TITLE: $l_1$ equipped with the sup norm is NOT a Banach Space
QUESTION [9 upvotes]: Prove that $l_1 = \{ x = (x_k)_{k\in\mathbb{N}}\subset \mathbb{R};\ \sum_{k\in\mathbb{N}}\ |x_k| < +\infty \}$ equipped with the norm
$\| x\| = \mathrm{sup}_{k\in\mathbb{N}} |x_k|$ is NOT a Banach Space.
I've tried to solve it considering the sequence
$x_n = (\frac{1}{k^{1 + 1/n}}),\ \forall n \in \mathbb{N}$
Since $\sum_{k\in\mathbb{N}}\frac{1}{k^{1 + 1/n}} < +\infty$ (because $1+1/n > 1$), $(x_n)_{n\in\mathbb{N}}$ is a sequence in $l_1$. We also have that $x_n \longrightarrow (1/k)_{k\in\mathbb{N}},$ as $n\longrightarrow +\infty$, and this one is not in $l_1$.
I would like to prove that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence, but I don't have a clue how to do it.

REPLY [4 votes]: Let $x_n$ be any sequence that converges to zero, but $\sum_n |x_n| = \infty$.
Let $[y_k]_n = \begin{cases} x_n & n \le k \\ 0 & \text{otherwise}\end{cases}$. Then $y_k \in l_1$ for all $k$, and $y_k$ is a Cauchy sequence (in the $l_\infty$ norm). Since $l_\infty$ is complete, $y_k$ converges to some $y \in l_\infty$ (it is easy to see that $[y]_n = x_n$), but clearly $y \notin l_1$.