Given a sequence $\{u_{n}\}$ so that $\left\{\begin{matrix} u_0=a>1\\u_{n+1}=\sqrt{1+\frac{1}{1+u_n}}\end{matrix}\right..$ Find $\lim u_n.$

Question

Given a sequence $\left \{ u_{n} \right \}$ so that $\left\{\begin{matrix} u_{0}= a> 1\\ u_{n+ 1}= \sqrt{1+ \frac{1}{1+ u_{n}}} \end{matrix}\right..$ Find $\lim u_{n}.$

If there exists a $\lim u_{n},$ I think with $n\rightarrow \infty\Rightarrow u_{n+ 1}= u_{n}= \sqrt{1+ \frac{1}{1+ u_{n}}}\Rightarrow 1< u_{n}< 2.$ What does that help for my problem here ? I need your thought on it ? Thank you.

Answer 1

Hint: Let $f(x) = \sqrt {1 + \frac{1}{{1 + x}}}$ for $x>0$. Then, for all $x>0$, $y>0$, \begin{align*} &\left| {f(x) - f(y)} \right| = \left| {\sqrt {1 + \frac{1}{{1 + x}}} - \sqrt {1 + \frac{1}{{1 + y}}} } \right| \\ &= \frac{{\left| {x - y} \right|}}{{(1 + x)(1 + y)\left( {\sqrt {1 + \frac{1}{{1 + x}}} + \sqrt {1 + \frac{1}{{1 + y}}} } \right)}} \le \frac{1}{2}\left| {x - y} \right|, \end{align*} i.e., $f$ is a contraction.

Answer 2

When $n \to \infty$

We have (if the limit does exist, we need to prove it is bounded from above/below and monotinicly increasing/decreasing):

$$ u_{n+1} \sim u_{n}$$

Thus we need to solve this equation:

$$ u_n = \sqrt{ 1 + \frac{1}{1+ u_n}} \\ \implies u_n \approx 1.206 $$