Convergence of recurrence sequence Let $a_1=1$ define $a_{n+1}=\frac{1}{2}(a_n+\frac{2}{a_n})$ . Show that the sequence converges to $\sqrt{2}$. The idea is to show the sequence is bounded and monotone . But how should I do it?
 A: Clearly $a_n>0$. And for $n\geq 2$, 
$$a_n=\frac12(a_{n-1}+\frac2{a_{n-1}})\geq \sqrt 2.$$
Hence 
$$2(a_n-a_{n+1})=a_n-\frac2 {a_n}=\frac{a_n^2-2}{a_n}\geq 0$$
so $a_n$ decreases when $n\geq2$. From this we konw $\lim_{n\to \infty}a_n$ exists, called $a$.
Finally, letting $n\to\infty$ in the recurrence relation we get $$a=\frac12(a+\frac2a),$$ 
so $\lim_{n\to \infty}a_n=a=\sqrt 2$($a\geq 0$).
A: *

*$a_2=\frac{3}{2}>\sqrt{2}$

*Show if $a_n>\sqrt{2}, \sqrt{2}<a_{n+1}<a_n$
$a_{n+1}-\sqrt{2}=\frac{1}{2}(a_n-\sqrt{2}+\frac{2-\sqrt{2}a_n}{a_n})=(a_n-\sqrt{2})\frac{1-\frac{\sqrt{2}}{a_n}}{2}$
so $0<a_{n+1}-\sqrt{2}<\frac{a_n-\sqrt{2}}2$
So $\lim_{n\to\infty}a_n=\sqrt{2}$
A: First notice que if $\lim_{n\to\infty} a_n = x$ then $x=\frac{1}{2}(x+\frac{2}{x})$ wchich is equivalent to $x^2 = 2$. Now we need only to prove the convergence. This follows from the fact the sequence $(a_n)_{n>1}$ is monotonic-decreasing and bounded below:
(1) $a_2 =1.5 > \sqrt 2$.
(2) if $a_n > \sqrt 2$ then $ a_n > a_{n+1} > \sqrt 2$: 
(2.a) $a_{n+1} - a_n = \frac 12(\frac{2}{a_n} - a_n) = \frac{2-a_n^2}{2a_n} < 0$, thus $ a_n > a_{n+1}$.
(2.b) $a_{n+1}^2 - 2 = \frac{(a_n^2 -2)^2}{4a_n^2} >0$.
