How do I prove that series given by $x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}$ converges to $ \sqrt{2}$?
$x_0 = 2$
We got the hint that the series is a non linear difference equation. My idea would be to prove that the series is decreasing and has the lower bound $\sqrt{2}$ but I've no luck doing that. How do I proceed?
Observe that inductively it is almost immediate to show the sequence is always positive, and now
$$(**)\;\;\;x_{n+1}-x_n=\frac{x_n}2+\frac1{x_n}-x_n=\frac1{x_n}-\frac{x_n}2=\frac{2-x_n^2}{2x_n}$$
and we now use induction to show $\;x_n\ge\sqrt2\;$ for all indexes: for $\;n=0\;$ it is trivial, so suppose it is true for $\;n\;$ and now we show for $\;n+1\;$ ...but the we do $(**)$ , and get:
$$x_{n+1}-x_n=\frac{2-x_n^2}{2x_n}\le0\,,\;\text{since by the ind. hypothesis}\;x_n\ge\sqrt2\implies 2-x_n^2\le 0$$
and $\;2x_n>0\;$.
Thus, $\;x_{n+1}-x_n\le 0\implies x_{n+1}\le x_n\;$ and this is thus a monotone descending sequence bounded from below ( say, by $\;\sqrt2\;$), so its limit exists (and equals is infimum)
You could also compute that $$ \frac{x_{n+1}-\sqrt2}{x_{n+1}+\sqrt2}=\left(\frac{x_{n}-\sqrt2}{x_{n}+\sqrt2}\right)^2 $$ which implies convergence, as the fraction is always smaller than $1$ for positive $x_n$.
