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I'm trying to prove this question. At the first grance it seems easy:

Given $a,b\in \mathbb R^+$, define the sequences $(x_n)$ and $(y_n)$ as $x_1=\sqrt {ab}$, $y_1=\frac{a+b}{2}$ and $x_{n+1}=\sqrt{x_n y_n}$, $y_{n+1}=\frac{x_n+y_n}{2}$. Prove that $(x_n)$ and $(y_n)$ converge to the same limit.

My approach:

construct these equations:

$\lim x_n=\sqrt{(\lim x_n)(\lim y_n)}$ and $\lim y_n=\frac {\lim x_n +\lim y_n}{2}$ then by the first one we can have $\lim x_n=\lim y_n$

I think it's not so easy, but where am I wrong?

Thanks a lot

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You first need to show that the limits exist. If you can do this, it follows from the definition of the $y_n$ that the sequences have the same limit (if you use "the first one", you also need to show $\lim x_n\ne 0$). –  David Mitra Jan 17 '13 at 17:00
    
The limit is known as the arithmetic-geometric mean of $a$ and $b$; see, for example, en.wikipedia.org/wiki/Arithmetic–geometric_mean –  Gerry Myerson Jan 17 '13 at 23:23
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1 Answer

Hint:

$x_n,y_n>0$ $\forall$ $n\in\mathbb N\implies x_{n}\leq y_{n}$ $\forall$ $n\in\mathbb N$ $[$since AM$\geq$ GM for positive numbers$]$

$y_{n+1}=\frac{x_n+y_n}{2}\leq y_n,x_{n+1}=\sqrt{x_ny_n}\geq x_n$ $\forall$ $n\in\mathbb N$

$\implies y_1\geq y_2\geq y_3\geq...\geq y_n...\geq x_n\geq...\geq x_3\geq x_2\geq x_1$

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