[proofplan] We prove completeness by starting with an arbitrary [Cauchy sequence](/page/Cauchy%20Sequence) in $\ell^p$ and constructing its coordinatewise limit. The coordinate sequences are Cauchy in $\mathbb{R}$, hence converge by [citetheorem:7986]. For $1 \leq p < \infty$, finite partial sums transfer the Cauchy estimate to the limit and then give convergence in $\ell^p$; for $p=\infty$, the same idea uses the supremum norm directly. These two cases show that every Cauchy sequence in $\ell^p$ converges in its norm. [/proofplan] [step:Take coordinatewise limits of a Cauchy sequence in $\ell^p$] Fix $1 \leq p \leq \infty$, and let $(x_m)_{m \in \mathbb{N}}$ be a Cauchy sequence in $(\ell^p,\|\cdot\|_{\ell^p})$. For each $m \in \mathbb{N}$, write \begin{align*} x_m = (x_{m,n})_{n \in \mathbb{N}}. \end{align*} For each fixed $n \in \mathbb{N}$ and all $m,k \in \mathbb{N}$, the coordinate estimate \begin{align*} |x_{m,n}-x_{k,n}| \leq \|x_m-x_k\|_{\ell^p} \end{align*} holds. If $1 \leq p < \infty$, this follows because $|x_{m,n}-x_{k,n}|^p$ is one term in the nonnegative series defining $\|x_m-x_k\|_{\ell^p}^p$. If $p=\infty$, it follows from the definition of the supremum norm. Since $(x_m)_{m \in \mathbb{N}}$ is Cauchy in $\ell^p$, the estimate above implies that, for each fixed $n \in \mathbb{N}$, the scalar sequence $(x_{m,n})_{m \in \mathbb{N}}$ is Cauchy in $\mathbb{R}$. By [citetheorem:7986], there exists a real number $x_n \in \mathbb{R}$ such that \begin{align*} \lim_{m \to \infty} x_{m,n} = x_n. \end{align*} Define the candidate limit sequence $x: \mathbb{N} \to \mathbb{R}$ by \begin{align*} x(n) = x_n. \end{align*} Equivalently, write $x=(x_n)_{n \in \mathbb{N}}$. [/step] [step:Pass finite $\ell^p$ estimates to the coordinatewise limit when $1 \leq p < \infty$] Assume $1 \leq p < \infty$. Let $\varepsilon>0$. Since $(x_m)_{m \in \mathbb{N}}$ is Cauchy in $\ell^p$, there exists $M \in \mathbb{N}$ such that \begin{align*} \|x_m-x_k\|_{\ell^p} < \varepsilon \end{align*} for all $m,k \geq M$. Fix $m \geq M$ and $N \in \mathbb{N}$. For every $k \geq M$, \begin{align*} \sum_{n=1}^{N} |x_{m,n}-x_{k,n}|^p \leq \sum_{n=1}^{\infty} |x_{m,n}-x_{k,n}|^p = \|x_m-x_k\|_{\ell^p}^p < \varepsilon^p. \end{align*} For this fixed finite set of indices $\{1,\dots,N\}$, coordinatewise convergence gives \begin{align*} \lim_{k \to \infty}\sum_{n=1}^{N} |x_{m,n}-x_{k,n}|^p = \sum_{n=1}^{N} |x_{m,n}-x_n|^p. \end{align*} Passing to the limit in the finite inequality yields \begin{align*} \sum_{n=1}^{N} |x_{m,n}-x_n|^p \leq \varepsilon^p. \end{align*} Because this holds for every $N \in \mathbb{N}$ and the partial sums are nondecreasing, we obtain \begin{align*} \sum_{n=1}^{\infty} |x_{m,n}-x_n|^p \leq \varepsilon^p. \end{align*} Thus $x_m-x \in \ell^p$ and \begin{align*} \|x_m-x\|_{\ell^p} \leq \varepsilon \end{align*} for every $m \geq M$. [guided] The key point is that the $\ell^p$ norm controls all finite coordinate blocks uniformly. Fix $\varepsilon>0$. Since $(x_m)_{m \in \mathbb{N}}$ is Cauchy in $\ell^p$, choose $M \in \mathbb{N}$ such that \begin{align*} \|x_m-x_k\|_{\ell^p} < \varepsilon \end{align*} whenever $m,k \geq M$. Now fix one index $m \geq M$ and one