# If every absolutely convergent series is convergent then $X$ is Banach

Show that

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

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.

• 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? – user326159 Feb 21 '19 at 17:12
• @user326159 The sequence $(y_n)$ is a convergent subsequence of a Cauchy sequence $(x_n)$. This implies that $(x_n)$ converges. – Thiago Alexandre May 1 '19 at 23:37
• How does $y_n$ convergent mean the space is complete? (how do we know it converges to an $x \in X$) – sma Feb 3 '20 at 22:19
• @ThiagoAlexandre By assumption, $y_n$ is an absolutely convergent sequence that converges in the sequence. – kam Apr 4 '20 at 17:20

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

$$\implies$$

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.

$$\Longleftarrow$$

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 $\|y_n - y_{n+1}\| < 2^{-n}$.