Cauchy sequence if two consecutive terms get closer

Question

A bounded sequence has the property that $\|a_{n+1}-a_n\|\to 0$ as $n\to\infty$ in a Banach space. Now, here comes the question that whether such a sequence is Cauchy. if not, we can add the assumption that the sequence has some monotonic behavior.

Answer 1

What kind of monotonicity conditions are you thinking about? Are they with respect to some order relation on the space? Just to note, those conditions mentioned are insufficient for convergence. For example, define the sequence $$\{a_n\}_{n \in \mathbb{N}} \subset \mathbb{R}^2 \ \text{ by }\ a_n = (\cos(\ln(n)), \sin(\ln(n))$$ This is a bounded sequence in $\mathbb{R}^2$ that fails to converge to a single value (it just continually traces around the unit circle), but we have $$ \lim_{n \rightarrow \infty}|a_{n+1}-a_{n}| = \lim_{n \rightarrow \infty} \sqrt{[\cos(\ln(n+1))-\cos(\ln(n))]^2 + [\sin(\ln(n+1))-\sin(\ln(n))]^2 } \\ = \lim_{n \rightarrow \infty} \sqrt{2 - 2[\cos(\ln(n+1))\cos(\ln(n))+\sin(\ln(n+1))\sin(\ln(n))]} \\ = \lim_{n \rightarrow \infty} \sqrt{2 -2\cos(\ln(n+1) - \ln(n))} \\ = \lim_{n \rightarrow \infty} \sqrt{2 - 2\cos(\ln(\frac{n+1}{n}))} \\ = 0 $$ That is to say the tracing is being done increasingly slow, so the points get closer together, just not to any particular value.