finite cutoff $N \in \mathbb{N}$. For every $k \geq M$, the first $N$ terms of the series are bounded by the whole series: \begin{align*} \sum_{n=1}^{N} |x_{m,n}-x_{k,n}|^p \leq \sum_{n=1}^{\infty} |x_{m,n}-x_{k,n}|^p = \|x_m-x_k\|_{\ell^p}^p < \varepsilon^p. \end{align*} Why do we keep the cutoff $N$ finite? Because finite sums commute directly with limits. For each $n \in \{1,\dots,N\}$, we already know $x_{k,n} \to x_n$ as $k \to \infty$. Therefore \begin{align*} \lim_{k \to \infty}|x_{m,n}-x_{k,n}|^p = |x_{m,n}-x_n|^p. \end{align*} Since there are only finitely many terms, we may pass the limit through the sum: \begin{align*} \lim_{k \to \infty}\sum_{n=1}^{N} |x_{m,n}-x_{k,n}|^p = \sum_{n=1}^{N} |x_{m,n}-x_n|^p. \end{align*} Taking the limit as $k \to \infty$ in the finite inequality gives \begin{align*} \sum_{n=1}^{N} |x_{m,n}-x_n|^p \leq \varepsilon^p. \end{align*} The cutoff $N$ was arbitrary. Since the partial sums \begin{align*} \sum_{n=1}^{N} |x_{m,n}-x_n|^p \end{align*} increase as $N$ increases, their supremum is the infinite series. Hence \begin{align*} \sum_{n=1}^{\infty} |x_{m,n}-x_n|^p \leq \varepsilon^p. \end{align*} This proves both that $x_m-x \in \ell^p$ and that \begin{align*} \|x_m-x\|_{\ell^p} \leq \varepsilon \end{align*} for every $m \geq M$. [/guided] [/step] [step:Conclude convergence in $\ell^p$ for $1 \leq p < \infty$] Still assume $1 \leq p < \infty$. Choose $m_0 \geq M$. The previous step gives $x_{m_0}-x \in \ell^p$. Since $x_{m_0} \in \ell^p$ and $\ell^p$ is a [vector space](/page/Vector%20Space), we have \begin{align*} x = x_{m_0} - (x_{m_0}-x) \in \ell^p. \end{align*} Moreover, the previous step proves that for every $\varepsilon>0$ there exists $M \in \mathbb{N}$ such that \begin{align*} \|x_m-x\|_{\ell^p} \leq \varepsilon \end{align*} for all $m \geq M$. Hence $x_m \to x$ in $\ell^p$. Therefore every Cauchy sequence in $\ell^p$ converges when $1 \leq p < \infty$. [/step] [step:Use the supremum norm to handle $\ell^\infty$] Assume $p=\infty$. Let $\varepsilon>0$. Since $(x_m)_{m \in \mathbb{N}}$ is Cauchy in $\ell^\infty$, there exists $M \in \mathbb{N}$ such that \begin{align*} \|x_m-x_k\|_{\ell^\infty} < \varepsilon \end{align*} for all $m,k \geq M$. Fix $m \geq M$. For every $n \in \mathbb{N}$ and every $k \geq M$, \begin{align*} |x_{m,n}-x_{k,n}| \leq \|x_m-x_k\|_{\ell^\infty} < \varepsilon. \end{align*} Letting $k \to \infty$ and using $x_{k,n}\to x_n$ gives \begin{align*} |x_{m,n}-x_n| \leq \varepsilon \end{align*} for every $n \in \mathbb{N}$. Taking the supremum over $n \in \mathbb{N}$ yields \begin{align*} \|x_m-x\|_{\ell^\infty} \leq \varepsilon. \end{align*} Choose $m_0 \geq M$. Since $x_{m_0}\in \ell^\infty$ and $x_{m_0}-x\in \ell^\infty$, we have \begin{align*} x = x_{m_0}-(x_{m_0}-x)\in \ell^\infty. \end{align*} Thus $x_m \to x$ in $\ell^\infty$. [/step] [step:Identify the normed space as Banach] We have shown that every Cauchy sequence in $(\ell^p,\|\cdot\|_{\ell^p})$ converges to an element of $\ell^p$ when $1 \leq p < \infty$, and also when $p=\infty$. Therefore $(\ell^p,\|\cdot\|_{\ell^p})$ is complete for every $1 \leq p \leq \infty$. Hence $\ell^p$ is a [Banach space](/page/Banach%20Space). [/step]