# Sequences $x_{n+1}=\frac{x_n+y_n}2$ and $y_{n+1}=\frac{2x_ny_n}{x_n+y_n}$

Given $x_1>y_1>0$ we define in a recursive way sequences $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ such that $$x_{n+1}=\frac{x_n+y_n}2$$

$$y_{n+1}=\frac{2x_ny_n}{x_n+y_n}$$

I think I can prove the first sequence is decreasing, the second one is decreasing and $x_n>y_n \space\forall n\in\Bbb{N}$. Then they both converge and both limits are equal. My doubt is : how can I calculate the value of the limit in terms of $x_1$ and $y_1$ ?

• Oct 20, 2020 at 9:07

Note the invariant $x_{n+1}y_{n+1}=x_{n}y_n$ (this comes from multiplying the two equations $x_{n+1}=\frac{x_n+y_n}2$ and $y_{n+1}=\frac{2x_ny_n}{x_n+y_n})$.  Let $x_0=a$ and $y_0=b$. Hence $x_ny_n=ab$ for all $n$.  Also as you have deduced, $x_n>y_n$ for all $n$, and this too remains invariant. This can be proved as follows:  Initially we have $x_0>y_0$. Suppose $x_n>y_n$ for some $n$. Then $x_{n+1}$ is the midpoint of the segment with end points $x_n,y_n$. Moreover $x_{n+1}>y_{n+1}$ since the harmonic mean is strictly less than the arithmetic mean.
 Hence for all $n$, $$0<x_{n+1}-y_{n+1}=\dfrac{x_n-y_n}{x_n+y_n}\times\dfrac{x_n-y_n}{2}<\dfrac{x_n-y_n}{2}$$  Thus we have $\lim_{n\to\infty} x_n=\lim_{n\to\infty} y_n=x$ where $x=\sqrt{ab}\Rightarrow\lim_{n\to\infty}\{x_n\}=\lim_{n\to\infty}\{y_n\}=\sqrt{ab}$

We have

$$x_{n+1}=\frac{x_n+y_n}2$$

$$y_{n+1}=\frac{2x_ny_n}{x_n+y_n}$$

Multiplying both members

$$x_{n+1}y_{n+1} = x_n y_n = k$$

Now adding both equations we obtain

$$x_{n+1}+y_{n+1} = \frac{(x_n+y_n)^2+4x_ny_n}{2(x_n+y_n)}$$

or

$$s_{n+1} = \frac{s_n^2+4k}{2s_n}$$

If $s_n$ converges to $s_{\infty}$ then

$$s_{\infty} = \pm 2\sqrt k$$

following with $s_{\infty} = 2\sqrt k$ and solving

$$x_{\infty}+y_{\infty} = 2\sqrt k\\ x_{\infty}y_{\infty} = k$$

we obtain

$$x_{\infty} = y_{\infty} = \sqrt k = x_1 = y_1$$

I hope this helps.

This is an example of a compound mean since $\,(x_{n+1},y_{n+1})\,$ is the compound of the arithmetic and harmonic means of $\,(x_n,y_n).\,$ Because the means are homogenous, without loss of generality, let $\, 1 = x_0 y_0. \,$ Also, let $\, t_0 := (x_0 - 1)/(x_0 + 1) \,$ which implies $\, x_0 = (1 + t_0)/(1- t_0). \,$ Using the compund mean iteration we find that $\, 1 = x_n y_n \,$ and if $\, t_n := (x_n - 1)/(x_n + 1) \,$ then $\, t_n = t_0^{2^n}. \,$ If we start with $\, |t_0| < 1, \,$ then $\, t_n \to 0 \,$ which implies that $\, x_n \,$ and $\, y_n \,$ both converge to $\,1.\,$ In the general case $\, x_n \,$ and $\, y_n \,$ have common limit $\, L. \,$ Since $\, x_0 y_0 = x_n y_n = L L \,$ we have $\, L = \sqrt{x_0 y_0}.\,$