As a follow-up to my previous post Prove that $\{x^n\}$ is Cauchy in $S\subseteq \ell_\infty$. I would like to ask the following question. Let $S$ be the set of sequences having only a finite number of non-zero terms. Clearly, $S\subseteq \ell_\infty$. Take $\{x^n\}$ in $S$ where $x^n=(1,\frac{1}{2},\frac{1}{3},\cdots,\frac{1}{n},0,0,\cdots).$ I want to prove that $\{x^n\}$ is convergent to $x=(1,\frac{1}{2},\frac{1}{3},\cdots)\notin S\subseteq \ell_\infty.$
MY TRIAL
Let $\epsilon>0$ be given and $n\in \Bbb{N}.$ Then, \begin{align}\Vert x^n-x\Vert_ {\infty}=\sup\limits_{n\in \Bbb{N}}\left|\frac{1}{n}-0 \right|\end{align} By convergence of $\{1/n\}_{n\in\Bbb{N}}$ to $0$, there exists $N\in\Bbb{N}$ such that \begin{align}\Vert x^n-x\Vert_ {\infty}=\sup\limits_{n\in \Bbb{N}}\left|\frac{1}{n}-0\right|=\frac{1}{n}<\epsilon,\;\;\forall\; n\geq N. \end{align}
I think I'm wrong but please, can you help? I'm new into Functional analysis. Thanks
Your idea is just fine, but you are not expressing it properly. You should say that, since $\left(\frac1n\right)_{n\in\mathbb N}$ converges to $0$, then, for each $\varepsilon>0$, there is a natural $N$ such that $n\geqslant N\implies\frac1n<\varepsilon$ (actually, this is what it means to assert that $\lim_{n\to\infty}\frac1n=0$) and so$$n\geqslant N\implies\lVert x-x^n\rVert_\infty=\sup\left\{0,\frac1{n+1},\frac1{n+2},\ldots\right\}=\frac1{n+1}<\varepsilon.$$However, $x\notin S$. So, this proves that in $S$ your sequence diverges.