1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.