$x_{n+1} = \frac{x_n+3}{x_n+1}$ converges to $\sqrt{3}$

Question

The sequence defined by

$$x_1 = 1, \qquad x_{n+1} = \frac{x_n+3}{x_n+1}$$

gives better and better approximations to $\sqrt{3}$

The first 3 terms are $x_1 = 1, x_2 = 2, x_3 = \frac{5}{3}$

Show that if the sequence converges, then it converges to $\sqrt{3}$

I don't see how this sequence converges to $\sqrt{3}$, can anyone shed some light on whether this question makes sense?

Answer 1

Suppose the proposed sequence converges to $L$, let us say. We shall prove first that $x_{n} > 0$ for every $n\in\mathbb{N}_{>0}$ using the induction principle. Clearly $x_{1} = 1 > 0$. Suppose that $x_{n} > 0$. Then we conclude the induction thesis by noticing that: \begin{align*} x_{n+1} = \frac{x_{n} + 3}{x_{n} + 1} > 0 \end{align*} since $x_{n} + 3 > 3 > 0$ and $x_{n} + 1 > 1 > 0$. Having said that, it yields that $L\geq 0$.

Consequently, on the assumption of convergence, one gets the desired result as next:

\begin{align*} \lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\frac{x_{n} + 3}{x_{n} + 1} & \Rightarrow L = \frac{L + 3}{L + 1} \Rightarrow L^{2} + L = L + 3 \Rightarrow L = \sqrt{3} \end{align*}

Answer 2

We can calculate the limit like so:

Let $L$ be the limit of the sequence. It follows that we must have $L=\frac{L+3}{L+1}$. We solve for $L$ to get \begin{align*}L(L+1)&=L+3 \\ L^2+L&=L+3\\ L^2&=3\\ L&=\pm\sqrt3.\end{align*} But obviously, the limit of the sequence must be positive, so $L=\sqrt3$ as stated.

Answer 3

The function $$f(x)= \frac{x+3}{x+1}$$ has two fixed points $\pm\sqrt{3}$ so we have $$\frac{f(x)-\sqrt{3}}{f(x)+\sqrt{3}} = k\cdot \frac{x-\sqrt{3}}{x+\sqrt{3}}$$ for some constant $k$ that turns out to be $k=\frac{1-\sqrt{3}}{1+\sqrt{3}}= -2+\sqrt{3}\in (-1,0)$.

From here we see that the sequence has limit $\sqrt{3}$.