Showing a recursive sequence is Cauchy

Try to prove the following: $|a_n-a_{n-1}|=1/2^{n-1}$.

If you prove that lemma, then for fixed $\epsilon>0$, choose an $N$ so that $1/2^{N-1}<\epsilon$.

Then, if $m,n\geq N$ (wlog say $m\geq n$), you have $$|a_m-a_n|\leq |a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+...+|a_{n+1}-a_n|$$ $$=\frac{1}{2^{m-1}}+...+\frac{1}{2^n}=\frac{1}{2^n}(\frac{1}{2^{m-1-n}}+...+1)<\frac{1}{2^n}(2)=\frac{1}{2^{n-1}}$$but since $n\geq N$, we have that $\frac{1}{2^{n-1}}\leq \frac{1}{2^{N-1}}<\epsilon$.

Do you notice that the distances between succeeding terms decay exponentially?

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 $0$.