Let $V \space$ be a vector space over $\mathbb{R}$, and $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ norms over $V$, which generate the same topology. Is it always true that if $v_n$ is a Cauchy sequence with respect to both norms, and $v_n$ converges in $(V, \Vert \cdot \Vert_1)$ then it converges in $(V, \Vert \cdot \Vert_2)$? If $dim(V)<+\infty$ the assertion is true, because there exist constants $c,C\in\mathbb{R}$ such that $\forall v\in V \space$ $c\Vert v \Vert_1 \leq \Vert v \Vert_2 \leq C\Vert v \Vert_1$, but I have a feeling it isn't in general (this would be strange, since completeness isn't a topological property; however maybe the additional structure of vector space might be used in some way).
EDIT
Thanks to everyone's answers I have realized that the above question was badly stated to begin with.
What I meant to ask was whether the property of a normed vector space of being complete only depends on the topology generated by the norm, and not by the norm itself. As stated indirectly by many users, convergence, unlike the property of being a Cauchy sequence, is a topological property. Therefore the answer to the question I actually asked is always affermative.
What I should have asked was: given two topologically equivalent norms $\Vert \cdot \Vert_1$, $\Vert \cdot \Vert_2$ on a vector space V, is it true that a sequence $v_n$ is a Cauchy sequence in $(V, \Vert \cdot \Vert_1)$ if and only if it is in $(V, \Vert \cdot \Vert_2)$?