Is $\ell^1$ complete with this norm?

Question

For $x \in \ell^1$ we set $\Vert x\Vert = \sup\limits_{N \in \mathbb{N}}|\sum\limits_{n=1}^{N}x_n|$.
One can easily see that this is a norm on $\ell^1$. I was wondering if this space is now complete. I tried finding an absolutely convergent series that does not converge, but I did not find anything.

Answer 1

The space $\ell^1$ is not complete with this norm. To find a counterexample, you may consider the sequence of sequences $x_n \in \ell^1$ defined for $n \ge 1$ by

$$x_n = (x_{n,k})_{k \ge 1} := \left( \frac{(-1)^k}{k}\right)_{1 \le k \le n}$$

and

$$x = (x_k)_{k \ge 1} := \left( \frac{(-1)^k}{k}\right)_{1 \le k}$$

Note that we have $x_n \in \ell^1$, but $x \notin \ell^1$.

We may check the sequence $(x_n)_{n \ge 1}$ converges to $x$ with regards to the norm $\| \cdot \|$. Indeed, let $\epsilon > 0$. We have

$$\left|\sum_{k = 1}^N x_{n,k} - x_{k}\right| = \left|\sum_{k = n+1}^{N} \frac{(-1)^k}{k} \right|$$

and since the series $\sum_k \frac{(-1)^k}{k}$ is convergent, we may find $n_0$ such that all those partial sums are smaller than $\epsilon$ for every $n \ge n_0$. Which gives us $\|x_{n}-x\| \le \epsilon$ for $n \ge n_0$. From this, we can deduce that $(x_n)_{n \ge 1}$ is a Cauchy sequence in $\ell^1$, but it has no limit in $\ell^1$.