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...

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?
