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).


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


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

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

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 $\;w=\ldots\ldots?$

  • $\begingroup$ 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

$$ a_{n+1}-2 = \sqrt{2+a_n}-2 $$ $$ a_{n+1}-2 = \frac{a_n-2}{\sqrt{2+a_n}+2} $$ $$ |a_{n+1}-2| \le \frac{1 }{2}.|a_n-2| $$ $$ |a_{n}-2| \le {(\frac{1 }{2})}^n.|a_0-2| $$ hence $$ a_n \to 2 $$

  • $\begingroup$ Nice solution, very concrete. $\endgroup$ – André Nicolas Sep 30 '13 at 21:31

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 $a_n$.

  • $\begingroup$ 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

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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