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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 $L = 2 - \sqrt{3}$.

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    $\begingroup$ All seems to check out but be careful with (a). To have that inequality you need $x_k < 4$, which is not hard to prove but it is important. $\endgroup$ – marlasca23 Dec 21 '20 at 9:46
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    $\begingroup$ sounds like you got this covered and sounds boring. i think you should skip on ahead to measure theory and lebesgue integrals. you seem more than ready. $\endgroup$ – BCLC Dec 21 '20 at 9:50
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    $\begingroup$ @BCLC, yeah I began looking at both Rosenthal's book & Axler's book. Those are real nice to get started alongside Williams. $\endgroup$ – Quasar Dec 21 '20 at 9:56
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    $\begingroup$ Just fyi, a similar question was asked before. $\endgroup$ – rtybase Dec 21 '20 at 10:07
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    $\begingroup$ Your proof for (b) works, but relies on properties of this particular sequence. But "$\lim_n x_{n+a} = \lim_n x_n$" holds for any sequence and any integer offset $a$ (with appropriate care about the start of the sequences). If you have not thought about it, you should consider why this is true. $\endgroup$ – Paul Sinclair Dec 21 '20 at 17:20
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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 $2-\sqrt{3}$.

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