TITLE: Prove or disprove: $\lim_{n\to \infty}\left(a_{n+1} - \frac 12 a_n\right)=0 \Rightarrow \lim_{n\to \infty}a_n=0$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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. $\square$