Q:

limit of recursively given seqeuence

Let  $0 < a < b$.
A seqeunce $x_n$ satisfying $x_1 = b$ , $x_{2n} = \frac{x_n + a}{2}$, $x_{2n+1} = \frac{x_n + \sqrt{a\,x_n} + a}{3}$. 
Consider if $\{x_n\}$ converges and if yes, find a limit.
As I can see, $x_2 = \frac{b+a}{2}<b$  and  $x_3=\frac{b+\sqrt{ab}+a}{3}<b$ so I assume that $x_n$  is monotonously decreasing and bounded from above but have no idea about the limit of $x_n$ itself.

A:

You have correctly shown that $x_n$ is decreasing and bounded above by b.
Can you show that it is bounded below as well?
To find the value it converges to, we assume that it converges to a value $x$, say.
Then $x_{2n}=x_{2n+1}=x$
$\frac{x+a}2=/frac{x+\sqrt{ax}+a}3$
$3(x+a)=2(x+\sqrt{ax}+a)$
$3x+3a=2x+2\sqrt{ax}+2a$
$x+a=2\sqrt{ax}$
$(x+a)^2=4ax$
$x^2+2ax+a^2=4ax$
$x^2-2ax+a^2=0$
$(x-a)^2=0$
$x=a$

