Showing a recursive sequence is Cauchy let $a_1=1$, $a_2=2$, and define $a_n=\frac{1}{2}(a_{n-1}+a_{n-2})$. How can I show that this sequence is Cauchy?
I began with $|a_n-a_{n-1}|=\frac{1}{2}(a_n-a_{n-1})$
which goes to $\frac{1}{2}(a_{n-1}+a_{n-2})-\frac{1}{2}(a_{n-2}+a_{n-3})$
I am have been toying with this for quite some time and it is just not clicking. Can somebody please help!

You said yourself that
$|a_n-a_{n-1}| = \frac{1}{2}(a_{n-1}+a_{n-2}) - \frac{1}{2}(a_{n-2}+a_{n-3})$. 
Cancelling the $a_{n-2}$ terms leaves $\frac{1}{2}(a_{n-1}-a_{n-3})$. Expanding $a_{n-1}$ gives
$\frac{1}{2}(a_{n-1}-a_{n-3}) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}+a_{n-3})-a_{n-3})\right) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}-a_{n-3}))\right)$.
This means $(a_n-a_{n-1})$ = $\frac{1}{4} (a_{n-2} - a_{n-3})$. If $a_1$ and $a_2$ are fixed then $a_n-a_{n-1}$ follows exponential decay and converges to $0$. 
