Determining the common limit between two recursive sequences

My Question reads: For any positive $a,b$, the arithmetic mean $A(a,b) =\frac{a+b}{2}$ is less than or equal to the geometric mean $G(a,b)=\sqrt{ab}$ .

Let $x_0=a$, $x_{n+1}=A(x_n,y_{n+1})$ and $y_0=b$, $y_{n+1}=G(x_n,y_n)$. We know that both sequences converge and further that they converge to the same number.

Now, determine the common limit in the instance that $a=2\cos\frac{\pi}{4}$ and $b=6\tan\frac{\pi}{3}$. It is possible to write out explicit formulas for $x_n$ and $y_n$.

I am having issues finding an explicit formula for $x_n$ and $y_n$. I started by plugging in some terms to see if I found a pattern, but it got much too messy to visualize. Is there another way to find these equations? Also, how would this then help me with finding the common limit?

• if $a = b, H(a,b) = a = b$ and that should be easy enough to show. – Doug M Oct 9 '17 at 21:24
• If $a = b$ then $y_1 = H(a,b) = a$ and $x_1 = G(a,y_1) = a$ and $x_n = a$ – Doug M Oct 9 '17 at 21:37

$y_1 = H(\sin x, \tan x) =$$\frac {2}{\frac 1{\sin x} + \frac {\cos x}{\sin x}}\\ \frac {2\sin x}{1+\cos x} = 2\tan \frac 12 x$
$x_1 = G(\sin x,2\tan \frac 12 x) = \sqrt {(2\sin \frac 12 x\cos\frac 12 x)(2\frac {\sin \frac {1}{2} x}{\cos\frac {1}{2} x})} = 2\sin\frac {1}{2}x$
$x_n = 2^n\sin \frac {x}{2^n}\\ y_n = 2^n\tan \frac {x}{2^n}$
$\lim_\limits{a\to\infty} a \sin \frac {x}{a}\\ \lim_\limits{a\to 0} \frac {\sin xa}{a}\\ \frac {\pi}{3}$