 Solution

We first justify the maximum in the proposed supremum norm. If a null sequence $x$ is nonzero, choose an index $j$ with $|x_j|>0$. All sufficiently late terms have absolute value smaller than $|x_j|/2$, since $x_i\to 0$. Hence the largest absolute value occurs among a finite initial set of terms, and is finite. For the zero sequence the maximum is zero.

The norm axioms now follow directly. The maximum is nonnegative and vanishes precisely when every coordinate vanishes. For a scalar $a$, $\Vert ax{\Vert }_{\infty }=|a|\Vert x{\Vert }_{\infty }$. For two null sequences, coordinatewise addition remains a null sequence andTaking the maximum proves the triangle inequality, so this really is a norm on $V$.

Let $e_n$ have a single $1$ in position $n$. If $m\ne n$, then $\Vert e_m-e_n{\Vert }_{\infty }=1$. A convergent sequence in a normed vector space must be a Cauchy sequence, by the triangle inequality. These distances violate the Cauchy condition with, for example, $\epsilon =1/2$. Thus $(e_n)$ does not converge in this norm, even though each fixed coordinate converges to zero.