Let $\{x_n\}$ be a sequence and let $r$ be a number such that $0 < r < 1$. Suppose that $$|x_{n+1} – x_n| \le r|x_n - x_{n-1}|$$ for all $n>1$. Prove that $\{x_n\}$ is a Cauchy sequence.
I assume this is going to involve the triangle inequality and a substitution somewhere, but the $r$ thrown in is really annoying. Does anyone know how to deal with this?
On
HINT: Show by induction on $m\ge 1$ that $$|x_{n+m}-x_n|\le\left(\sum_{k=0}^{m-1}r^k\right)|x_{n+1}-x_n|\;.$$
On
To show that the sequence is Cauchy, we will be looking at $|x_n-x_m|$, where say $n\gt m$, and trying to get an estimate of its size.
Let $|x_1-x_0|=a$
We have $|x_2-x_1|\le r|x_1-x_0|=ar$
Similarly, $|x_3-x_2|\le r|x_2-x_1|\le ar^2$, and in general $|x_{k+1}-x_k|\le ar^k$.
Now $$x_n-x_m=(x_{m+1}-x_{m})+(x_{m+2}-x_{m+1}) +(x_{m+3}-x_{m+2})+\cdots +(x_{n}-x_{n-1}).$$ Taking absolute values, and using the Triangle Inequality, we get $$|x_n-x_m|\le |x_{m+1}-x_{m}|+|x_{m+2}-x_{m+1}| +|x_{m+3}-x_{m+2}|+\cdots +|x_{n}-x_{n-1}|.$$
Using the inequalities previously established, we conclude that $$|x_n-x_m|\le ar^m +ar^{m+1}+ar^{m+2}+\cdots.$$ The infinite geometric series has sum $\frac{a}{1-r}r^m$.
Let $\epsilon \gt 0$ be given. By picking $N=N(\epsilon)$ large enough, and $N\le m\lt n$, we can make $\frac{a}{1-r}r^m\lt \epsilon$. Thus for such $m$ and $n$, we have $|a_n-a_m|\lt \epsilon$.
Hint: find a bound for $|x_m - x_n|$ that involves a geometric series.