Please how could I possibly show that the sequence defined by
$U_{n+1} = \frac{1}{2} (U_n + \frac{a}{U_n})$ converges Given that $U_0 > 0$ and $a>0$
I have calculated $\frac{U_{n+1}}{U_n}$ which is positive but $U_{n+1}-U_n$ doesn't yield any result and i can't show it is bounded.
Thanks in advance
First note that if $u_0>0$ and $a>0$, then it is easy to show that $$u_n>0$$ Then by AM-GM we have $$u_{n+1} =\frac{1}{2}\left(u_n+\frac{a}{u_n}\right) \ge \sqrt{u_n\frac{a}{u_n}}=\sqrt{a}$$
Hence $$u_{n+1}-u_n=\frac{1}{2}\left(\frac{a}{u_n}-u_n\right)\le\frac{1}{2}\left(\frac{a}{\sqrt{a}}-\sqrt{a}\right)=0$$
Hence, $u_n$ is decreasing and bounded from below. Therefore it converges.
After proving that it converges, one can obtain the limit by taking limit on both sides of the recursion relation, which gives $$l=\frac{1}{2}\left(l+\frac{a}{l}\right)$$ $$l^2=a$$ $$l=\sqrt{a}$$ where the negative root is rejected since the it is a positive sequence.
If the {$u_{n}$} sequence tends to some u then it must be $u={1\over2}(u+{a\over u})$ which results in $u=\pm\sqrt a$ but this is very imprecise. Embarking on this and knowing that $u_{n+1}={1\over2}(u_{n}+{a\over u_{n}})={\sqrt a\over2}({{u_n\over\sqrt a}+{\sqrt a \over u_{n}})}>\sqrt a$ the limit must probably be $\sqrt a$,therefore by defining error function $e_{n}=u_{n}-\sqrt a$ we get $e_{n+1}={1\over2}(e_{n}+{a\over u_{n}}-\sqrt a)$. Since $u_{n}>\sqrt a$ we have ${a\over u_{n}}-\sqrt a<0$ and $e_{n+1}<{1\over2}e_{n}$ which tends to zero and the sequence is convergent to $\sqrt a$
