That's interesting I suppose...

Let's check the extrema of $x_{n+1} = \dfrac {x_n^2+1}{2x_n-1}$ .

$\begin{aligned} \frac {d x_{n+1}}{dx_n} & = \frac {2x_n(2x_n-1)-2(x_n^2+1)}{(2x_n-1)^2} = \frac {2(x_n^2-x_n-1)}{(2x_n-1)^2} \end{aligned}$

Implying that $\dfrac {d x_{n+1}}{dx_n} = 0$ , when $x_n = \dfrac {1\pm \sqrt 5}2 = \begin{cases} \varphi \\ 1-\varphi \end{cases}$ , where $\varphi$ is the gold ratio .

Since $\begin{cases} x_n > 0 & \text{for }x_1 > \frac 12 \\ x_n < 0 & \text{for }x_1 < \frac 12 \end{cases} \implies \begin{cases} x_n \in [\varphi, \infty) & \text{for }x_1 > \frac 12 \\ x_n \in (-\infty, 1-\varphi] & \text{for }x_1 < \frac 12 \end{cases}$ .

Since $x_n$ is bounded, let's assume $\displaystyle \lim_{n \to \infty} x_n$ converges to a limit $\ell$ , then

$\begin{aligned} \lim_{n \to \infty} x_n & = \lim_{n \to \infty} x_{n+1} \\ & = \lim_{n \to \infty} \frac {x_n^2+1}{2x_n-1} & \small \blue{\text{Since }\lim_{n \to \infty} x_n = \ell} \\ \implies \ell & = \frac {\ell^2+1}{2\ell-1} \\ \ell^2 - \ell - 1 & = 0 \\ \ell & = \begin{cases} \alpha = \varphi & \text{if }x_1 > \frac 12 \\ \beta = 1 - \varphi & \text{if }x_1 < \frac 12 \end{cases} \end{aligned}$

We get the same results by finding the extrema of $x_{n+1}$ with $x_n = \varphi, 1-\varphi$

$\begin{cases} \text{For }x_n = \varphi, & x_{n+1} = \dfrac {\varphi^2+\blue 1}{2\varphi -1} = \dfrac {\varphi^2+\blue{ \varphi^2-\varphi}}{2\varphi -1} = \varphi & = x_n \\ \text{For }x_n = 1-\varphi, & x_{n+1} = \dfrac {(1-\varphi)^2+1}{2(1-\varphi) -1} = \dfrac {2\varphi - 1 - \varphi^2-1}{2\varphi -1} = 1-\varphi & = x_n \end{cases}$

This means that $x_{n+1} = x_n$ at the extrema of $\alpha = \varphi$ and $\beta = 1-\varphi$ . That is $x_n$ converges to $\alpha$ and $\beta$ .

Now consider $\displaystyle \sum_{k=0}^\infty \left(\alpha^k - \beta^k\right)^2 x^k$ for $x=0.1$ or

$\begin{aligned} \sum_{k=0}^\infty \frac {\left(\varphi^k - (1-\varphi)^k\right)^2}{10^k} & = \sum_{k=1}^\infty \frac {5F_k^2}{10^k} \approx \boxed{0.567} & \small \blue{\text{where }F_n \text{ is the }n\text{th Fibonacci number.}} \end{aligned}$

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Reference: Fibonacci number