Absolutely convergent but not convergent

Question

Here, Lemma $2.1$ states that

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

I would like to know a series which is absolutely convergent but not convergent. Can someone give such example? I always thought that absolutely convergent series implies the series converges.

Answer 1

If the space is not complete, absolute convergence actually doesn't imply convergence. Take for example $X = \mathbb{Q}$ (with the norm inherited from $\mathbb{R}$) and choose a sequence of rational numbers $a_n$ such that $$ \sum_{n=1}^\infty a_n $$ is irrational, but $$ \sum_{n=1}^\infty |a_n| = 1. $$

See this question for details. Thus, in $X$, the series is absolutely convergent, but not convergent.

Answer 2

Let $E$ be the space of all continuous functions $f:[0,1]\to\mathbb{R}$, equipped with the following norm : $$\forall f\in E,\,\Vert f\Vert_1=\int_0^1\vert f(t)\vert\,dt$$

For each $n\in\mathbb{N}^\star$, let $f_n\in E$ defined by :

$f_n(t)=1$ if $t\in[0,\frac12]$

$f_n(t)=1-n(t-\frac12)$ if $t\in[\frac12,\frac12+\frac1n]$

$f_n(t)=0$ if $t\in[\frac12+\frac1n,1]$

Now put $g_n=f_n-f_{n+1}$. It is not difficult to show that $\Vert g_n\Vert=\frac1{2n(n+1)}$, which implies the convergence of the series $\sum_{n\ge1}\Vert g_n\Vert$

But the sequence $(f_n)$ is a divergent one (with respect to the norm $\Vert\,\,\Vert_1$) and so the series $\sum_{n\ge1}g_n$ is also divergent in $E$, in the same sens.

The reason why such an example exists is the fact that the normed space $(E,\Vert\,\,\Vert_1)$ is not complete (it is not a Banach space).