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 $\lim u_n.$ 
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.

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.
