If a sequence of summable sequences converges to a sequence, then that sequence is summable. Let $(a_i)^n$ be a sequence of complex sequences each of which are summable (they converge).  Then if they have a limit, the limit sequence $(b_i)$ is also summable.  All under the sup norm for sequences.
Let $(a_i)^n$ sum to $c_n$.  I.e. for all $\epsilon \gt 0$, there's $N$ such that $m \gt N \implies |\sum_{i=1}^m a_i^n - c_n| \lt \epsilon$.  I want to show that there's $b$ such that $|\sum_{i=1}^m b_i - b| \lt \epsilon$ similarly.  Let $b = \lim c_n$.   Where to?

(edited) This doesn't work, the example is pointwise convergent:  Consider the alternating sequence $x=(-1,1,-1,1,\ldots)$ and let $x_n$ take the first $n$ terms of $x$ and be zero afterwards. $x_n$ converges pointwise to $x$, and is summable, but the (pointwise) limit $x$ is not summable.

Here is a uniform convergence counterexample: Take the sequence $x(i) = 1/i$, which is not summable. Now take $x_n$ to match $x$ for the first $n$ terms and be zero after. $x_n$ converges to $x$ uniformly, but the limit is not summable. 
