Is it v1,v2,v3,.. converging sequence of space $\mathbb{E}$?

Question

$(\mathbb{E}, \Vert{.}\Vert)$ - Banach space over scalars $\mathbb{R}$ and if $v_1, v_2, v_3,..$ it's the sequence of this space that with $\forall v\in\mathbb{E}$ a sequence $ \Vert v - v_1\Vert,\Vert v - v_2\Vert, \Vert v - v_3\Vert,...$ is real converging sequence. Is it necessary that $v_1, v_2, v_3,..$ converging sequence of space $\mathbb{E}$? Why?

Answer 1

Take your favourite two elements $u, v_1 \in \mathbb E$ and define for $n > 1$ $$ v_n = v_1+nu. $$ Then the sequence $\Vert v - v_1 \Vert, \Vert v_1 - v_2 \Vert, \Vert v_2 - v_3 \Vert$ is constantly equal to $\Vert u \Vert$ (except perhaps for the first term), and hence convergent. But $$ \Vert v_n \Vert = \Vert v_1 + nu \Vert \geq n\Vert u \Vert - \Vert v_1 \Vert, $$ so that $(v_n)_{n=1}^\infty$ is unbounded and hence cannot converge.

Answer 2

Take $\mathbb{E}=\mathbb{R}$ and $v_n=\sum_{i=1}^n 1/i$. Then $|v_n-v_{n-1}|=1/n\rightarrow 0$. But $\{v_n\}$ is not a convergent sequence.