# Absolute convergence

In mathematics, an infinite series of numbers is said to converge absolutely (or to be absolutely convergent) if the sum of the absolute value of the summand is finite. More precisely, a real or complex series $\textstyle\sum_{n=0}^\infty a_n$ is said to converge absolutely if $\textstyle\sum_{n=0}^\infty \left|a_n\right| = L$ for some real number $\textstyle L$. Similarly, an improper integral of a function, $\textstyle\int_0^\infty f(x)\,dx$, is said to converge absolutely if the integral of the absolute value of the integrand is finite—that is, if $\textstyle\int_0^\infty \left|f(x)\right|dx = L.$

Absolute convergence is important for the study of infinite series because its definition is strong enough to have properties of finite sums that not all convergent series possess, yet is broad enough to occur commonly. (A convergent series that is not absolutely convergent is called conditionally convergent.)

## Background

One may study the convergence of series $\sum_{n=0}^{\infty} a_n$ whose terms an are elements of an arbitrary abelian topological group. The notion of absolute convergence requires more structure, namely a norm, which is a real-valued function $\|\cdot\|: G \to \mathbb{R}$ on abelian group G (written additively, with identity element 0) such that:

1. The norm of the identity element of G is zero: $\|0\| = 0.$
2. For every x in G, $\|x\| = 0$ implies $x = 0.$
3. For every x in G, $\|-x\| = \|x\|.$
4. For every x, y in G, $\|x+y\| \leq \|x\| + \|y\|.$

In this case, the function $d(x,y) = \|x-y\|$ induces on G the structure of a metric space (a type of topology). We can therefore consider G-valued series and define such a series to be absolutely convergent if $\sum_{n=0}^{\infty} \|a_n\| < \infty.$

In particular, these statements apply using the norm |x| (absolute value) in the space of real numbers or complex numbers.

## Relation to convergence

If G is complete with respect to the metric d, then every absolutely convergent series is convergent. The proof is the same as for complex-valued series: use the completeness to derive the Cauchy criterion for convergence—a series is convergent if and only if its tails can be made arbitrarily small in norm—and apply the triangle inequality.

In particular, for series with values in any Banach space, absolute convergence implies convergence. The converse is also true: if absolute convergence implies convergence in a normed space, then the space is a Banach space.

If a series is convergent but not absolutely convergent, it is called conditionally convergent. An example of a conditionally convergent series is the alternating harmonic series. Many standard tests for divergence and convergence, most notably including the ratio test and the root test, demonstrate absolute convergence. This is because a power series is absolutely convergent on the interior of its disk of convergence.

### Proof that any absolutely convergent series of complex numbers is convergent

Since a series of complex numbers converges if and only if both its real and imaginary parts converge, we may assume with equal generality that the $a_n$ are real numbers. Suppose that $\sum |a_n|$ is convergent. Then, $2\sum |a_n|$ is convergent.

Since $0 \le a_n + |a_n| \le 2|a_n|$, we have

$0 \le \sum_{n = 1}^m (a_n + |a_n|) \le \sum_{n = 1}^m 2|a_n|\le \sum_{n = 1}^\infty 2|a_n|$.

Thus, $\sum_{n = 1}^m (a_n + |a_n|)$ is a bounded monotonic sequence (in m), which must converge.

$\sum a_n = \sum(a_n+|a_n|) - \sum |a_n|$ is a difference of convergent series; therefore, it is also convergent, as desired.

### Proof that any absolutely convergent series in a Banach space is convergent

The above result can be easily generalized to every Banach space (X, ǁ ⋅ ǁ). Let xn be an absolutely convergent series in X. As $\scriptstyle\sum_{k=1}^n\|x_k\|$ is a Cauchy sequence of real numbers, for any ε > 0 and large enough natural numbers m > n it holds:

$\left|\sum_{k=1}^m\|x_k\|-\sum_{k=1}^n\|x_k\|\right| = \sum_{k=n+1}^m\|x_k\|< \varepsilon.$

By the triangle inequality for the norm ǁ ⋅ ǁ, one immediately gets:

$\left\|\sum_{k=1}^m x_k-\sum_{k=1}^nx_k\right\| = \left\|\sum_{k=n+1}^m x_k\right\| \le \sum_{k=n+1}^m\|x_k\|<\varepsilon,$

which means that $\scriptstyle\sum_{k=1}^nx_k$ is a Cauchy sequence in X, hence the series is convergent in X.[1]

## Rearrangements and unconditional convergence

In the general context of a G-valued series, a distinction is made between absolute and unconditional convergence, and the assertion that a real or complex series which is not absolutely convergent is necessarily conditionally convergent (meaning not unconditionally convergent) is then a theorem, not a definition. This is discussed in more detail below.

Given a series $\sum_{n=0}^{\infty} a_n$ with values in a normed abelian group G and a permutation σ of the natural numbers, one builds a new series $\sum_{n=0}^\infty a_{\sigma(n)}$, said to be a rearrangement of the original series. A series is said to be unconditionally convergent if all rearrangements of the series are convergent to the same value.

When G is complete, absolute convergence implies unconditional convergence:

Theorem. Let $\sum\limits_{i=1}^\infty a_i=A \in G$, $\sum\limits_{i=1}^\infty \|a_i\|<\infty$ and let σ be a permutation of $\mathbb{N}$, then $\sum\limits_{i=1}^\infty a_{\sigma(i)}=A$.

The issue of the converse is interesting. For real series it follows from the Riemann rearrangement theorem that unconditional convergence implies absolute convergence. Since a series with values in a finite-dimensional normed space is absolutely convergent if each of its one-dimensional projections is absolutely convergent, it follows that absolute and unconditional convergence coincide for Rn-valued series.

