Question

\[\text{fix }a>0, \text{ and let }x_1>\sqrt{a}\] Define
\[x_{n+1}=\frac{1}{2}(x_n+\frac{a}{x_n})\]
Show
\[(x_n)\text{ converges and}\lim_{n\rightarrow \infty} x_n=\sqrt{a}\]

Answer 1

You do this in two steps.

First show that the sequence (x_n) is monotonically decreasing and the then show that sqrt(a) is the greatest lower bound, glb (also sometimes called the inf), of the set of values { x_n }.

Then you invoke a standard theorem in limits which says in such circumstances (x_n) converges and that the limit of the sequence is the glb{ x_n }.

Does that get you going?

Answer 2

yes thank you. glb is root is root of a because arithmetic mean is bigger than geometric mean.