If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:
Attempt 1:
Write out the terms of $b_n$:
$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$
Adding up the terms you get:
$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$
But a dead end here.
Attempt 2:
For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$
Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges. 
For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end. 

Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then
$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$
Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any fixed $m$ and for any $n > m$, we have
$$ |a_n| \leq \Bigg| \underbrace{b_{n-1} + \cdots + \frac{b_{n-m}}{2^{m-1}}}_{\text{(1)}} \Bigg| + \underbrace{\frac{|b_{n-m-1}|}{2^m} + \cdots \frac{|a_1|}{2^{n-1}}}_{(2)}.$$
Note here that


*

*$\text{(1)}$ consists of fixed number of terms, each tending to zero as $n\to\infty$.

*$\text{(2)}$ is uniformly bounded by $\frac{M}{2^m} + \frac{M}{2^{m+1}} + \cdots = \frac{M}{2^{m-1}}$.
So, taking limsup as $n\to\infty$ yields
$$ \limsup_{n\to\infty} |a_n| \leq \frac{M}{2^{m-1}}. $$
Since the LHS is a fixed number and $m$ is arbitrary, letting $m\to\infty$ proves the claim.
