[proofplan] We prove both implications directly from the definitions. If $X$ is complete and $\sum \|x_n\|_X$ converges, then the partial sums of $\sum x_n$ are Cauchy because their tails are bounded by scalar norm tails, so completeness gives convergence. Conversely, assuming every norm-summable series converges, we take an arbitrary [Cauchy sequence](/page/Cauchy%20Sequence), extract a subsequence whose successive increments have summable norms, and realize that subsequence as the sequence of partial sums of an absolutely convergent telescoping series. The original Cauchy sequence then converges to the same limit as the subsequence. [/proofplan] [step:Use completeness to turn absolute summability into convergence of partial sums] Assume first that $X$ is a [Banach space](/page/Banach%20Space). Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$ such that the scalar series $\sum_{n=1}^{\infty}\|x_n\|_X$ converges in $\mathbb{R}$. For each $m\in\mathbb{N}$, define the vector partial sum $s_m\in X$ by \begin{align*} s_m := \sum_{n=1}^{m} x_n. \end{align*} Also define the scalar partial sum $a_m\in[0,\infty)$ by \begin{align*} a_m := \sum_{n=1}^{m}\|x_n\|_X. \end{align*} Since the scalar series $\sum_{n=1}^{\infty}\|x_n\|_X$ converges, the sequence $(a_m)_{m\in\mathbb{N}}$ converges in $\mathbb{R}$. Let $L\in\mathbb{R}$ denote its limit. We show that $(s_m)_{m\in\mathbb{N}}$ is Cauchy in $X$. Let $\varepsilon>0$. Since $a_m\to L$, there exists $N\in\mathbb{N}$ such that for every $m\geq N$, \begin{align*} |a_m-L|<\frac{\varepsilon}{2}. \end{align*} If $q>p\geq N$, then by the triangle inequality in $X$, \begin{align*} \|s_q-s_p\|_X \leq \sum_{n=p+1}^{q}\|x_n\|_X. \end{align*} By the definition of the scalar partial sums, \begin{align*} \sum_{n=p+1}^{q}\|x_n\|_X = a_q-a_p. \end{align*} Since $a_q\leq L$ and $a_p\leq L$ for the increasing sequence $(a_m)$, and since both $a_q$ and $a_p$ are within $\varepsilon/2$ of $L$, we have \begin{align*} a_q-a_p \leq |L-a_q|+|L-a_p|<\varepsilon. \end{align*} Thus $\|s_q-s_p\|_X<\varepsilon$ for all $q>p\geq N$. The case $p=q$ gives $\|s_q-s_p\|_X=0$, and the case $p>q$ follows by symmetry of the norm. Hence $(s_m)$ is a Cauchy sequence in $X$. Because $X$ is complete, there exists $s\in X$ such that $s_m\to s$ in $X$. By the definition of convergence of a vector series, this means that $\sum_{n=1}^{\infty}x_n$ converges in $X$. [guided] Assume that $X$ is a Banach space, meaning that every Cauchy sequence in $X$ converges to an element of $X$. Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $X$ for which the scalar series of norms $\sum_{n=1}^{\infty}\|x_n\|_X$ converges. To prove that the vector series $\sum_{n=1}^{\infty}x_n$ converges, we must prove that its sequence of partial sums converges. For each $m\in\mathbb{N}$, define \begin{align*} s_m := \sum_{n=1}^{m}x_n \in X. \end{align*} The goal is to show that $(s_m)$ converges in $X$. Since $X$ is assumed complete, it is enough to show that $(s_m)$ is Cauchy. We compare vector tails with scalar tails. Define, for each $m\in\mathbb{N}$, \begin{align*} a_m := \sum_{n=1}^{m}\|x_n\|_X \in [0,\infty). \end{align*} The hypothesis that $\sum_{n=1}^{\infty}\|x_n\|_X$ converges means exactly that the scalar sequence $(a_m)$ converges in $\mathbb{R}$. Let $L\in\mathbb{R}$ be its limit. Let $\varepsilon>0$. Because $a_m\to L$, there exists $N\in\mathbb{N}$ such that for every $m\geq N$, \begin{align*} |a_m-L|<\frac{\varepsilon}{2}. \end{align*} Now take integers $q>p\geq N$. The difference of the vector partial sums is the finite tail \begin{align*} s_q-s_p = \sum_{n=p+1}^{q}x_n. \end{align*} Applying the triangle inequality in the [normed vector space](/page/Normed%20Vector%20Space) $X$ gives \begin{align*} \|s_q-s_p\|_X \leq \sum_{n=p+1}^{q}\|x_n\|_X. \end{align*} The right-hand side is the corresponding scalar tail: \begin{align*} \sum_{n=p+1}^{q}\|x_n\|_X = a_q-a_p. \end{align*} Since each term $\|x_n\|_X$ is nonnegative, the scalar partial sums are increasing, so $a_p\leq a_q\leq L$. Therefore \begin{align*} a_q-a_p \leq |L-a_q|+|L-a_p|<\varepsilon. \end{align*} Combining