Prove that $\{x^n\}$ is Cauchy in $S\subseteq \ell_\infty$

Question

I'm kind of new into Functional analysis. So, I have the following question bothering me. 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 Cauchy in $S.$

MY TRIAL

Let $m\geq n.$ Then, \begin{align}\Vert x^n-x^m \Vert_ {\ell_\infty}=\sup\limits_{m,n\in \Bbb{N}}\left|\frac{1}{m}-\frac{1}{n} \right|\end{align} By convergence of $\frac{1}{n}$ and $\frac{1}{m}$ for each $m$ and $n$, there exists $N\in\Bbb{N}$ such that \begin{align}\Vert x^n-x^m \Vert_ {\ell_\infty}=\sup\limits_{m,n\in \Bbb{N}}\left|\frac{1}{m}-\frac{1}{n} \right|=\frac{1}{n}-\frac{1}{m}<\epsilon,\;\;\forall\;m\geq n\geq N. \end{align}

Please check if I'm right or wrong. If I'm wrong, kindly help out. Thanks

Answer 1

Actually.$$m\geqslant n\implies\lVert x^n-x^m\rVert_\infty=\begin{cases}0&\text{ if }m=n\\\dfrac1{n+1}&\text{ otherwise,}\end{cases}$$since, if $m>n$,$$\lVert x^n-x^m\rVert_\infty=\sup\left\{0,\frac1{n+1},\frac1{n+2},\ldots,\frac1m\right\}=\frac1{n+1}.$$So, given $\varepsilon>0$, take $N\in\mathbb{N}$ such that $\frac1N<\varepsilon$ and then$$m\geqslant n\geqslant N\implies\lVert x^n-x^m\rVert_\infty<\varepsilon.$$