Subject 4. Let $\left(x_{n}\right)_{n \geq 0}$ and $\left(y_{n}\right)_{n \geq 0}$ be two sequences of real numbers defined by $x_{0}>1$ and $x_{n+1}^{2}-2 x_{n+1}=x_{n}-2$, and $y_{n}=1-\frac{1}{2}-\frac{1}{2^{2}}-\cdots-\frac{1}{2^{n}}$. Calculate $\lim _{n \rightarrow \infty} \frac{x_{n}-2}{y_{n}}$.
Solution: $\left(x_{n+1}-1\right)^{2}=x_{n}-1 \Rightarrow x_{n+1}-1= \pm \sqrt{x_{n}-1}$.

If $x_{n+1}-1=-\sqrt{x_{n}-1} \Rightarrow x_{n+2}-1=-\sqrt{x_{n+1}-1}=-i \sqrt[4]{x_{n}-1} \notin \boldsymbol{R}$. $.1 \mathrm{p}$

If $x_{n+1}-1=\sqrt{x_{n}-1} \Rightarrow x_{n}-1=\sqrt[2^{n}]{x_{0}-1}$ (inductively).

$2 \mathrm{p}$

Since $y_{n}=1-\frac{1}{2}-\frac{1}{2^{2}}-\cdots-\frac{1}{2^{n}}=\frac{1}{2^{n}}$ we have

$\lim _{n \rightarrow \infty} \frac{x_{n}-2}{y_{n}}=\lim _{n \rightarrow \infty} \frac{2^{n} \sqrt{x_{0}-1}-1}{\frac{1}{2^{n}}}=\lim _{n \rightarrow \infty} \frac{\left(x_{0}-1\right)^{\frac{1}{2^{n}}}-1}{\frac{1}{2^{n}}}=\ln \left(x_{0}-1\right)$ $3 \mathrm{p}$