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


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*Prove that $0<y_n<x_n$ $\forall n\in\Bbb{N}$.

*Prove that $y_n$ is increasing and bounded above, and $x_n$ is decreasing and bounded below.

*Prove that $0<x_{n+1}-y_{n+1}<\frac{(x_1-y_1)}{2^n} $ for $n\in\Bbb{N}$.

*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 $x_n$. 

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