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

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?

• In the formula $x_{n+1} = (x_n+3)/(x_n+1)$, replace the $x_n$ and $x_{n+1}$ with the limit, $x$. Then, solve for $x$.
– Doug
Commented Jul 12, 2022 at 2:59
• Also note that $x_n>0$, so the limit must be non-negative.
– Feng
Commented Jul 12, 2022 at 2:59

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*}

• Thanks for your answer! but why is it that when we plot the graph of $\frac{x+3}{x+1}$, we get a value that tends to "1" rather than $\sqrt{3}$?
– john
Commented Jul 12, 2022 at 3:10
• @john you are welcome! That is because it is a recurrence equation, not a function: each term $x_{n+1}$ is related to the term $x_{n}$ according to the proposed relation. Commented Jul 12, 2022 at 3:12
• @john the graph of the sequence is not necessarily $\frac{x+3}{x+1}$. Commented Jul 12, 2022 at 3:12
• @ÁtilaCorreia Ah I see.. thanks
– john
Commented Jul 12, 2022 at 3:16

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.

• It is correct that this method evaluates the limit if there was one, and you do determine the correct limit in this way. Unfortunately it does not prove that a limit exists at all. Its possible to have divergence and to still crank out a number this way. Youre only proving what the limit WOULD be IF it existed; you literally start with the assumption of convergence in order to evaluate it. Commented Jul 12, 2022 at 3:30
• @SquishyRhode That's exactly what the OP asked for: "Show that if the sequence converges, then it converges to $\sqrt 3$" Commented Jul 12, 2022 at 4:06

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 $$\sqrt{3}$$.