Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy 
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy.

My work :
Need to show that for every $\epsilon \gt 0$ there exist $N$ such that $n,m\ge N \implies  | a_n - a_m|  \lt\epsilon$.
$$|a_n-a_m| = \dfrac{1}{2}|(a_{n-1} + a_{n-2}) - ( a_{m-1} + a_{m-2})|$$
I feel triangle inequality might be helpful here, but really not sure how to link it to the $\epsilon$. Appreciate any help...
 A: Note that for $n\geq 0$,
$$a_{n+2}-a_{n+1} =\frac{ (a_{n+1}+a_n)}{2}-a_{n+1}=\frac{a_n-a_{n+1}}{2}.$$
Hence
$$|a_{n+2}-a_{n+1}|= \frac{1}{2}|a_{n+1}-a_n|= \frac{1}{2^2}|a_{n}-a_{n-1}|= \frac{1}{2^n}|a_{2}-a_{1}|=\frac{1}{2^n}.$$
Now if $n>m\geq 1$ then, by the triangle inequality,
$$|a_{n}-a_{m}|\leq |a_{n}-a_{n-1}|+\dots+|a_{m+1}-a_{m}|= 
\frac{1}{2^{n-2}}+\dots+\frac{1}{2^{m-1}}.$$
Can you take it from here?
A: A slightly unorthodox approach to this problem (as I have nothing else to add to Robert's spot on answer) is to solve the recurrence using (for example) characteristic polynomials, which for 
$a_{n+2}=\frac{a_{n+1}+a_n}{2}$ is:
$$2x^2-x-1=0$$
which has the following roots $x_1=1$ and $x_2=-\frac{1}{2}$ and the general form
$$a_n=C_1\cdot x_1^n+C_2\cdot x_2^n=C_1+C_2\cdot \left(-\frac{1}{2}\right)^n$$
Using the initial condition $a_1=1, a_2=2$ we have
$$\left\{\begin{matrix}
1=C_1-\frac{C_2}{2}\\ 
2=C_1+\frac{C_2}{4}
\end{matrix}\right.$$
leading to $C_1=\frac{5}{3}$ and $C_2=\frac{4}{3}$ or
$$a_n=\frac{5}{3}+\frac{4}{3}\cdot \left(-\frac{1}{2}\right)^n$$
and
$$\lim\limits_{n\rightarrow\infty}a_n=\frac{5}{3}$$
