Vercongence?

Calculus Level 3

Consider the following strange definition:

We say a sequence $(x_n)$ verconges to $x$ if there exist an $\epsilon>0$ such that for all $N\in \Bbb{N}$ , $n\ge N \implies |x_n-x|<\epsilon$

Which of the following conclusions is wrong ?

Inspired by Abbot's Understanding Analysis

All bounded sequences are vercongent All convergent sequences are vercongent All vercongent sequences are convergent All vercongent sequences are bounded

1 solution

A Former Brilliant Member
Jul 4, 2016

Note that the definition of verconvergent sequences is equivalent to the following:

$(x_n)$ is verconvergent $\iff$ $(x_n)$ is bounded.

One way to prove this is proving that bounded sequences are verconvergent and unbounded sequences are not. I leave the details to the reader.

Consider $x_n = (-1)^n$ . Note that this sequence is bounded ( $|x_n| \leq 1$ ) and hence is verconvergent. But $(x_n)$ doesn't converge.

Hence, verconvergent sequences are not convergent in general.

Vercongence?

Inspired by Abbot's Understanding Analysis

1 solution

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