The sequence $\langle u_n\rangle$ is not decreasing sequence : $u_0=1$ and $u_1=2>1$. However, you can prove following facts:
- $\langle u_{2n}\rangle$ is decreasing.
- $\langle u_{2n-1}\rangle$ is increasing.
- $1\le u_n \le 2$ for all $n$.
- $u_{2n+1}-u_{2n}\to 0$ as $n$ goes to infinity.
By the 1, 2 and 3, you can prove the convergence of $\langle u_{2n}\rangle$ and
$\langle u_{2n-1}\rangle$. By 4, you can check the limit of $\langle u_{2n}\rangle$ and $\langle u_{2n-1}\rangle$ have same value (and $\langle u_n\rangle$ converges.) By 3, the limit of $\langle u_n\rangle$ is greater or equal than $1$. Take $u=\lim_{n\to\infty} u_n$, then the relation between $u_n$ and $u_{n+1}$, you get $u=1+\frac{1}{u}$. Therefore $u=\frac{1}{2}(1\pm\sqrt{5})$. However $u\ge 1$ so $u=\frac{1}{2}(1+\sqrt{5})$.