function by constraining its inputs. $\endgroup$ – AJMansfield Sep 30 '13 at 23:18
  • $\begingroup$ See math.stackexchange.com/questions/115501/… and math.stackexchange.com/questions/449592/… (Also other post are linked there.) $\endgroup$ – Martin Sleziak Mar 4 '14 at 18:29
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Hints:

$$a_1\le a_2\;\;\text{and}\;\;a_{n+1}:=\sqrt{2+a_n}\stackrel{\text{induction}}\le\sqrt{2+a_{n+1}}=:a_{n+2}$$

$$a_1\le 2\;\;\text{and}\;\; a_{n+1}:=\sqrt{2+a_n}\stackrel{\text{induction}}\le\sqrt{2+2}=2$$

The above shows your sequence is a monotone ascending one and bounded above, so its limit exists, say it is $\;w\;$, and now a little arithmetic of limits:

$$w\leftarrow a_{n+1}=\sqrt{2+a_n}\rightarrow\sqrt{2+w}$$

so $\;w=\ldots\ldots?$

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  • $\begingroup$ nice! Is there any idea to find quickly whether recursively defined sequences are bounded? $\endgroup$ – math Sep 30 '13 at 20:51
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    $\begingroup$ Not really, imo...inspection, I guess: take good, long look at the sequence and try to make an educated guess what an upper/lower bound of it could be. Now prove it rigorously. $\endgroup$ – DonAntonio Sep 30 '13 at 20:52
  • $\begingroup$ I was also having trouble to compute the limit of $\frac{a_{n+1}}{a_n}$ where $a_n$ are the terms in the Fibonacci sequence. Your idea was simple and elegant and really helped a lot! Thanks. $\endgroup$ – math Sep 30 '13 at 21:16
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    $\begingroup$ The last formula depends on continuity of $\sqrt{}$, which the OP stated above (in a response to Daniel) was disallowed? $\endgroup$ – copper.hat Sep 30 '13 at 21:19
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$$ a_{n+1}-2 = \sqrt{2+a_n}-2 $$ $$ a_{n+1}-2 = \frac{a_n-2}{\sqrt{2+a_n}+2} $$ $$ |a_{n+1}-2| \le \frac{1 }{2}.|a_n-2| $$ $$ |a_{n}-2| \le {(\frac{1 }{2})}^n.|a_0-2| $$ hence $$ a_n \to 2 $$

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  • $\begingroup$ Nice solution, very concrete. $\endgroup$ – André Nicolas Sep 30 '13 at 21:31
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Here is an alternate way to approach this problem.

Set $f(x) = \sqrt{2 + x}$, then we notice $f(x) : [0,\infty)\rightarrow[0,\infty)$

and $f'(x) = \frac{1}{2\sqrt{2 + x}} \le \frac{1}{2\sqrt{2}}<1$ for $x\in (0,\infty)$

So by the contraction mapping principle $f$ has a unique fixed point in $(0,\infty)$ and any iteration will converge to to this fixed point. This rigorously justifies just solving the equation

$$x = \sqrt{2 + x}$$ to get the limit of the sequence $a_n$.

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  • $\begingroup$ Very minor point: To directly use the contraction mapping principle, you need to work on a complete space. So, you should have $[0,\infty)$ above rather than $(0,\infty)$. Nothing else needs to be changed. $\endgroup$ – copper.hat Sep 30 '13 at 23:11
  • $\begingroup$ On the other hand, the proof of the contraction mapping principle in this case is very concrete, and amounts to JaCk091's answer. $\endgroup$ – Andrés E. Caicedo Sep 30 '13 at 23:55
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Note that $(2-\sqrt{2+x}) (2+\sqrt{2+x} )= 2-x$. We assume $x \ge 0$ in the following.

In particular, $\sqrt{2+x} \le 2$ iff $x \le 2$.

Then, if $x \le 2$, we have $2-\sqrt{2+x} = \frac{1}{2+\sqrt{2+x}} (2-x) \le \frac{1}{2+\sqrt{2}}(2-x)$.

Hence $2-a_{n+1} \le \frac{1}{2+\sqrt{2}} (2-a_n)$, from which we get $0 \le 2-a_n \le \frac{1}{(2+\sqrt{2})^{n-1}} (2-a_1)$. Hence $a_n \uparrow 2$.

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