Here is a problem I've been working on. I am stuck and wondered if you guys could shed any light.
Let $a>0$ and $u_{1}>a$.
Consider the sequence $(u_{n})_{n=1}^{\infty }$ defined by:
$$ u_{n+1}\: :=\: \frac{1}{2}(u_{n}\:+\:\frac{a^2}{u_{n}})\;\;\;\;\forall n\in \mathbb{N} $$
Show that the sequence is convergent and that:
$$ \lim_{n \to \infty}u_{n}=a $$
Thank you in advance guys,
Deborah
$$u_2:=\frac12\left(u_1+\frac{a^2}{u_1}\right)\le u_1\iff u_1^2\ge a^2\iff u_1\ge a\;\;\color{green}\checkmark$$
Inductively:
$$u_{n+1}=\frac12\left(u_n+\frac{a^2}{u_n}\right)\le u_n\iff u_n\ge a$$
But the last inequality can also be proved inductively:
$$u_{n+1}:=\frac12\left(u_n+\frac{a^2}{u_n}\right)\ge a\iff u_n^2+a^2\ge au_n\iff$$
$$u_n^2-au_n+a^2\ge 0\;\;\text{, and this is true iff the quadratic's discriminant is non-positive:}$$
$$\Delta:=a^2-4a^2=-3a^2<0$$
Thus, the sequence $\;\{u_n\}\;$ is monotone non-increasing and bounded below by $\;a\;$ so its limit exists, call it $\;U\;$ (0bserve that $\;U\ge a>0\;$ and, in particular, $\;U\neq 0\;$ ), and now a little arithmetic of limits:
$$U\leftarrow u_{n+1}:=\frac12\left(u_n+\frac{a^2}{u_n}\right)\rightarrow \frac12\left(U+\frac{a^2}{U}\right)\implies \ldots$$