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?

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

*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 $\lim_{n\to\infty}a_n=\sqrt{2}$
