How to show that this sequence really converges How I'm supposed to show that the sequence $x_i$ , (i is an integer number) that is given below converges to $2$. 
$x_0=x_1=1$ and for $i\geq1$:
$$x_{i+1}=x_i+x_{i-1}-(x_i \cdot x_{i-1})/2$$
By substituting $i=2,3,\ldots$ I see that it converges to $2$ but formally I don't know how to accomplish it.

Let $u_i=1-x_i/2,\ \forall i\ge 0$. Then, we have the equivalent sequence $\{u_i\}_{i\ge 0}$ where $u_0=u_1=1/2$, and for $i\ge 1$, $$2-2u_{i+1}=2-2u_i+2-2u_{i-1}-(2-2u_i)(2-2u_{i-1})/2\\\implies u_{i+1}=u_i+u_{i-1}-1+(1-u_i)(1-u_{i-1})=u_iu_{i-1}.$$ Note that since $u_0,u_1\in (0,1)$, $u_i\in (0,1)$ for all $i\ge 1$. Therefore, defining $v_i=-\ln_2 u_i$, one obtains the recursion $$v_{i+1}=v_i+v_{i-1},$$ with $v_0=v_1=1$. Therefore, the sequence $\{v_i\}$ is the celebrated Fibonacci sequence with the explicit expression for $v_i$ as below: $$v_i=\frac{\phi^{i+1}-\psi^{i+1}}{\phi- \psi},$$ where $$\phi = \frac{1+\sqrt{5}}{2},\ \psi = \frac{1-\sqrt{5}}{2}.$$ Since $v_n\to \infty$ as $n\to \infty$, it follows that $u_n\to 0$ as $n\to \infty$, which implies that $x_n\to 2$ as $n\to \infty$. In fact, one can find an explicit expression of $x_n$ as $$x_n=2-2^{1-\frac{\phi^{n+1}-\psi^{n+1}}{\phi- \psi}},\ n\ge 0.$$
