<s>[INST] Convergence of the recursive sequence $x_{n+1} = \frac{1}{4-x_n}$ I'm doing exercise 2.4.1 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, if my proof is rigorous and technically correct.

(a) Prove that the sequence defined by $x_1 = 3$ and
\begin{align*}
    x_{n+1} = \frac{1}{4 - x_n}
\end{align*}
converges.
(b) Now that we know $\lim x_n$ exists, explain why $\lim x_{n+1}$ must also exist and equal the same value.
(c) Take the limit of each side of the recursive equation in part (a) to explicitly compute the $\lim x_n$.

Proof.
(a) By direct computation, we find that $x_1 = 3$, $x_2 = 1$. Let us prove that $(x_n)$ is a decreasing sequence, that is $x_{n+1} < x_{n}$ for all $n \in \mathbf{N}$. This is true for $n=1$. By induction, let's assume that $x_{k+1} < x_k$. Therefore,
\begin{align*}
 x_{k+1} &< x_{k} \implies \frac{1}{4 - x_{k+1}} < \frac{1}{4 - x_k} \implies x_{k+2} < x_{k+1}
\end{align*}
So, $(x_{n})$ is a monotonically decreasing sequence.
Moreover, we can show that $(x_n)$ is bounded. We are interested to show that $x_n > 0$ for all $n \in \mathbf{N}$. This holds for $n=1$. Assume that $x_k > 0$, then
\begin{align*}
 x_{k+1} = \frac{1}{4 - x_k} > \frac{1}{4} > 0
\end{align*}
Thus, the sequence $(x_n)$ has a lower bound $0$. By the Monotone Convergence Theorem, the sequence $(x_n)$ converges.
(b) The sequence $(x_{n+1})=x_2,x_3,x_4,\ldots$ also converges and has the same limiting value because the $1$-tail of $(x_n)$ is a (i) monotonically decreasing sequence (ii) has the same lower bound $0$. It is the infinite tail of the sequence, that ultimately determines the convergence of a sequence.
(c) We have:
\begin{align*}
 \lim (x_n) &= \frac{1}{4 - \lim x_{n+1}}\\
 L &= \frac{1}{4 - L} \\
 4L - L^2 &= 1\\
 L^2 - 4L + 1 &= 0\\
 (L - 2)^2 - 3 &= 0\\
 L&= 2 \pm \sqrt{3}
\end{align*}
As $0 < L < 1$, we have $L = 2 - \sqrt{3}$.
 [/INST] An alternative solution.
It is clear from the form of the recursion that there is a sequence $\{y_n\}_{n=1}^\infty$ such that $x_n=\frac{y_n}{y_{n+1}}$.
$$\frac{y_{n+1}}{y_{n+2}}=\frac{1}{4-\frac{y_n}{y_{n+1}}}=\frac{y_{n+1}}{4y_{n+1}-y_{n+2}}\implies y_{n+2}=4y_{n+1}-y_n$$.
The base case is $y_1=3$ and $y_2=1$.
$$\lambda^2=4\lambda-1\implies \lambda=2\pm\sqrt{3}\implies y_n=C_+(2+\sqrt{3})^n+C_-(2-\sqrt{3})^n$$
We don't need to solve for the two constants, but their values are $C_{\pm}=\frac{33\mp 19\sqrt{3}}{6}$.
Nevertheless, $2-\sqrt{3}<1\implies\lim_{n\to\infty}(2-\sqrt{3})^n=0$ and so:
$\lim_{n\to\infty}x_n=\lim_{n\to\infty}\frac{C_+(2+\sqrt{3})^n+C_-(2-\sqrt{3})^n}{C_+(2+\sqrt{3})^{n+1}+C_-(2-\sqrt{3})^{n+1}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$.
Notice that this essentially solves all three parts together.
(a) The computation shows that the limit exists.
(b) We are just shifting the index.
(c) The limit's existence was established by actually finding it to be $2-\sqrt{3}$.
</s>