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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?

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    $\begingroup$ 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 $x$. $\endgroup$
    – Doug
    Commented Jul 12, 2022 at 2:59
  • $\begingroup$ Also note that $x_n>0$, so the limit must be non-negative. $\endgroup$
    – Feng
    Commented Jul 12, 2022 at 2:59

3 Answers 3

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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*}

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  • $\begingroup$ 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 $\sqrt{3}$? $\endgroup$
    – john
    Commented Jul 12, 2022 at 3:10
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    $\begingroup$ @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. $\endgroup$ Commented Jul 12, 2022 at 3:12
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    $\begingroup$ @john the graph of the sequence is not necessarily $\frac{x+3}{x+1}$. $\endgroup$
    – bobeyt6
    Commented Jul 12, 2022 at 3:12
  • $\begingroup$ @ÁtilaCorreia Ah I see.. thanks $\endgroup$
    – john
    Commented Jul 12, 2022 at 3:16
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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.

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    $\begingroup$ 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. $\endgroup$ Commented Jul 12, 2022 at 3:30
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    $\begingroup$ @SquishyRhode That's exactly what the OP asked for: "Show that if the sequence converges, then it converges to $\sqrt 3$" $\endgroup$
    – jjagmath
    Commented Jul 12, 2022 at 4:06
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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 $\sqrt{3}$.

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