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Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in a Banach space $X$. The series $\sum_{n \in \mathbb{N}} x_n$ converges in $X$ if $\sum_{n \in \mathbb{N}} \|x_n\| < \infty$.

Is this statement true?

Since $X$ is a Banach space, if we prove the series $\sum_{n \in \mathbb{N}} x_n$ a Cauchy sequence then it converges. That is, for any $m \ge n > N$, there is $\|\sum_{m \in \mathbb{N}} x_m - \sum_{n \in \mathbb{N}} x_n\| = \|\sum_{k = n}^m x_k\| < \epsilon$ for any $\epsilon > 0$. But how can this be derived from the boundedness of $\sum_{n \in \mathbb{N}} \|x_n\|$?

Analysis Newbie
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3 Answers3

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Triangle inequality: that follows from the fact that$$\left\lVert\sum_{k=n}^mx_k\right\rVert\leqslant\sum_{k=n}^m\lVert x_k\rVert.$$

José Carlos Santos
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In Banach spaces, absolute convergence implies convergence. Since $s_{n} = \sum_{k=0}^{n}\lVert x_{k}\rVert$ in monotonic and bounded, it converges. Therefore the given series $\sum_{n\in\mathbb{N}}x_{n}$ converges:

About Banach Spaces And Absolute Convergence Of Series

user0102
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Since $X$ is a Banach space, it is complete. Notice that $\sum_{k=1}^{\infty} ||u_k|| < \infty$ implies that for every $\epsilon>0$ there is some index $N$ such that $\sum_{k=N}^{\infty} ||u_k|| < \epsilon$.

Let $x_n = \sum_{k=1}^n u_k$. Then for any $m \ge n \ge N$, we have: $||x_n - x_m|| = ||\sum_{k=1}^n u_k - \sum_{k=1}^m u_k|| = ||\sum_{k=n+1}^m u_k|| \le \sum_{k=n+1}^m ||u_k|| \le \sum_{k=N}^{\infty} ||u_k|| < \epsilon$. Therefore, this is a Cauchy sequence and since $X$ is complete there exists an $x$ that is the limit of the series.

user71118
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