Let $0<y_1<x_1 $ and set $x_{n+1} = \frac{x_n+y_n}{2}$ and $y_{n+1}=\sqrt{x_ny_n}$, $n\in\Bbb{N}$.

  1. Prove that $0<y_n<x_n$ $\forall n\in\Bbb{N}$.
  2. Prove that $y_n$ is increasing and bounded above, and $x_n$ is decreasing and bounded below.
  3. Prove that $0<x_{n+1}-y_{n+1}<\frac{(x_1-y_1)}{2^n} $ for $n\in\Bbb{N}$.
  4. Prove that $\lim_{x\to \infty}{x_n}=\lim_{y\to \infty}{y_n}$.

For the first one I tried to prove by induction but I found $x_n<y_n$.

For the second one:
$y_{n+1}=\sqrt{x_ny_n}\implies y_{n+2}=\sqrt{x_{n+1}y_{n+1}}=\sqrt{\frac{x_n+y_n}{2}.\sqrt{x_ny_n}}$
So, $y_{n+2}^2>y_{n+1}^2$ which implies $y_n$ is increasing??? I did same thing for $x_n$.

  • 1
    $\begingroup$ I don't see how you deduce $y_{n+2}^2>y_{n+1}^2$. $\endgroup$ – Bernard Dec 7 '15 at 22:35
  • $\begingroup$ I did a mistake. I took $y_{n+2}^2=\frac{x_n+y_n}{2}.{x_ny_n}$. Thank you. $\endgroup$ – diogenes Dec 7 '15 at 22:43
  • $\begingroup$ The limit is known as the arithmetic-geometric mean. $\endgroup$ – John Bentin Dec 7 '15 at 23:12
  1. By AM > GM, $x_n$ > $y_n$. Or, $$4x_{n+1}^2-4y_{n+1}^2 = (x_n-y_n)^2 > 0$$

2-1. $$x_{n+1} > y_{n+1}$$ $$\frac{x_n+y_n}{2} > \sqrt{x_ny_n}$$ $$\frac{x_n+y_n}{2}\sqrt{x_ny_n} > {x_ny_n}$$ $$\sqrt{\frac{x_n+y_n}{2}\sqrt{x_ny_n}} > \sqrt{x_ny_n}$$ $$\sqrt{x_{n+1}y_{n+1}} > \sqrt{x_ny_n}$$ $$\therefore y_{n+2}>y_{n+1}$$

2-2. $$x_{n+1}=\frac{x_{n}+y_{n}}{2}<\frac{x_{n}+x_{n}}{2}=x_{n}$$

$y_n$ increases and $x_n$ decreases, but $y_n<x_n$. Therefore $y_n$ is upper bounded and $x_n$ is lower bounded.

3. $$x_{n+1}-y_{n+1}=\frac{x_{n}+y_{n}}{2}-\sqrt{x_ny_n}=\frac{x_{n}-2\sqrt{x_ny_n}+y_{n}}{2}<\frac{x_{n}-2\sqrt{y_ny_n}+y_{n}}{2}=\frac{x_n-y_n}{2}$$


4. $$0<x_{n+1}-y_{n+1}<\cdots<\frac{x_1-y_1}{2^n}$$ $$\lim_{n\to\infty}0\le \lim_{n\to\infty}x_{n+1}-y_{n+1}\le \lim_{n\to\infty}\frac{x_1-y_1}{2^n}=0$$ $$\therefore \lim_{n\to\infty}x_{n+1}-y_{n+1} = 0$$

  • $\begingroup$ Your last equality in (3) is wrong - $(x_n+x_n)/2-\sqrt{y_ny_n}$ is just $x_n-y_n$, not $\frac12$ of that quantity. This shows that the difference decreases, but it's not enough to show that it goes to zero. $\endgroup$ – Steven Stadnicki Dec 7 '15 at 22:55
  • $\begingroup$ @StevenStadnicki I fixed it. Thanks. $\endgroup$ – Kay K. Dec 7 '15 at 23:01

You can prove $y_n<x_n$, $(x_n)$ is decreasing and $(y_n)$ is increasing in one inductive step:

  • $x_{n+1}=\dfrac{x_n+y_n}2<\dfrac{2x_n}2=x_n$ by the inductive hypothesis.
  • $y_{n+1}=\sqrt{x_ny_n}>\sqrt{y_n y_n}=y_n$ by the same inductive hypothesis.
  • $y_{n+1}=\sqrt{x_ny_n}\le\frac{x_n+y_n}2=x_{n+1}$ by the A.G.M. inequality. The inequality is strict by the case of equality in the A.G.M. inequality: it happens only in case $x_n=y_n$, which is false by the inductive hypothesis.

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