But there are unconditionally and non-absolutely convergent series with values in Hilbert space2, for example:

$a_n = n^{-1} e_n,\ \ \text{where}\ \ \{e_n\}_{n=1}^{\infty}$

is an orthonormal basis. A theorem of A. Dvoretzky and C. A. Rogers[2] asserts that every infinite-dimensional Banach space admits an unconditionally convergent series that is not absolutely convergent.

### Proof of the theorem

For any ε > 0, we can choose some $\kappa_\varepsilon,\lambda_\varepsilon \in \mathbb{N}$, such that

$\forall N>\kappa_\varepsilon\ ,\sum\limits_{n=N}^\infty \|a_n\| < \frac{\varepsilon}{2}$

and

$\forall N>\lambda_\varepsilon\ ,\left\|\sum\limits_{n=1}^N a_n-A\right\| < \frac{\varepsilon}{2}$.

Let $N_\varepsilon:=\max(\kappa_\varepsilon, \lambda_\varepsilon)$, $M_{\sigma,\varepsilon}:= \max \left\{ \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right) \right\}$.

For any $N > M_{\sigma,\varepsilon}\ \ (N \in \mathbb{N})$ let

$I_{\sigma,\varepsilon}:=\left\{ 1,\ldots,N \right\}\setminus \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right),$
$S_{\sigma,\varepsilon} := \min\{ \sigma(k) \mid k \in I_{\sigma,\varepsilon} \}$ (note that $\ S_{\sigma,\varepsilon} \geq N_{\varepsilon}+1$), and
$L_{\sigma,\varepsilon} := \max\{ \sigma(k) \mid k \in I_{\sigma,\varepsilon} \}.$

Then

$\left\|\sum\limits_{i=1}^N a_{\sigma(i)}-A \right\|= \left\| \sum_{i \in \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right)} a_{\sigma(i)} - A + \sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)} \right\|$
$\leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \left\| \sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)} \right\| \leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \sum_{i\in I_{\sigma,\varepsilon}} \| a_{\sigma(i)} \|$
$\leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \sum_{j= S_{\sigma,\varepsilon} }^{ L_{\sigma,\varepsilon} } \| a_j \| \leq \left\| \sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \sum_{j= N_\varepsilon + 1}^{\infty} \| a_j \| < \varepsilon.$

This shows that

$\forall\varepsilon > 0\ , \exists M_{\sigma,\varepsilon}\, \forall N > M_{\sigma,\varepsilon} \,\, , \left\|\sum\limits_{i=1}^N a_{\sigma(i)}-A \right\|< \varepsilon,$

that is, $\sum\limits_{i=1}^\infty a_{\sigma(i)}=A$. Q.E.D.

## Products of series

The Cauchy product of two series converges to the product of the sums if at least one of the series converges absolutely. That is, suppose that

$\sum_{n=0}^\infty a_n = A$ and $\sum_{n=0}^\infty b_n = B$.

The Cauchy product is defined as the sum of terms cn where:

$c_n = \sum_{k=0}^n a_k b_{n-k}.$

Then, if either the an or bn sum converges absolutely, then

$\sum_{n=0}^\infty c_n = AB.$

## Absolute convergence of integrals

The integral $\int_A f(x)\,dx$ of a real or complex-valued function is said to converge absolutely if $\int_A \left|f(x)\right|\,dx < \infty.$ One also says that f is absolutely integrable.

When A = [a,b] is a closed bounded interval, every continuous function is integrable, and since f continuous implies |f| continuous, similarly every continuous function is absolutely integrable. It is not generally true that absolutely integrable functions on [a,b] are integrable: let $S \subset [a,b]$ be a nonmeasurable subset and take $f = \chi_S - 1/2,$ where $\chi_S$ is the characteristic function of S. Then f is not Lebesgue measurable but |f| is constant. However, it is a standard result that if f is Riemann integrable, so is |f|. This holds also for the Lebesgue integral; see below. On the other hand a function f may be Kurzweil-Henstock integrable (or "gauge integrable") while |f| is not. This includes the case of improperly Riemann integrable functions.

Similarly, when A is an interval of infinite length it is well known that there are improperly Riemann integrable functions f which are not absolutely integrable. Indeed, given any series $\sum_{n=0}^\infty a_n$ one can consider the associated step function $f_a: [0,\infty) \rightarrow \mathbf{R}$ defined by $f_a([n,n+1)) = a_n$. Then $\int_0^\infty f_a \, dx$ converges absolutely, converges conditionally or diverges according to the corresponding behavior of $\sum_{n=0}^\infty a_n.$

Another example of a convergent but not absolutely convergent improper Riemann integral is $\int_{\mathbf{R}} \frac{\sin x}{x} \, dx$.

On any measure space A, the Lebesgue integral of a real-valued function is defined in terms of its positive and negative parts, so the facts:

1. f integrable implies |f| integrable
2. f measurable, |f| integrable implies f integrable

are essentially built into the definition of the Lebesgue integral. In particular, applying the theory to the counting measure on a set S, one recovers the notion of unordered summation of series developed by Moore–Smith using (what are now called) nets. When S = N is the set of natural numbers, Lebesgue integrability, unordered summability and absolute convergence all coincide.

Finally, all of the above holds for integrals with values in a Banach space. The definition of a Banach-valued Riemann integral is an evident modification of the usual one. For the Lebesgue integral one needs to circumvent the decomposition into positive and negative parts with Daniell's more functional analytic approach, obtaining the Bochner integral.