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?

Solution 1:

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 $\lim_{n\to \infty}a_n=a=\sqrt 2$($a\geq 0$).