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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.


3 Answers 3


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.

  • 3
    $\begingroup$ Hey, I am facing the same problem and your solution was really good! However, I don't know why proving $(y_n)$ is convergent finished the problem... Now, for every $\varepsilon >0$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ we have that $\sum_{n=N}^\infty y_n <\varepsilon$, but what does conclude the proof? $\endgroup$
    – user326159
    Feb 21, 2019 at 17:12
  • 3
    $\begingroup$ @user326159 The sequence $(y_n)$ is a convergent subsequence of a Cauchy sequence $(x_n)$. This implies that $(x_n)$ converges. $\endgroup$ May 1, 2019 at 23:37
  • $\begingroup$ How does $y_n$ convergent mean the space is complete? (how do we know it converges to an $x \in X$) $\endgroup$
    – sma
    Feb 3, 2020 at 22:19
  • $\begingroup$ @ThiagoAlexandre By assumption, $y_n$ is an absolutely convergent sequence that converges in the sequence. $\endgroup$
    – kam
    Apr 4, 2020 at 17:20
  • $\begingroup$ @ThiagoAlexandre I don't really get it. A Cauchy sequence is bounded and isn't that we can always find a convergent subsequence but that doesn't mean that if there exists a convergent subsequence then the sequence converges. Otherwise, we can conclude that every Cauchy sequence is convergent. What do I miss here? $\endgroup$
    – Nabs
    Dec 12, 2023 at 22:54

A normed space is complete if and only if every absolutely convergent series converges.


We will prove this by proving absolutely convergent series is a Cauchy series.

We define a absolutely convergent series. Suppose $x_n\in E$ and $\sum_{n=1}^\infty||x_n||<\infty$ and denote $$ s_n=\sum_{k=1}^nx_k $$ Because the sequence of partial sums converges in E, for every $\epsilon >0 $, there exists $k>0$ such that $$ \sum_{n=k+1}^\infty ||x_n||<\epsilon $$ To show $(s_n)$ is a Cauchy sequence, $\forall \epsilon>0,\exists M,\forall m,n>M$ such that $$ ||s_m-s_n||=||x_{n+1}+x_{n+2}+\dotsm+x_m||\le \sum_{r=n+1}^\infty ||x_r||<\epsilon $$ (without loss of generality, we assume $m>n$)

Since E is complete, $s_n$ converges.


We need to prove if every absolutely convergent series in a normed space converges, then the normed space is complete.

Let $(x_n)$ be an Cauchy sequence in E and therefore $\forall \epsilon>0,\exists p_k\in N,\forall m,n>p_k$ such that $$ ||x_m-x_n||<2^{-k} $$ without loss of generality, we can assume $(p_k)$ is strictly increasing.

Then the series $\sum_{k=1}^\infty (x_{p_{k+1}}-x_{p_k})$ is absolutely convergent and therefore, convergent and therefore, the sequence $$ x_{p_k}=x_{p_1}+(x_{p_2}-x_{p_1})+(x_{p_3}-x_{p_2})+\dotsm+(x_{p_k}-x_{p_{k-1}}) $$ converges to an element $x\in E$

Then $$ ||x_n-x||\le ||x_n-x_{p_n}||+||x_{p_n}-x||\rightarrow 0 $$ Q.E.D.


Hint: show that there is a subsequence $y_n$ such that $\|y_n - y_{n+1}\| < 2^{-n}$.


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