Do Cauchy sequences in $(\ell^1, \|\cdot\|_\infty)$ converge in $\ell^1$?
I feel that the answer is No, but I am not able to find a counterexample. If $x^{(n)}$ is a sequence in $\ell^1$, then we have $$\sum_{i=1}^\infty |x^{(n)}_i| < \infty$$ for every $n\in\mathbb N$. Also if $x^{(n)}$ is Cauchy w.r.t the sup-norm, then $$\forall\epsilon>0\exists N\in\mathbb N\forall m,n>N (\|x^{(m)} - x^{(n)}\|_\infty < \epsilon)$$
That is the definition, and that is all I can see. Any ideas?
Let us take $x^{(n)} \in \ell^1$ such that $$ x^{(n)}(k) = \begin{cases} 1/k & \text{if } k \le n,\\ 0 & \text{otherwise}. \end{cases} $$ For $m \ge n+1$, $$\|x^{(m)} - x^{(n)}\|_\infty = \frac{1}{n + 1} \to 0$$ but $x^{(n)}$ does not converges in $\ell^1$ because $$\sum_{k \in \mathbb N} \frac{1}{k} = +\infty.$$