Let $x_1 = 2$, $x_{n+1}=2+\frac{1}{x_{n}}$, $n \in \mathbb{N}$. Does the sequence converge or diverge?

Question

Let $x_1 = 2$, $x_{n+1}=2+\dfrac{1}{x_{n}}$, $n \in \mathbb{N}$. Does the sequence converge or diverge?

How can I solve this question using Monotone Convergence Theorem or any other better method? I can't think of a way to show it as monotone increasing or decreasing?

Answer 1

Hint: Let $r$ be the positive root of the equation $x=2+\frac{1}{x}$ (which is $1+\sqrt{2}$).

$|x_{n+1}-r|=|(2+\frac{1}{x_n})-(2+\frac{1}{r})|=\frac{|x_n-r|}{r|x_n|}$

Answer 2

Let $p_{1}=2$, $q_{1}=1$.

\begin{align*} p_{n+1} &= 2p_{n}+q_{n} \\ q_{n+1} &= p_{n} \\ p_{n} & > q_{n} \\ p_{n} &\ge 2^{n} \\ q_{n} &\ge 2^{n-1} \end{align*}

Also, \begin{align*} p_{1}^{2}-2p_{1}q_{1}-q_{1}^{2} &= -1 \\ p_{n+1}^{2}-2p_{n+1}q_{n+1}-q_{n+1}^{2} &= -(p_{n}^{2}-2p_{n}q_{n}-q_{n}^{2}) \\ &= (-1)^{n+1} \\ \frac{p_{n+1}^{2}}{q_{n+1}^{2}}-\frac{2p_{n+1}}{q_{n+1}}-1 &= \frac{q_{n}^{2}}{p_{n}^{2}}+\frac{2q_{n}}{p_{n}}-1 \\ &= \frac{(-1)^{n+1}}{p_{n+1}^{2}} \\ x_{n} &= \frac{p_{n}}{q_{n}} \\ x_{n+1}-x_{n} &= \frac{2p_{n}+q_{n}}{p_{n}}-\frac{p_{n}}{q_{n}} \\ &= \frac{2p_{n}q_{n}+q_{n}^{2}-p_{n}^2}{p_{n}q_{n}} \\ &= \frac{(-1)^{n+1}}{p_{n}q_{n}} \\ &= \frac{(-1)^{n+1}}{q_{n}q_{n+1}} \\ \end{align*}

Now \begin{align*} L &= \lim_{n\to \infty} x_{n} \\ 0 &= L^2-2L-1 \\ L &= \sqrt{2}+1 \tag{reject $1-\sqrt{2}$} \end{align*}

Further notes:

Solving the recurrence relations explicitly, we have $$ \begin{pmatrix} p_{n} \\ q_{n} \end{pmatrix}= \begin{pmatrix} \dfrac{(1+\sqrt{2})^{n+1}+(1-\sqrt{2})^{n+1}}{2} \\[3pt] \dfrac{(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}}{2\sqrt{2}} \end{pmatrix}$$

Answer 3

Hint For any $c>1$, $f(x) = 2 + \frac1x$ is contractive in $[c,\infty)$ (why?). Apply Banach fixed-point theorem to the recurrence $x_{n+1} = f(x_n), x_1 = 2$.

Answer 4

First, show that $$x_1 + \sum_{i=1}^{N-1} |x_{k+1}-x_k|$$ converges (This is simple, since we can show by induction that $x_n\ge 2$, for all $n$ and $|x_{n+2}-x_{n+1}|\le\frac{|x_{n+1}-x_{n}|}{4}$). Then, from the fact that "absolute convergence implies convergence" and also $$ x_N = x_1 + \sum_{i=1}^{N-1} x_{k+1}-x_k $$ we have the result.