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.
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.