<s>[INST] Convergence behavior of a rational iterative procedure of the form $\frac{1}{2}(z+1/z)$. I am trying to prove the following result.
First, we have an auxiliary sequence satisfying 
$E_{n+1}=\dfrac{E_n^2}{(1+\sqrt{1-E_n^2})^2}$ with $E_0<1$. 
It is called the Landen transformation.
Let $S_0=1+r\exp(i\theta)$ with $r\le F_0$, and $F_0=E_0$.
The interested iteration relation writes
$$S_{n+1}=\frac{\sqrt{1-E_n^2}}{1+\sqrt{1-E_n^2}}(\frac{S_n}{\sqrt{1-E_n^2}}+\frac{\sqrt{1-E_n^2}}{S_n})$$
I intend to find out how the $S_{n}$ converges to 1. 
Since $E_n$ satisfies the Landen transformation, it decreases to zero rapidly.
Thus, the iteration is approximately to be $\frac{1}{2}(z+1/z)$ and in this case, the problem can be handled if $|S_n-1|<0.1$ for some $n$.
However, I have no idea how to analysis the first several terms when $E_n$ and $F_n$ is near $1$.
Notice that if $S_n=1+E_n$, then $S_{n+1}=1+E_{n+1}$. Thus，we set 
$$z_n=\frac{S_n}{1+E_n}$$
From the iteration relation
\begin{equation*}
S_{n+1}=\frac{\sqrt{1-E_n^2}}{1+\sqrt{1-E_n^2}}(\frac{S_n}{\sqrt{1-E_n^2}}+\frac{\sqrt{1-E_n^2}}{S_n}),
\end{equation*}
we deduce that
\begin{align*}
\frac{S_{n+1}}{1+E_{n+1}}&=\frac{\frac{\sqrt{1-E_n^2}}{1+\sqrt{1-E_n^2}}}{1+E_{n+1}}\left( 
\frac{S_n}{1+E_n}\sqrt{\frac{1+E_n}{1-E_n}}+\frac{\sqrt{\frac{1-E_n}{1+E_n}}}{\frac{S_n}{1+E_n}}
\right)\\
&=\frac{\sqrt{1-E_n^2}}{2}\left( 
\frac{S_n}{1+E_n}\sqrt{\frac{1+E_n}{1-E_n}}+\frac{\sqrt{\frac{1-E_n}{1+E_n}}}{\frac{S_n}{1+E_n}}
\right),
\end{align*}
which can be reformulated as
\begin{equation*}
z_{n+1}=\frac{1+E_n}{2}
\left[
z_n+\frac{{\frac{1-E_n}{1+E_n}}}{z_n}\right].
\end{equation*}
Then, the problem is to show how $z_n\to 1$.
Next, we map $z_n$ into the unit disk by setting
\begin{equation*}
w_n=\frac{z_n-1}{z_n+1}.
\end{equation*}
Then the inverse transform writes 
\begin{equation*}
z_n=\frac{1+w_n}{1-w_n}
\end{equation*}
Then we have 
\begin{align*}
w_{n+1}&=\frac{z_{n+1}-1}{z_{n+1}+1}=\frac{
\frac{1}{2}\left[(1+E_n)z_n+\frac{1-E_n}{z_n}\right]-1}{
\frac{1}{2}\left[(1+E_n)z_n+\frac{1-E_n}{z_n}\right]+1}\\
&=\frac{
\left[(1+E_n)z_n^2+1-E_n\right]-2z_n}{
\left[(1+E_n)z_n^2+1-E_n\right]+2z_n}\\
&=\frac{(z_n-1)(z_n-1+E_n(z_n+1))}{(z_n+1)(z_n+1+E_n(z_n-1))}\\
&=w_n\frac{w_n+E_n}{1+E_nw_n}.
\end{align*}
Next, we prove that how $w_n\to 0$.
For the initial data, we have
\begin{equation*}
z_0=\frac{S_0}{1+E_0},\quad w_0=\frac{z_0-1}{z_0+1}.
\end{equation*}
We may take the assumption that $\Re(S_0)>1-E_0$.
Thanks in advance.
 [/INST] Suppose that
$|w_n|<F_n$, then we set 
\begin{equation*}
w_n=r\exp(i \theta), \quad r\le F_n.
\end{equation*}
Thus, 
\begin{align*}
w_{n+1}=w_n\frac{w_n+E_n}{1+E_nw_n}=
r\exp(i \theta)\frac{r\exp(i \theta)+E_n}{1+E_nr\exp(i \theta)}
\end{align*}
Then 
\begin{align*}
|w_{n+1}|&=r\left|\frac{r\exp(i \theta)+E_n}{1+E_nr\exp(i \theta)} \right|=r
\sqrt{\frac{E_n^2+r^2+2E_nr\cos\theta}{1+E_n^2r^2+2E_nr\cos\theta}}\\
&\le \frac{r(E_n+r)}{(1+E_nr)}
\le \frac{F_n(E_n+F_n)}{(1+E_nF_n)}.
\end{align*}
The inequality maight be coarse. 
Moreover, 
$$\frac{E_n+F_n}{1+E_nF_n}<1.$$ 
and the fractional tends to zero.
Then we conclude that $F_n\to 0$.
However, if we know for which $n$, $F_n\le \epsilon$, the result will be better. But I cannot accomplish this task.
</s>