Prove or disprove: $\lim_{n\to \infty}\left(a_{n+1} - \frac 12 a_n\right)=0 \Rightarrow \lim_{n\to \infty}a_n=0$ I am stuck on a simple exercise. Let $(a_n)_{_n\in\mathbb N}$ be a sequence of real numbers. Prove or disprove the following statement:
$$\lim_{n\to \infty}\left(a_{n+1} - \frac 12 a_n\right)=0 \Rightarrow \lim_{n\to \infty}a_n=0.$$
I tried to prove it directly but didn't make any progress. So I considered proving the contraposition: If you assume $(a_n)_{_{n\in\mathbb N}}$ is a convergent sequence, the contraposition of this statement is easy to prove. Let $\lim_{n\to \infty}a_n=:a\neq0$, then $\lim_{n\to \infty}(a_{n+1} - \frac 12 a_n)=a-\frac a2= \frac a2 \neq 0$, and therefore $\lim_{n\to \infty}(a_{n+1} - \frac 12 a_n)=a-\frac a2= \frac a2 \neq 0$. But to correctly prove the contraposition, I also have to consider divergent series. Does $$(a_n)_{_{n\in\mathbb N}}\;divergent \Rightarrow \left( a_{n+1} - \frac 12 a_n \right)_{_{n\in\mathbb N}}\;divergent\; \lor\lim_{n\to \infty}\left(a_{n+1} - \frac 12 a_n\right)\neq0$$ hold? If yes, it would prove the mentioned statement above.
 A: You have that
$$\lim_{n\to \infty}\left(a_{n + 1} - \frac{1}{2}a_n\right) = 0 \tag{1}\label{eq1A}$$
This means that, by the definition of limits, for any $\epsilon \gt 0$, there's an integer $n_0$ such that for all $n \ge n_0$ you have
$$\left|a_{n + 1} - \left(\frac{1}{2}\right)a_n\right| \lt \epsilon \implies -\epsilon \lt a_{n+1} - \left(\frac{1}{2}\right)a_n \lt \epsilon \tag{2}\label{eq2A}$$
The procedure below is somewhat similar to what user2661923's question comment suggests, which I read while I was writing this answer. In addition, as stated in the comment, I also don't see any way offhand to finish the proof using contraposition as you tried.
For $n = n_{0}$, \eqref{eq2A} gives
$$-\epsilon \lt a_{n_{0} + 1} - \left(\frac{1}{2}\right)a_{n_{0}} \lt \epsilon \tag{3}\label{eq3A}$$
Next, for $n = n_{0} + 1$, you also have
$$-\epsilon \lt a_{n_{0} + 2} - \left(\frac{1}{2}\right)a_{n_{0} + 1} \lt \epsilon \tag{4}\label{eq4A}$$
Multiplying all $3$ parts of \eqref{eq3A} by $\frac{1}{2}$ and adding the results to \eqref{eq4A} gives
$$-\left(1 + \frac{1}{2}\right)\epsilon \lt a_{n_{0} + 2} - \left(\frac{1}{4}\right)a_{n_{0}} \lt \left(1 + \frac{1}{2}\right)\epsilon \tag{5}\label{eq5A}$$
Now, for $n = n_0 + 2$, \eqref{eq2A} gives
$$-\epsilon \lt a_{n_{0} + 3} - \left(\frac{1}{2}\right)a_{n_{0} + 2} \lt \epsilon \tag{6}\label{eq6A}$$
Multiplying all $3$ parts of \eqref{eq5A} by $\frac{1}{2}$ and adding the results to \eqref{eq6A} gives
$$-\left(1 + \frac{1}{2} + \frac{1}{4}\right)\epsilon \lt a_{n_{0} + 3} - \left(\frac{1}{8}\right)a_{n_{0}} \lt \left(1 + \frac{1}{2} + \frac{1}{4}\right)\epsilon \tag{7}\label{eq7A}$$
You can repeat this procedure $k$ times to get
$$-2\epsilon \lt -\left(\sum_{i=0}^{k-1}\frac{1}{2^{i}}\right)\epsilon \lt a_{n_0 + k} - \left(\frac{1}{2^{k}}\right)a_{n_0} \lt \left(\sum_{i=0}^{k-1}\frac{1}{2^{i}}\right)\epsilon \lt 2\epsilon \tag{8}\label{eq8A}$$
This can be fairly easily proven, e.g., by induction, which I'll leave to you to do.
Next, choose a $k_{0}$ large enough so that $\left|\left(\frac{1}{2^{k_0}}\right)a_{n_0}\right| \lt \epsilon$, e.g., $k_{0} = 1$ if $a_{n_0} = 0$, else $k_0 = \max(\left\lfloor\log_{2}{|a_{n_0}|} - \log_{2}{\epsilon}\right\rfloor + 1, 1)$. Then, since $\left|\left(\frac{1}{2^{k_0}}\right)a_{n_0}\right|$ is a non-negative decreasing function in $k_{0}$ for all $k \ge k_{0}$, you have
$$-3\epsilon \lt a_{n_0 + k} \lt 3\epsilon \implies \left|a_{n_0 + k}\right| \lt 3\epsilon \tag{9}\label{eq9A}$$
I trust you can finish the rest of the proof to show that
$$\lim_{n \to \infty}a_n = 0 \tag{10}\label{eq10A}$$
A: We prove the following general claim, since doing so does not harm the essence of the idea:

Tauberian Theorem for Nørlund means. Let $(b_n)$ and $(c_n)$ be sequences such that
  
  
*
  
*$b_n > 0$ for all $n \geq 1$ and $\frac{b_n}{b_1 + \cdots + b_n} \to 0$ as $n\to\infty$.
  
