$a_1 > 0$ and for all $n$, $a_{n+1}=1+\frac{2}{a_n}$

Question

We have sequence $(a_n)$, where $a_1 > 0$ and for all $n$, $a_{n+1}=1+\frac{2}{a_n}$. I made two subsequences of ${a_n}$ such as one is decreasing, but is bigger than $2$ and the other one is increasing, but is smaller than $2$. I have proved that part. How can I show that the limit of those two subsequences is $2$?

Answer 1

For the existence of the limit see this duplicate. If it exists, let's call it $x$.

Then, $$x = \lim_{n\to\infty} x_n = \lim_{n\to\infty} \left(1 + \frac{2}{x_{n-1}}\right) = 1 + \frac{2}{\lim_{n\to\infty} x_{n-1}} = 1 + \frac{2}{x},$$ so that $$x^2-x-2=0.$$

Answer 2

HINT: use that $$a_n=-\frac{\left(-\frac{1}{2}\right)^n+4C}{\left(-\frac{1}{2}\right)^n-2C}$$

Answer 3

By setting $a_n=\sqrt{2}\,b_n$ we get $$ \sqrt{2}\,b_{n+1} = 1 + \frac{\sqrt{2}}{b_n},\qquad b_{n+1} = \frac{1}{\sqrt{2}}+\frac{1}{b_n} $$ hence with the notation of (generalized) continued fractions $$ b_{n+1} = \left[\tfrac{1}{\sqrt{2}}, b_n\right]=\left[\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}, b_{n-1}\right]=\ldots=\Big[\underbrace{\tfrac{1}{\sqrt{2}},\ldots,\tfrac{1}{\sqrt{2}}}_{n+1\text{ times}},\tfrac{a_0}{\sqrt{2}}\Big] $$ and for any $a_0>0$ we have that $b_n\to \left[\,\overline{\tfrac{1}{\sqrt{2}}}\,\right]=\sqrt{2}$ as $n\to +\infty$ and $a_n\to \color{red}{2}$.