Define $x_1:=\sqrt{2}$ and recursively $x_n=\sqrt{x_{n-1}+2}$ for every $n\geq 2$.

How can I show $(x_n)_{n\in\mathbb N}$ is bounded? What is the limit $\displaystyle\lim_{n\to \infty} x_n$?

Indeed, I was able to show $(x_n)_{n\in\mathbb N}$ is bounded showing it's a Cauchy sequence. For this I first showed the inequality: $$|x_{n+2}-x_{n+1}|<\frac{1}{2}|x_{n+1}-x_n|.$$ However, that was too much work. Can I show the sequence is bounded directly?


  • $\begingroup$ Try proving by induction that $x_n \leq 2$ for all $n$. $\endgroup$ Jun 16, 2015 at 12:47
  • 1
    $\begingroup$ To show the sequence is bounded, just use the fact the function $x \mapsto \sqrt{x+2}$ is strictly increasing (i.e. order preserving) and $2$ is a fixed point of the function. So if you start from any number $y < 2$, you will have $\sqrt{y+2} < 2$. Apply this to you sequence start from $x_1$, followed by $x_2$, $x_3$, etc. $\endgroup$ Jun 16, 2015 at 13:11

3 Answers 3


You can prove this following this steps:

  1. Prove that the sequence is bounded above by 2.

  2. Prove that the sequence is increasing.

This two conditions imply that the sequence has a limit.

  1. Calculate the limit $x=\lim x_n$ to get the equation $$x=\sqrt{x+2}$$

  2. Find the limit value.


You probably want $\lim_{n\to\infty}x_n$, rather than $\lim_{n\to\infty}\sqrt{x_n+2}$ (but they're actually equal).

Let's show that $x_n<2$ for every $n$. The base step is clear, so assume $x_{n-1}<2$; then $$ x_n^2=x_{n-1}+2<4 $$ which shows $x_n<2$ as desired.

Now let's prove the sequence is increasing. We want $x_{n+1}\ge x_{n}$, that is $\sqrt{x_n+2}\ge x_n$ or $$ x_n+2\ge x_n^2 $$ that becomes $$ x_n^2-x_n-2\le0 $$ or $$ (x_n+1)(x_n-2)\le0 $$ which is true because of what we showed before.

  • $\begingroup$ @robjohn I substitute inequalities with equivalent ones. $\endgroup$
    – egreg
    Jun 16, 2015 at 13:04

By inspection we can see that it will be $< \sqrt{2}$ and that because we are always adding something on the sequience will be monotonically increasing.

The final element in our infinate series is $x = \sqrt{x + 2}$

solve for $x$

$$ x^2 = x+2 \Rightarrow x^2 - x -2 = 0$$

From the well known formula

$$x =\dfrac{-(-1)\pm \sqrt{(-1)^2 - 4 \cdot (1) \cdot(-2)}}{2 \cdot (1)}$$

$$ x = \frac{1 \pm \sqrt{9}}{2} = 2 \text{ or } -1$$

We have already ruled out anything less than $\sqrt{2}$ so the answer is $2$


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