Bounded sequence in $R^n$ with norm of difference between two consecutive elements converges to 0 If I have a sequence ${v^i}_{i=1,2,\cdots} \subseteq \mathbb R^n$, satisfying

*

*$\|v^i\| \leq B$  for all $i=1,2,\cdots$

*$\lim_{i\rightarrow \infty} \|v^{i+1} - v^i\|_2 = 0$
Can I say that the sequence $\{v^i\}_{i=1,2,\cdots}$ is convergent?
 A: No.  Let $(a_n)$ be any non-negative decreasing sequence such that $\lim_{n \to \infty} a_n = 0$ but $\sum_{n \geq 1} a_n = \infty$.  For concreteness let's just take $a_n = 1/n$.  Now we can construct a counterexample $(v_n)$ recursively.
Let $v_0 = 0$, and start adding the terms of the sequence $1/n$ one by one (so $v_1 = 1$, $v_2 = 1+1/2$, $v_3 = 1+1/2+1/3$, etc).  Since $\sum 1/n$ diverges, eventually this sequence of partial sums will land in the interval $[2,3]$.  Then turn around start subtracting the $1/n$ terms instead.  Again, since $\sum 1/n$ diverges, after subtracting terms for long enough you will land back in the interval $[0,1]$.  Then turn around again and start adding again.  Since $\sum 1/n$ diverges, you can always repeat this process of adding terms until landing in the interval $[2,3]$ and then subtracting terms until landing in the interval $[0,1]$.  Let $(v_n)$ be the sequence of values obtained by this process.  Then it's clear that $(v_n)$ is bounded and does not converge by construction, but $|v_{n} - v_{n-1}| = 1/n \to 0$.
In case it helps, here is a more concrete description of the first few terms of $(v_n)$:
\begin{align*}
v_1 &= 1 \\
&\vdots \\
v_4 &= 1 + \frac12 + \frac13 + \frac14 \approx 2.083 \\
v_5 &= 1 + \frac12 + \frac13 + \frac14 - \frac15 \\
&\vdots \\
v_{13} &= 1 + \frac12 + \frac13 + \frac14 - \frac15 - \dots - \frac{1}{13} \approx 0.986 \\
v_{14} &= 1 + \frac12 + \frac13 + \frac14 - \frac15 - \dots - \frac{1}{13} + \frac{1}{14} \\
&\vdots
\end{align*}
