Prove that if $ |a_n-a_{n-1}| < \frac{1}{2^{n+1} }$ and $a_0=\frac12$, then $\{a_n\}$ converges to $0I try to solve this question but I don't know how.
given $ a_0 = \frac12 $ and for each $n\geq 1$:
$$ |a_n-a_{n-1}| < \frac{1}{2^{n+1}} $$
show that $\{a_n\}$ converges and the limit is $a$ such that $0<a<1$
Update (Edited):
I showed by cauchy that $ |a_m-a_n| < |a_m-a_{m-1}+a_{m+1}-...+a_{n+1}-a_n| < \frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}}$
by the sum of Geometric series, $q=2, a_1=\frac{1}{2^{m+1}}$ then $s_n=\frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}]$, so $$\frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}} = \frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}] = \frac{1}{2^{n+1}}-\frac{1}{2^{m+1}}\leq\frac{1}{2^{n+1}}$$
now, it converges! 
Can someone help me please to show that the limit is $a$ with $0<a<1$?
Thank you!

You can also prove this generalization the same way:
if $|a_{n+1}-a_n| < c_n$
where $c_n$ is decreasing
and $\sum_{n=1}^{\infty} c_n$ converges,
then $\lim_{n \to \infty} a_n$ exists
and is less than
$a_1+\sum_{n=1}^{\infty} c_n$.
