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$x_{n} = \sqrt{1+\dfrac{3}{x_{n-1}}}$ with initial condition $x_1 = 2$.

Prove that $\lim_{n\to\infty}{x_n}$ exist.

I've calculated the first few values and it seems to be an oscillating series with tending towards $L$, where $L$ is the real root to the cubic $x^3 - x - 3 = 0$.

I've tried expressing the sequence as two monotone sub-sequences that converges to the same limit, but to no avail. Would appreciate any help! Thank you!

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Hint:

Prove by induction that:

$$x_{2n+2}<x_{2n}$$

$$x_{2n+1}>x_{2n-1}$$


$$x_{2n+2}=\sqrt{1+\frac3{\sqrt{1+\frac3{x_{2n}}}}}<\sqrt{1+\frac3{\sqrt{1+\frac3{x_{2n-2}}}}}=x_{2n}$$

$$x_{2n+1}=\sqrt{1+\frac3{\sqrt{1+\frac3{x_{2n-1}}}}}>\sqrt{1+\frac3{\sqrt{1+\frac3{x_{2n-3}}}}}=x_{2n-1}$$

Thus, it is monotone and each subsequence can likewise be shown to be bounded and therefore convergent. Let $x^\star$ be the proposed limit. Show by induction in the same manner as above that

$$\sqrt{\frac52}<x_{2n}<x^\star<x_{2n+1}<2$$

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To prove that the sequence defined by $x_1=2$ and $x_{n+1}=\sqrt{1+\frac{3}{x_n}}$ is convergent you do not need to find an explicit form for $x_n$. As you noticed, if $x_n\to L$ then $L$ is a root of the polynomial $x^3-x-3$, that has a unique real root $\alpha\approx 1.6717$. So if such a sequence is convergent, it is convergent to $\alpha$. In order to prove it is convergent, you may apply the Banach fixed point theorem. If $f(x)=\sqrt{1+\frac{3}{x}}$, we have $f'(x)=-\frac{3}{2 x^2 f(x)}$, hence $\left|f'(x)\right|\leq\frac{3}{4}<1$ for any $x\geq 1$ and $\alpha$ is an attractive fixed point.

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  • $\begingroup$ "+1" for a slick argument. $\endgroup$ – Sahiba Arora Jun 5 '17 at 14:42
  • $\begingroup$ @SahibaArora slick? I'd just call it heavy duty. $\endgroup$ – Simply Beautiful Art Jun 5 '17 at 14:45
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    $\begingroup$ I've accepted the other answer for now since I'm not familiar with Banach fixed point theorem and the other answer seems more tractable for me right now. I will read it up and try to understand what it is doing! Thanks nonetheless! $\endgroup$ – David Jun 5 '17 at 14:46
  • $\begingroup$ @SimplyBeautifulArt Well then we have an example of a heavy duty slick argument. $\endgroup$ – Sahiba Arora Jun 5 '17 at 14:46

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