Prove that $|a_{n+1}-a_{n}| \leq \frac{1}{2}|a_{n}-a_{n-1}|$ implies $\{a_n\}$ bounded and convergent. I am trying to prove that every sequence for which
$|a_{n+1}-a_{n}| \leq \frac{1}{2}|a_{n}-a_{n-1}|$
is true, is bounded and convergent.
I tried defining $b_n = a_n$ and $c_n = a_{n-1}$ and then use Stolz, but it didn't work. 
How could I prove this?
 A: Define $c:=|a_1-a_0|\ge 0$. Then by induction $|a_{n+1}-a_{n}|\le \frac{c}{2^n}$, therefore $(x_n)_n$ is a Cauchy sequence indeed
$$
|x_n-x_m|\le |x_n-x_{n-1}|+|x_{n-1}-x_{n-2}|+\cdots+|x_{m+1}-x_m| \le \frac{c}{2^{m-1}}
$$
for all $n>m$. Then it is enough to choose $m$ sufficiently large so that $|x_n-x_m|<\varepsilon$.
A: For the boundedness:
You can write, using the triangle inequality and your assumption,$^{(\dagger)}$ $$
\lvert a_n - a_0 \rvert = \left\lvert\sum_{k=0}^{n-1} (a_{k+1}-a_k)\right\rvert
\leq \sum_{k=0}^{n-1} \left\lvert a_{k+1}-a_k\right\rvert
\leq \sum_{k=0}^{n-1} \frac{1}{2^{k}}\left\lvert a_{1}-a_0\right\rvert
\leq  \sum_{k=0}^{\infty} \frac{1}{2^{k}}\left\lvert a_{1}-a_0\right\rvert
= 2\left\lvert a_{1}-a_0\right\rvert
$$
so we get $\lvert a_n\rvert \leq \lvert a_n - a_0 \rvert + \lvert a_0\rvert \leq 2\lvert a_1 - a_0 \rvert + \lvert a_0\rvert$ for all $n$.
$(\dagger)$ Indeed, by a straightforward induction we get $\left\lvert a_{n+1}-a_n\right\rvert
\leq \frac{1}{2^{n}}\left\lvert a_{1}-a_0\right\rvert$ for all $n\geq 0$.
(for the convergence, the same "trick" but starting the sum at some $m$ (i.e., with $a_m$ instead of $a_0$) will enable you to show the sequence is Cauchy.)
A: Hint: Try
$$ |a_n|=|a_1+(a_2-a_1)++\cdots+(a_{n-1}-a_{n-2})+(a_n-a_{n-1})|\le|a_1|+|a_2-a_1|+\cdots+|a_{n-1}-a_{n-2}|+|a_n-a_{n-1}|$$
