I 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
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?
Thank you!