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A Normed Linear Space $X$ is a Banach Space iff every absolutely convergent series is convergent.
My try:
Let $X$ is a Banach Space .Let $\sum x_n$ be an absolutely convergent series .Consider $s_n=\sum_{i=1}^nx_i$. Now $\sum \|x_n\|<\infty \implies \exists N$ such that $\sum_{i=N}^ \infty \|x_i\|<\epsilon$ for any $\epsilon>0$
Then $\|s_n-s_m\|\le \sum _{i=m+1}^n \|x_i\|<\epsilon \forall n,m>N$
So $s_n$ is Cauchy in $X$ and hence converges $s_n\to s$ (say).
Thus $\sum x_i$ converges.
Conversely, let $x_n$ be a Cauchy Sequence in $X$. Here I can't proceed how to use the given fact.
Any help will be great.
Best Answer
For the converse argument; let $X$ be a normed linear space in which every absolutely convergent series converges, and suppose that $\{x_n\}$ is a Cauchy sequence.
For each $k \in \mathbb{N}$, choose $n_k$ such that $||(x_m − x_n)|| < 2^{−k}$ for $m, n \geq n_k$. In particular, $||x_{n_{k+1}}−x_{n_k}||< 2^{-k}$. If we define $y_{1} = x_{n_{1}}$ and $y_{k+1} = x_{n_{k+1}} − x_{n_{k}}$ for $k \geq 1$, it follows that $\sum ||y_{n}|| ≤ ||x_{n_{1}} || + 1$ i. e., ($y_{n}$) is absolutely convergent, and hence convergent.