Cauchy sequences and absolutely convergent series

1. Oct 18, 2014

Fredrik

Staff Emeritus
1. The problem statement, all variables and given/known data

I want to prove that if X is a normed space, the following statements are equivalent.

(a) Every Cauchy sequence in X is convergent.
(b) Every absolutely convergent series in X is convergent.

I'm having difficulties with the implication (b) ⇒ (a).

2. Relevant equations

Only some standard definitions.

3. The attempt at a solution

I have no problems with the implication (a) ⇒ (b). I'm including the proof only because it can perhaps provide some inspiration for the proof of the converse.

(a) ⇒(b): Let $\sum_{k=1}^\infty x_k$ be an arbitrary absolutely convergent series in X. For each $n\in\mathbb Z^+$, define $s_n=\sum_{k=1}^n x_k$ and $t_n=\sum_{k=1}^n \|x_k\|$. The desired result follows from
$$\|s_n-s_m\|=\left\|\sum_{k=m+1}^n x_k\right\| \leq\sum_{k=m+1}^n\|x_k\| =|t_n-t_m|.$$ The argument is simple: $(t_n)$ is convergent and therefore Cauchy. This and the inequality above together imply that $(s_n)$ is Cauchy. Now (a) implies that $(s_n)$ is convergent.

Here's the part I'm struggling with:

(b) ⇒ (a): Let $(s_n)$ be an arbitrary Cauchy sequence in X. I'm using the notation $s_n$ rather than $x_n$ because I'm going to define a sequence $(x_n)$ such that $(s_n)$ is its sequence of partial sums. We define $x_1=s_1$ and for each $n\geq 2$, $x_n=s_n-s_{n-1}$. We have
$$\sum_{k=1}^n x_k =(s_n-s_{n-1})+(s_{n-1}-s_{n-2})+\cdots+(s_2-s_1)+s_1 =s_n.$$ If we can show that $\sum_{k=1}^\infty x_k$ is absolutely convergent, we're done. Unfortunately that doesn't look possible. I think we have to continue something like this: For each $k\in\mathbb Z^+$, let $n_k$ be a positive integer such that
$$i,j\geq n_k\ \Rightarrow\ \|s_i-s_j\|<2^{-k}.$$ Let $k\in\mathbb Z^+$ be arbitrary. Define $m_k=n_k+1$. We have
$$\|x_{m_k}\|=\|s_{m_k}-s_{m_k-1}\|=\|s_{n_k+1}-s_{n_k}\|<2^{-k}.$$ Does this help at all? It tells us that $\sum_{k=1}^\infty x_{m_k}$ is absolutely convergent, and therefore convergent. But how does knowing that $\sum_{k=1}^\infty x_{m_k}$ is convergent help us? It means that the sequence $\big(\sum_{k=1}^p x_{m_k}\big)_{p=1}^\infty$ is convergent, but this isn't a subsequence of $(s_n)$. $(s_{m_p})_{p=1}^\infty$ is a subsequence of $(s_n)$, but I don't see why it would be convergent.

2. Oct 18, 2014

PeroK

There's a proof here:

https://www.math.ucdavis.edu/~thomases/201A_F10_hw3_sol.pdf [Broken]

Last edited by a moderator: May 7, 2017
3. Oct 18, 2014

Fredrik

Staff Emeritus
Thanks. I understand it now. I need to write $\|s_{n_{k+1}}-s_{n_k}\|<2^{-k}$ instead of $\|s_{n_k+1}-s_{n_k}\|<2^{-k}$. I tried that earlier, but somehow I convinced myself that if I do this, the sum fails to telescope.