Let $0<a<b$, $x_1=a>0$, $y_1=b>0$, $x_{n+1}=\sqrt{x_ny_n}$, and $y_{n+1}={x_n+y_n \over 2}$, $n\geq 2$. Show that $\lim _{n\rightarrow \infty}x_n=\lim_{n\rightarrow\infty} y_n$.

I have considered the fact that $(y_n-x_n)^2>0$ implies that ${x_n+y_n \over 2}>\sqrt{x_ny_n}$. I'm not entirely sure how to prove this result. Any solutions/hints are greatly appreciated.


First, see that $x_1=\frac{x_1+x_1}2<y_2<\frac{y_1+y_1}2=y_1$. Thus, $x_1<y_2<y_1$ since $x_1<y_1$.

In the same manner, see that $x_1<x_2<y_2$.

You have already shown that $x_{n+1}=\sqrt{x_ny_n}<\frac{x_n+y_n}2=y_{n+1}$.

Putting all this together, you get


Prove with induction that we have the stronger statement:


Clearly, you can see $x$ is monotonically increasing while $y$ is monotonically decreasing and that they are bounded. From this, it is clear that there must exist a limit.

Let us call the limits $X$ and $Y$. From this, we see

$$Y=\frac{X+Y}2\implies2Y=X+Y\implies X=Y\\\mathsf{or}\\X=\sqrt{XY}\implies X^2=XY\implies X=Y$$

  • $\begingroup$ Congrats, you understood my answer! $\endgroup$ – mathguy Sep 25 '16 at 1:22
  • $\begingroup$ @mathguy ;) I figured it would be worth writing it all out. $\endgroup$ – Simply Beautiful Art Sep 25 '16 at 9:41
  • $\begingroup$ @SimpleArt Thank you for your answer. Forgive me, but I'm not entirely sure how to show that $x_1<x_2<y_2$ ("in the same manner"). Could you explain please? $\endgroup$ – hungryformath Dec 1 '16 at 23:42
  • $\begingroup$ @hungryformath ($x_2<y_2$ as you have already shown)$$x_1=\sqrt{x_1x_1}<x_2<y_2$$Can you take it from there? $\endgroup$ – Simply Beautiful Art Dec 2 '16 at 1:27
  • 1
    $\begingroup$ @hungryformath In essence, you are trying to show that $x_2<y_2$. It requires showing that, from their definitions,$$\sqrt{x_1y_1}<\frac{x_1+y_1}2$$which is easily shown by squaring both sides. After some manipulation, I think you should end up with the statement $(x_1-y_1)^2\stackrel?>0$ which is obvious. This can easily be reproduced to the general case. $\endgroup$ – Simply Beautiful Art Dec 2 '16 at 2:14

First show that $x_n < y_n$ for all $n$, then use this to prove that $x_n$ is increasing and $y_n$ is decreasing. Also, they are bounded (why?) so both converge to finite limits, call them $X$ and $Y$. Taking limits, $Y = \dfrac{X+Y}{2}$ so $X=Y$.


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.