linear algebra - Norm equivalence and the sequence limit of normalized vectors

Consider finite-dimensional vector spaces, on which two norms $\|\|_1$ and $\|\|_2$ are always equivalent. Then a sequence of vectors ${\bf v}_k \to {\bf v}$ w.r.t. norm 1 iff ${\bf v}_k \to {\bf v}$ w.r.t. norm 2.

The question is about sequence $\frac{{{{\mathbf{v}}_k}}}{{{{\left\| {{{\mathbf{v}}_k}} \right\|}_1}}} \to \frac{{\mathbf{v}}}{{{{\left\| {\mathbf{v}} \right\|}_1}}}$ w.r.t. norm 1, will this limit imply $\frac{{{{\mathbf{v}}_k}}}{{{{\left\| {{{\mathbf{v}}_k}} \right\|}_2}}} \to \frac{{\mathbf{v}}}{{{{\left\| {\mathbf{v}} \right\|}_2}}}$ w.r.t. norm 2 (or maybe a weaker claim that one convergence implies the other but the limit might be different)? I have trouble showing this is true. If this is false, anyone can help provide a counterexample? Thanks!

For example, $(k,\sqrt k),k=1,2,...$ satisfies $\frac{{(k,\sqrt k )}}{{{{\left\| {(k,\sqrt k )} \right\|}_p}}} = \frac{{(k,\sqrt k )}}{{{{({k^p} + {k^{\frac{p}{2}}})}^{\frac{1}{p}}}}} \to (1,0)$ for any $p$-norm.

Answer

Supposedly $v\ne0$. Let $u_k=v_k/\|v_k\|_1$ and $u=v/\|v\|_1\ne0$. By assumption, $u_k\to u$ (with respect to both norms, because all norms are equivalent) and hence $\|u_k\|_2\to \|u\|_2$ (because the norm function is continuous). Therefore
$$
\left\|\frac{u_k}{\|u_k\|_2}-\frac{u}{\|u\|_2}\right\|_2
\le \frac{1}{\|u_k\|_2}\|u_k-u\|_2
+\left|\frac{1}{\|u_k\|_2}-\frac{1}{\|u\|_2}\right|\|u\|_2 \to0.
$$
Consequently, $v_k/\|v_k\|_2=u_k/\|u_k\|_2\to u/\|u\|_2=v/\|v\|_2$.