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. 
 A: 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.
A: Let $c_n = 2^n a_n$. The assumption reads 
$$ \frac{c_{n+1}-c_n}{2^{n+1}-2^n} \to 0. $$
The sequence $2^n$ tends monotonically to $+\infty$, hence by the Stolz–Cesàro theorem 
$$ a_n = \frac{c_n}{2^n} \to 0. $$
A: Let $\epsilon > 0$.
Since $a_{n+1}-a_n/2$ converges to $0$, there is an integer $m$ such that
for any $n \ge m$, $|a_{n+1}-a_n/2| \le \epsilon/4$. 
For such an $n$, you have
$|a_{n+1}| - \epsilon/2 \\
\le |a_{n+1} - a_n/2| + |a_n/2| - \epsilon/2 \\
\le \epsilon/4 + |a_n|/2 - \epsilon/2 \\
= |a_n|/2 - \epsilon/4 \\
= (|a_n| - \epsilon/2)/2$
Intuitively you can interpret this as something saying that $|a_n|$ has to decrease somewhat exponentially at least until $|a_n|$ gets too close to $\epsilon/2$.
Then let us show there is an $m' \ge m$ such that $|a_{m'}| \le \epsilon$.  
If $|a_m| \le \epsilon$ then we are already done by picking $m'=m$, so suppose $|a_m| > \epsilon$.
Now, $(|a_m|- \epsilon/2) / (\epsilon/2) > 1 > 0$ so there is an integer $k$ such that $2^k \ge (|a_m|- \epsilon/2) / (\epsilon/2)$.
Looking at $m' = m+k$ we get
$|a_{m+k}| - \epsilon/2 \le (|a_m| - \epsilon/2) 2^{-k} \le \epsilon/2$,
and so $|a_{m'}| \le \epsilon$.  
Then we can prove by induction that for any $n \ge m'$, $|a_n| \le \epsilon$ :  
This is true for $n=m'$.
Suppose $n \ge m'$ and $|a_n| \le \epsilon$.
Then $|a_n|-\epsilon/2 \le \epsilon/2$, and so because $n \ge m$,
$|a_{n+1}| - \epsilon/2 \le (|a_n| - \epsilon/2)/2 \le \epsilon/4 < \epsilon/2$, and finally $|a_{n+1}| \le \epsilon$.  
Therefore, for all $n \ge m', |a_n| \le \epsilon$, and we have shown that the sequence $a_n$ converges to $0$.
A: Let $b_n=a_{n+1}-(a_{n}/2)$ so that $b_{n} \to 0$ as $n\to\infty$. Now we have $a_{2}=b_{1}+(a_{1}/2),a_{3}=b_{2}+b_{1}/2+a_{1}/4$ and continuing in this manner we get $$a_{n+1}=b_{n}+b_{n-1}/2+\cdots+b_{1}/2^{n-1}+a_{1}/2^{n}$$ Let $\epsilon>0$ be given then we can see that there is a positive integer $m$ such that $|b_n|<\epsilon$ for all $n\geq m$. Thus if $n>m$ then we have $$|a_{n+1}|\leq |b_{n} |+|b_{n-1}|/2+\cdots+|b_m|/2^{n-m}+|b_{m-1}|/2^{n-m+1}+\cdots +|b_{1}|/2^{n-1}+|a_{1}|/2^{n}$$ First $(n-m+1) $ terms on right are together less than $2\epsilon$ and the rest of $m$ terms tend to $0$. Thus taking $\limsup$ on both sides we see that $\limsup|a_{n+1}|\leq 2\epsilon$ and since $\epsilon$ was arbitrary this means that $a_{n} \to 0$ as $n\to\infty$. 
A: Hint :
If $L_t n > \infty$ $\frac{A_{n+1}}{A_{n}}$ is less than $1$. Then the sequence $A_n$ must converge.
