Is $(l^1 ,\|.\|)$ a Banach space?

Question

Suppose $x=\{x_n\}\in l^1$ and $\|x\|=\sup|\sum_{k=1}^{n}x_k|$, let $\|x\|_1=\sum_{n=1}^{\infty}|x_n|$ is a norm for $l^1$ . Is $(l^1 ,\|.\|)$ a Banach space?

Answer 1

Let $x^{(n)}:=\sum_{j=1}^n\frac{(-1)^j}je^{(j)}\in\ell^1$, where $e^{(n)}$ is the sequence whose $n$-th term is $1$, the others $0$. Then $$\left\lVert x^{(m+n)}-x^{(n)}\right\rVert=\max_{1\leqslant j\leqslant m}\left|\sum_{k=n+1}^{n+j}\frac{(-1)^k}k\right|$$ which converges to $0$ as $m,n\to +\infty$ (because we can control the remainder of such a series).

However, we cannot have convergence to an element of $\ell^1$, because convergence for $\lVert\cdot\rVert$ implies coordinate-wise convergence, and the only candidate is not in $\ell^1$.