So the monotone convergence theorem states that if a sequence is bounded and monotone then it converges.

Now I am trying to prove that the sequence defined recursively as $x_1 = \sqrt 2$, $x_{n+1}=\sqrt {2x_n}$ converges and to find it's limit.

I am able to show that the sequence is monotone by determining if the ratio of $\frac{x_{n+1}}{x_n}$ is greater than and equal to or less than or equal to 1. I am even know what the limit is.

My dilemma is I am not sure how to determine if this sequence is bounded and I am not quite sure what I need to do to show that its bounded. Can anyone offer a few hints or strategies? I know that it is bounded below by $\sqrt2$; however the sequence is increasing so I need to show that there is an upperbound as well.


Hint: Use Induction. Suppose $x_n < 2$, what can you say about $x_{n+1}$?

  • $\begingroup$ I get it, but one question how did you choose $x_n < 2$? Common sense or some other method? $\endgroup$
    – spitfiredd
    Feb 25 '14 at 1:47
  • $\begingroup$ Maybe you could use the derivative of $f(x)= (2x)^{1/2}$; show it is positive, and use the fact that $x_1>2$. $\endgroup$
    – user99680
    Feb 25 '14 at 1:59
  • $\begingroup$ I solved $x=\sqrt{2x}$ to get $2$. Any number larger than the limit of the sequence would also work, essentially I skipped to finding the limit. $\endgroup$
    – Macavity
    Feb 25 '14 at 2:13
  • $\begingroup$ @user99680 $x_1 <2$ and I am not sure if derivatives are useful here. $\endgroup$
    – Macavity
    Feb 25 '14 at 2:31

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