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!
 A: Hint: $\vert a_n - a_0 \vert = \vert a_n - a_{n-1} + a_{n-1} - \dots - a_1 + a_1 - a_0 \vert \leq \sum_{k=1}^n \vert a_k - a_{k-1} \vert$
A: Hint: Fix any $m$, and then use triangle inequality and induction to show that $$|a_n-a_m|<\frac1{2^{m+1}}-\frac1{2^{n+1}}$$ for all $n>m$. It follows from this that $\{a_n\}$ is Cauchy, and so converges, say to $a$. In particular, setting $m=0$, we have $$|a_n-a_0|<\frac12-\frac1{2^{n+1}}$$ for all $n$, and since $a_0=\frac12$, we have $0<a_n<1$ for all $n>0$. Thus, $0\leq a\leq 1$.
It remains only to show that $\{a_n\}$ cannot converge to $0$ or to $1$. Note that if we set $b_n=1-a_n$, then $\{b_n\}$ has all the same characteristics as $\{a_n\}$. If we can show that $a\neq 0$, then identical arguments show $b\neq 0$, and so $a=1-b\neq 1$, completing the proof. Thus, we need only show that $a>0$. I recommend noting that $a_1=\frac14+c$ for some $c>0$, and use the work above to conclude that $a\geq c$.
Addendum: Don't waste your time trying to determine the precise values of all the $a_n$s, nor of $a$--there simply isn't enough information given for us to determine this. Fortunately, we don't need that much information to prove the desired results.
A: 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
$a_1+\sum_{n=1}^{\infty} c_n$.
