If $|x_{n+2} – x_{n+1}| \le c|x_{n+1} – x_n|$ for all $n \in \mathbb{N}$ with $0 \le c < 1$, then $(x_n)$ has a bounded variation.

Question

A sequence $(x_n)$ is said to have a bounded variation in case the sequence $(v_n)$ given by $v_n=\sum_{i=1}^n |x_{i+1} – x_i|$ is bounded.

The task itself is to prove the following:

• If $|x_{n+2} – x_{n+1}| \le c|x_{n+1} – x_n|$ for all $n \in \mathbb{N}$ with $0 \le c < 1$, then $(x_n)$ has a bounded variation.

My attempt:

First, let the sequence $(y_n)$ be given by $y_n = |x_{n+1} – x_n|$. Then, I broke it down in two cases: $0<c<1$ (I) and $c=0$ (II).

• For (I), we have that $\lim \frac{|y_{n+1}|}{|y_n|}<1$.Hence, by the ratio test, the series $\sum y_n = \sum |x_{n+1}-x_n|$ is convergent, and for the sequence of its partial sums, which is $(v_n)$, we have that it is bounded. We conclude $(x_n)$ has a bounded variation.
• For (II), I'm not sure. Perhaps taking the limit of the absolute value $|x_{n+2} – x_{n+1}| \le 0$ would imply that the series $\sum y_n$ converges to $0$?

Is it Ok to break this problem down in the two former cases like I did?

Answer 1

HINT

I would start with noticing that:

\begin{align*} |x_{n + 1} - x_{n}| \leq c|x_{n} - x_{n-1}| \leq c^{2}|x_{n - 1} - x_{n - 2}|\leq \ldots \leq c^{n}|x_{1} - x_{0}| \end{align*}

Can you take it from here?