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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.

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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

$$x_1<x_2<y_2<y_1$$

Prove with induction that we have the stronger statement:

$$x_1<x_2<x_3<\dots<y_3<y_2<y_1$$

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

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

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $X=Y$.

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