I am aiming to prove that with the norm on $\ell^{1}$ given by $\left\|\left(a_{1}, a_{2}, \ldots\right)\right\|=\sup \left\{\left|a_{n}\right|: n \geq 1\right\}$, $\left(\ell^{1},\|\cdot\|\right)$ is not a Banach space.
So I need to prove that every Cauchy sequence in $\left(\ell^{1},\|\cdot\|\right)$ converges to some element of $\left(\ell^{1},\|\cdot\|\right)$.
I tried by starting with a sequence $(x^{(n)})_{n=1}^{\infty}$ with $x^{n}=(0,0,\dots,1/n,0,0,\dots)$ where the value $1/n$ is the $n$-th element. But, i didn't manage to get anywhere with this. Any help appreciated. Thanks
Take $x^n=(1,1/2,\ldots,1/n,0,\ldots)$.
FIRST: This sequence is $\|\cdot\|$-Cauchy.
Indeed, assuming $n>k$, we have that $$\|x^n-x^k\|=\|(0,\ldots,0,1/(k+1),\ldots,1/n,0,\ldots)\|=$$ $$=\sup\{1/j,\ k+1\le j\le n\}=\frac{1}{k+1}\to0\ (k\to\infty). $$
SECOND: We can prove that $x^n\overset{\|\cdot\|}{\longrightarrow} x:=(1,1/2,1/3,\ldots)$.
Indeed $\|x^n-x\|=\|(0,\ldots,0,1/(n+1),1/(n+2),\ldots)\| =\sup_{j>n}\frac{1}{j}=\frac{1}{n+1}\to0\ (n\to\infty)$.
THIRD It is clear that $x^n\in\ell^1$ but $x\notin\ell^1$.
So you have a $\|\cdot\|$-Cauchy sequence $(x^n)_n\subset\ell^1$ that is $\|\cdot\|$-convergent to something not belonging to $\ell^1$.