If $\lim_{n\to\infty} x_{n+1}-\frac12 x_n = 0$ then $x_n\to 0$

Let $x_n$ be a sequence of real numbers such that $\displaystyle\lim_{n\to\infty} x_{n+1}-\frac12 x_n = 0$

Prove that $\lim_{n\to\infty} x_n = 0$

I have a proof if one assumes that $(x_n)$ is bounded. In that case, $(x_n)$ has as accumulation point, say $\alpha$, and the hypothesis $\lim_{n\to\infty} x_{n+1}-\frac12 x_n = 0$ implies that $2\alpha$ is also an accumulation point of $(x_n)$.

Hence $\forall p \;, \; 2^p\alpha$ is an accumulation point, which contradicts the boundedness assumption, unless $\alpha =0$. $0$ is consequently the only accumulation point of $(x_n)$ and the claim is proved.

It remains to show that $(x_n)$ is bounded. I've tried contradiction for that matter, without success.

Use that $|a+b|\ge ||a|-|b||$, so that $$ \lim_{n\to \infty} \left||x_{n+1}|-\left|\frac{x_n}{2}\right|\right|=0. $$ For each $\varepsilon>0$ there exists $n_0$ such that if $n\ge n_0$ then $$ \left||x_{n+1}|-\left|\frac{x_n}{2}\right|\right|\le \varepsilon \, \implies \, |x_{n+1}| \le \left|\frac{x_n}{2}\right|+\varepsilon $$ Assume wlog $n_0=0$, since we do not care a finite number of terms. Then by induction $$ |x_{n}| \le \left|\frac{x_0}{2^n}\right|+2\varepsilon. $$ The claim follows by the arbitrariness of $\varepsilon$.

Alternatively, rewrite the hypothesis as $$\lim_{n\to\infty}x_n\left(\frac{x_{n+1}}{x_n}-\frac{1}{2}\right)=0.$$ Even if we can't split the limit, as we don't know a priori that $x_n$ converges, we do know we must have either $x_n\to0$ or $x_{n+1}/ x_n \to 1/2 $, which still implies $x_n\to0$.

For the record, here the out-of-the-blue proof outlined by the authors of Selected problems in real analysis.

Let $\displaystyle y_n=x_n-\frac{x_{n-1}}{2}$.

Note that $$\displaystyle x_n-\frac{x_1}{2^{n-1}}=\sum_{1<k\leq n} \frac{y_k}{2^{n-k}}=\sum_{1<k\leq m} \frac{y_k}{2^{n-k}}+ \sum_{m<k\leq n} \frac{y_k}{2^{n-k}}$$

Choose $m$ large enough to make the second sum arbitrarily close to $0$, then choose $n\geq m$ large enough to make the first sum go to $0$.

Looking back at this question a few years later, here is a natural way to prove boundedness of $(x_n)$.

Let $\displaystyle z_n=x_{n+1}-\frac{x_{n}}{2}$ and let $M_z$ be a bound for the sequence $(z_n)$. Starting with small indexes $|x_0|\leq |x_0|$,
$|x_1| = |z_0+x_0/2|\leq M_z + 1/2|x_0|$,
$|x_2| = |z_1+x_1/2|\leq M_z + 1/2(M_z + 1/2|x_0|) = (1+1/2)M_z + 1/2^2|x_0|$,
$|x_3| = |z_2+x_2/2|\leq M_z + 1/2[(1+1/2)M_z + 1/2^2|x_0|] = (1+1/2+1/2^2)M_z + 1/2^3|x_0|$ and it is easily shown by induction that $$|x_n|\leq M_z(\sum_{k=0}^{n-1} 1/2^k) + 1/2^n |x_0|,$$ which is less than $2M_z + |x_0|$.