Think about Fibonacci sequence

Relevant wiki: Fibonacci Sequence

We note that

\(\begin{array} {} x_0 = 1 = \dfrac 11 & = \dfrac {F_2}{F_1} \\ x_1 =1 + \dfrac 1{1+1} = \dfrac 32 & = \dfrac {F_4}{F_3} \\ x_2 =1 + \dfrac {\frac 32}{1+\frac 32} = \dfrac 85 & = \dfrac {F_6}{F_5} \end{array} \)

The first few $x_n$ implies the claim that $x_n = \dfrac {F_{2n+2}}{F_{2n+1}}$ , where $F_n$ is the $n$ th Fibonacci number. Let us prove by induction that the claim is true for all $n \ge 0$ . Assuming the claim is true for $n$ , then:

$\begin{aligned} x_{n+1} & = 1+ \frac {x_n}{1 + x_n} = 1 + \frac {\frac {F_{2n+2}}{F_{2n+1}}}{1+\frac {F_{2n+2}}{F_{2n+1}}} = 1 + \frac {F_{2n+2}}{F_{2n+1}+F_{2n+2}} = 1 + \frac {F_{2n+2}}{F_{2n+3}} = \frac {F_{2n+3} + F_{2n+2}}{F_{2n+3}} = \frac {F_{2n+4}}{F_{2n+3}} \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 0$ . Then we have $\displaystyle \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac {F_{2n+4}}{F_{2n+3}} = {\color{#3D99F6}\varphi} \approx \boxed{1.618}$ , where $\color{#3D99F6} \varphi$ denotes the golden ratio.