these estimates gives \begin{align*} \|s_q-s_p\|_X<\varepsilon. \end{align*} Thus the partial sums $(s_m)$ are Cauchy in $X$. Completeness of $X$ now supplies an element $s\in X$ with $s_m\to s$ in $X$. By definition, this is precisely convergence of the series $\sum_{n=1}^{\infty}x_n$ in $X$. [/guided] [/step] [step:Extract a subsequence with summable increments from an arbitrary Cauchy sequence] Assume conversely that every sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ satisfying convergence of $\sum_{n=1}^{\infty}\|x_n\|_X$ has a convergent vector series $\sum_{n=1}^{\infty}x_n$ in $X$. Let $(y_n)_{n\in\mathbb{N}}$ be a Cauchy sequence in $X$. For each $k\in\mathbb{N}$, since $(y_n)$ is Cauchy, there exists $N_k\in\mathbb{N}$ such that for all $m,n\geq N_k$, \begin{align*} \|y_m-y_n\|_X \leq 2^{-k}. \end{align*} Choose a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ of natural numbers recursively as follows: set $n_1\geq N_1$, and once $n_k$ has been chosen, choose $n_{k+1}\in\mathbb{N}$ such that \begin{align*} n_{k+1}\geq \max\{N_k,N_{k+1},n_k+1\}. \end{align*} This recursion ensures two facts: the sequence $(n_k)_{k\in\mathbb{N}}$ is strictly increasing, and $n_k\geq N_k$ for every $k\in\mathbb{N}$. Therefore, for every $k\in\mathbb{N}$, both $n_k$ and $n_{k+1}$ are at least $N_k$, so the Cauchy estimate defining $N_k$ gives \begin{align*} \|y_{n_{k+1}}-y_{n_k}\|_X \leq 2^{-k}. \end{align*} [/step] [step:Represent the selected subsequence as partial sums of an absolutely convergent telescoping series] Define a sequence $(z_k)_{k\in\mathbb{N}}$ in $X$ by \begin{align*} z_1 := y_{n_1} \end{align*} and, for every $k\geq 2$, \begin{align*} z_k := y_{n_k}-y_{n_{k-1}}. \end{align*} We first verify that $\sum_{k=1}^{\infty}z_k$ is absolutely convergent. For every $M\geq 2$, using the increment estimate with index $k-1$ gives \begin{align*} \sum_{k=1}^{M}\|z_k\|_X = \|y_{n_1}\|_X+\sum_{k=2}^{M}\|y_{n_k}-y_{n_{k-1}}\|_X. \end{align*} Hence \begin{align*} \sum_{k=1}^{M}\|z_k\|_X \leq \|y_{n_1}\|_X+\sum_{k=2}^{M}2^{-(k-1)}. \end{align*} Since \begin{align*} \sum_{k=2}^{M}2^{-(k-1)} \leq 1, \end{align*} the scalar partial sums of $\sum_{k=1}^{\infty}\|z_k\|_X$ are increasing and bounded above by $\|y_{n_1}\|_X+1$. An increasing bounded real sequence is Cauchy, and by the [[Completeness of the Real Line](/theorems/7986)][citetheorem:7986] it converges in $\mathbb{R}$. Therefore the scalar series $\sum_{k=1}^{\infty}\|z_k\|_X$ converges in $\mathbb{R}$. By the assumed absolute convergence property, the vector series $\sum_{k=1}^{\infty}z_k$ converges in $X$. Let $y\in X$ denote its sum, so its partial sums converge to $y$. For each $M\in\mathbb{N}$, define \begin{align*} t_M := \sum_{k=1}^{M}z_k. \end{align*} By the definition of $(z_k)$, the finite sum telescopes: \begin{align*} t_M = y_{n_M}. \end{align*} Thus $y_{n_M}\to y$ in $X$ as $M\to\infty$. [/step] [step:Use the Cauchy property to pass convergence from the subsequence to the full sequence] It remains to prove that the original sequence $(y_n)$ converges to $y$. Let $\varepsilon>0$. Since $(y_n)$ is Cauchy, there exists $N\in\mathbb{N}$ such that for all $m,n\geq N$, \begin{align*} \|y_m-y_n\|_X<\frac{\varepsilon}{2}. \end{align*} Since $y_{n_M}\to y$ in $X$, there exists $M_0\in\mathbb{N}$ such that for all $M\geq M_0$, \begin{align*} \|y_{n_M}-y\|_X<\frac{\varepsilon}{2}. \end{align*} Choose $M_1\geq M_0$ such that $n_{M_1}\geq N$, which is possible because $(n_k)$ is strictly increasing and hence unbounded in $\mathbb{N}$. For any $n\geq N$, the triangle inequality gives \begin{align*} \|y_n-y\|_X \leq \|y_n-y_{n_{M_1}}\|_X+\|y_{n_{M_1}}-y\|_X. \end{align*} The first term is less than $\varepsilon/2$ because $n\geq N$ and $n_{M_1}\geq N$. The second term is less than $\varepsilon/2$ because $M_1\geq M_0$. Therefore \begin{align*} \|y_n-y\|_X<\varepsilon. \end{align*} Thus $y_n\to y$ in $X$. Since every Cauchy sequence in $X$ converges in $X$, the normed [vector space](/page/Vector%20Space) $X$ is complete, and hence $X$ is a Banach space. [/step]