*$c_n \to \ell$ as $n\to\infty$ for some $\ell$.
  
  
  Then we have
  $$ \lim_{n\to\infty} \frac{b_1 c_n + b_2 c_{n-1} + \cdots + b_n c_1}{b_1 + \cdots + b_n} = \ell. $$

Before proving this theorem, we check that this indeed implies the desired claim. Choose
$$b_n = 2^{1-n} \qquad\text{and} \qquad c_n = a_{n+1} - \frac{1}{2}a_n.$$
If $c_n$ converges to some limit $\ell$, then
$$ \frac{a_{n+1} - 2^{-n}a_1}{2 - 2^{-n}} = \frac{b_1 c_n + b_2 c_{n-1} + \cdots + b_n c_1}{b_1 + b_2 + \cdots + b_n} \xrightarrow{n\to\infty} \ell = 0. $$
This implies that $a_n \to 2\ell$ as $n\to\infty$. In OP's case, we have $\ell = 0 $ and therefore the desired conclusion follows.

Proof of Theorem. Fix an arbitrary $N \geq 1$. Then for any $n \geq N$,
\begin{align*}
&\left| \frac{b_1 c_n + b_2 c_{n-1} + \cdots + b_n c_1}{b_1 + \cdots + b_n} - \ell \right| \\
&\leq \sum_{k=1}^{n} \frac{b_{n+1-k}}{b_1+\cdots+b_n} |c_k - \ell| \\
&\leq \sum_{k=1}^{N} \frac{b_{n+1-k}}{b_1+\cdots+b_n} |c_k - \ell| + \biggl( \sup_{k > N} |c_k - \ell| \biggr) \sum_{k=N+1}^{n} \frac{b_{n+1-k}}{b_1+\cdots+b_n} \\
&\leq \sum_{k=1}^{N} \frac{b_{n+1-k}}{b_1+\cdots+b_{n+1-k}} |c_k - \ell| +\sup_{k > N} |c_k - \ell|.
\end{align*}
Taking $\limsup$ as $n\to\infty$, the first sum in the last step converges to $0$ since each of the $N$ terms converges to $0$. So we obtain a bound
$$ \limsup_{n\to\infty} \left| \frac{b_1 c_n + b_2 c_{n-1} + \cdots + b_n c_1}{b_1 + \cdots + b_n} - \ell \right| \leq \sup_{k > N} |c_k - \ell|. $$
But since the left-hand side is independent of $N$, letting $N\to\infty$ shows that the limsup is in fact zero. This implies the desired convergence. $\square$
A: Let $ \varepsilon >0 \cdot $
There exists some $ n_{1}\in\mathbb{N} $ such that $ \left(\forall n\geq n_{1}\right),\ \left|a_{n+1}-\frac{1}{2}a_{n}\right|<\varepsilon \cdot $
And we have for every $ n> n_{1} $ :
\begin{aligned} \left|a_{n}\right|=\left|\sum_{k=0}^{n-1}{\frac{1}{2^{n-1-k}}\left(a_{k+1}-\frac{1}{2}a_{k}\right)}+\frac{a_{0}}{2^{n}}\right|\\ \leq\sum_{k=0}^{n-1}{\frac{1}{2^{n-1-k}}\left|a_{k+1}-\frac{1}{2}a_{k}\right|}+\frac{\left|a_{0}\right|}{2^{n}}\ \ \ &=\sum_{k=0}^{n_{0}-1}{\frac{1}{2^{n-1-k}}\left|a_{k+1}-\frac{1}{2}a_{k}\right|}+\sum_{k=n_{0}}^{n-1}{\frac{1}{2^{n-1-k}}\left|a_{k+1}-\frac{1}{2}a_{k}\right|}+\frac{\left|a_{0}\right|}{2^{n}}\\ &\leq\frac{1}{2^{n}}\sum_{k=0}^{n_{0}-1}{2^{k+1}\left|a_{k+1}-\frac{1}{2^{k}}a_{k}\right|}+\varepsilon\sum_{k=n_{0}}^{n-1}{\frac{1}{2^{n-1-k}}}+\frac{\left|a_{0}\right|}{2^{n}} \end{aligned}
Since $ \sum\limits_{k=0}^{n_{0}-1}{2^{k+1}\left|a_{k+1}-\frac{1}{2^{k}}a_{k}\right|} $ and $ \left|a_{0}\right| $ aren't depending of $ n $, we have : $ \lim\limits_{n\to +\infty}{\frac{1}{2^{n}}\left(\left|a_{0}\right|+\sum_{k=0}^{n_{0}-1}{\frac{1}{2^{n-1-k}}\left|a_{k+1}-\frac{1}{2}a_{k}\right|}\right)}=0 $, thus there exists some $ n_{2} $ such that $$ \left(\forall n\geq n_{1}\right),\ \frac{1}{2^{n}}\left(\left|a_{0}\right|+\sum_{k=0}^{n_{0}-1}{\frac{1}{2^{n-1-k}}\left|a_{k+1}-\frac{1}{2}a_{k}\right|}\right)<\varepsilon $$
Hence, for every $ n> \max\left(n_{1},n_{2}\right) $, we have : $$ \left|a_{n}\right|<\varepsilon +\varepsilon\sum_{k=n_{0}}^{n-1}{\frac{1}{2^{n-1-k}}}=\varepsilon\left(3-2^{n_{0}-n+1}\right)\leq 3\varepsilon $$
Thus : $$ \lim_{n\to +\infty}{a_{n}}=0 $$
