If $\{x_n\}$ is a Cauchy sequence then there exists an $M$ such that for all $n \ge M$ we have $|x_{n+1}-x_n| \le |x_n-x_{n-1}|$.

Question

Question: True /False prove or find a counterexample: If $\{x_n\}$ is a Cauchy sequence then there exists an $M$ such that for all $n \ge M$ we have $|x_{n+1}-x_n| \le |x_n-x_{n-1}|$.

I think this is true because $|x_{n+1}-x_n| < \varepsilon$ is true for $n\ge M$, but $|x_n-x_{n-1}|< \varepsilon$ is not necessarily true for $n\ge M$. This means that there is chance that $|x_{n+1}-x_n|<|x_n-x_{n-1}|$.

Is this correct? and if it is, is it enough to justify that the statement is true?

Thank you in advance.

Answer 1

Hint:   consider the sequence $\,\dfrac{1}{1},\dfrac{1}{1},\dfrac{1}{2},\dfrac{1}{2},\ldots\dfrac{1}{n},\dfrac{1}{n},\ldots\,$