What is the limit of x(1)=0, x(2)=1, x(n) = (x(n-1)+x(n-2))/2 [duplicate]

The sequence is defined by recurrence: $x_1=1$, $x_2=2$ \begin{equation} x_{n+1} = \frac{1}{2}(x_{n}+x_{n-1}) \end{equation} It is asked to calculate the limit. I have seen that the sequence is bounded by 1 and 2. On the other hand, I have seen that the sequence of the pairs, $x_{2n}$, is decreasing and the sequence of the odd, $x_{2n-1}$, is increasing. Therefore both are convergent. I have seen that both limits are equal, therefore the original sequence is convergent with the same limit.

But I won't be able to prove what the limit is.

$x_{n+2}$ is the middle of $[x_n,x_{n+1}]$, therefore the distance between $x_n$ and $x_{n+1}$ is divided by $2$ at each iteration, namely $$ \frac{x_{n+2}-x_{n+1}}{2}=\frac{x_{n+1}+x_n-2x_{n+1}}{4}=-\frac{1}{2}\frac{x_{n+1}-x_n}{2} $$ (there is a $-$ because $(x_{2n})$ and $(x_{2n+1})$ are of opposite monotony). Therefore $$ \frac{x_{n+1}-x_n}{2}=\left(-\frac{1}{2}\right)^{n-1}\frac{x_2-x_1}{2}=\frac{(-1)^{n-1}}{2^n} $$ Thus $x_{n+1}-x_n=\frac{(-1)^{n-1}}{2^{n-1}}$, summing this gives that $$ x_n=x_1+\sum_{k=1}^{n-1}(x_{k+1}-x_k)=1-\sum_{k=1}^{n-1}\left(-\frac{1}{2}\right)^{k-1}=1+\frac{1-(-1/2)^{n-1}}{1+1/2}=1+2\frac{1-(1/2)^{n-1}}{3} $$ Therefore $\lim\limits_{n\rightarrow +\infty}x_n=\frac{5}{3}$ (you could also solve the characterisic equation of $(x_n)$ and do the standard calculations.)