I am working on the following exercise for an introductory Real Analysis course.

Suppose that $x_0 \geq 2$ and $x_n = 2 + \sqrt{x_{n-1}-2}$ for $n \in \mathbb{N}$. Use the Monotone Convergence Theorem to prove that either $\lim_{n \to \infty}x_n=2$ or $3$ or $\infty$.

I need some help to move in the right direction. I'll start with what I know so far. My first approach was to break the question into different cases.

I've found that if $x_0=2$, then $x_n =2$ for all $n \in \mathbb{N}$. So, if this is the case, the series is neither decreasing nor increasing? Furthermore, I've found that if $x_0 = 3$ then $x_n =3$ for all $n \in \mathbb{N}$. For $2 < x_0 < 3$, the series is monotone increasing, and converges to $3$. For $x_0 >3$, the series is monotone decreasing and converges to $3$. You will notice none of the cases I've given result in $\lim_{n \to \infty}x_n= \infty$. I have not proven any of the above.

Now, the Monotone Convergence Theorem states, according to my textbook:

A monotone increasing sequence that is bounded above converges.

A monotone decreasing sequence that is bounded below converges.

Now, if I can show that the series is monotone increasing/decreasing and bounded above/below for the different cases, I believe I can use the following.

$$L=\lim_{n \to \infty} x_{n+1}=\lim_{n \to \infty} 2+\sqrt{x_n -2}=2 + \sqrt{\lim_{n \to \infty}x_n-2}=2+ \sqrt{L-2}$$



Which, solving for $L$, yields the solutions $L=2$ and $L=3$. This makes sense according to what I claimed earlier. Once again, I have not found under what conditions the series is divergent. Any help would be appreciated.

Also, any advice as to proving that the series is either monotone increasing and bounded above, or monotone decreasing and bounded below, under the different conditions for $x_0$, would be appreciated.

  • 2
    $\begingroup$ You are certainly right that the sequence cannot have limit "$\infty$." $\endgroup$ – André Nicolas Jan 24 '13 at 19:15

You can get a lot of mileage from just plotting the graph of $f(x)=2+\sqrt{x-2}$. Note that $f$ is increasing and concave, the graph crosses the diagonal $y=x$ at $x=2$ and $x=3$. Now draw some pictures showing how the iteration $x_n=f(x_{n-1})$ works in the intervals $(2,3)$ and $(3,\infty)$. Finally, translate your newfound intuition into rigorous proof. You will not find that difficult, once you understand what you need to prove.

(Iteration, graphically: Start at $(x_{n-1},x_{n-1})$ on the diagonal, move vertically to the graph, hitting $(x_{n-1},f(x_{n-1}))=(x_{n-1},x_n)$, move horizontally to the diagonal, hitting $(x_n,x_n)$, repeat.)


The sequence is non increasing when: $$x_{n+1}\le x_n\iff 2+\sqrt{x_n-2}\le x_n\iff \sqrt{x_n-2}\le x_n-2\iff x_n-2\le x_n^2-4x_n+4\iff x_n^2-5x_n+6\ge 0\iff (x_n-2)(x_n-3)\ge 0\iff x_n\ge 3$$ If $x_0\ge 3$ then $x_n\ge 3$ (use induction) and so $(x_n)$ is non decreasing. As $x_n\ge 3$, $(x_n)$ is bounded below. By the Monotone Convergence Theorem, $(x_n)$ converges to a real number $L$.

If $2\le x_n<3$ then $x_n<3$ (another simple induction). Therefore, $(x_n)$ is non decreasing and as $x_n<3$, $(x_n)$ is bounded above. By the Monotone Convergence Theorem, $(x_n)$ converges to a real number $L$.

In either case, as you say, $$L=\lim_{n\to \infty}x_{n+1}=2+\sqrt{L-2}\iff L=2\text{ or }L=3$$

  • $\begingroup$ This is always the case... No. (You made a mistake when expanding $(x_n-2)^2$ and, in fact, the sequence $(x_n)_n$ is increasing if $2\lt x_0\lt3$ and decreasing if $x_0\gt3$.) $\endgroup$ – Did Jan 24 '13 at 19:35
  • $\begingroup$ @Did I think it's fixed now $\endgroup$ – Nameless Jan 24 '13 at 19:43
  • $\begingroup$ You could deal with the cases $x_0=2$ and $x_0=3$ separately since then the sequence is neither (strictly) increasing nor (strictly) decreasing. There might also be a problem with your last argument, which I fail to get since $L=+\infty$ is a solution of the equation $L=2+\sqrt{L-2}$. $\endgroup$ – Did Jan 24 '13 at 19:52
  • $\begingroup$ @Did I never said strictly so I dealt only with non-strict inequalities. The MCT implies in that case as well. As I said before the last argument, $L$ is a real number. $\endgroup$ – Nameless Jan 24 '13 at 19:56
  • $\begingroup$ Of course, but in-/de-creasing are often used to mean strictly in-/de-creasing. $\endgroup$ – Did Jan 24 '13 at 19:58

Focus on showing that sqrt(x - 2) is decreasing and bounded below. This should give you what you're looking for.

  • $\begingroup$ I would not focus on showing that sqrt(x-2) is decreasing since the function is increasing. $\endgroup$ – Did Jan 25 '13 at 6:07

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