# Absolute convergence

(Redirected from Absolutely convergent series)

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

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.) Absolutely convergent series behave "nicely". For instance, rearrangements do not change the value of the sum. This is not true for conditionally convergent series: The alternating harmonic series ${\textstyle 1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots }$ converges to ${\displaystyle \log 2}$, while its rearrangement ${\textstyle 1+{\frac {1}{3}}-{\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{7}}-{\frac {1}{4}}+\cdots }$ (in which the repeating pattern of signs is two positive terms followed by one negative term) converges to ${\textstyle {\frac {3}{2}}\log 2}$.

## Background

One may study the convergence of series ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ 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 ${\displaystyle \|\cdot \|:G\to \mathbb {R} }$ on abelian group G (written additively, with identity element 0) such that:

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

In this case, the function ${\displaystyle d(x,y)=\|x-y\|}$ 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 ${\displaystyle \sum _{n=0}^{\infty }\|a_{n}\|<\infty .}$

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

Suppose that ${\textstyle \sum |a_{k}|,a_{k}\in \mathbb {C} }$ is convergent. Then equivalently, ${\textstyle \sum [\mathrm {Re} (a_{k})^{2}+\mathrm {Im} (a_{k})^{2}]^{1/2}}$ is convergent, which implies that ${\textstyle \sum |\mathrm {Re} (a_{k})|}$ and ${\textstyle \sum |\mathrm {Im} (a_{k})|}$ converge by termwise comparison of non-negative terms. It suffices to show that the convergence of these series implies the convergence of ${\textstyle \sum \mathrm {Re} (a_{k})}$ and ${\textstyle \sum \mathrm {Im} (a_{k})}$, for then, the convergence of ${\textstyle \sum a_{k}=\sum \mathrm {Re} (a_{k})+i\sum \mathrm {Im} (a_{k})}$ would follow, by the definition of the convergence of complex-valued series.

The preceding discussion shows that we need only prove that convergence of ${\textstyle \sum |a_{k}|,a_{k}\in \mathbb {R} }$ implies the convergence of ${\textstyle \sum a_{k}}$.

Let ${\textstyle \sum |a_{k}|,a_{k}\in \mathbb {R} }$ be convergent. Since ${\displaystyle 0\leq a_{k}+|a_{k}|\leq 2|a_{k}|}$, we have

${\displaystyle 0\leq \sum _{k=1}^{n}(a_{k}+|a_{k}|)\leq \sum _{k=1}^{n}2|a_{k}|\leq \sum 2|a_{k}|}$.

Since ${\textstyle \sum 2|a_{k}|}$ is convergent, ${\textstyle s_{n}=\sum _{k=1}^{n}(a_{k}+|a_{k}|)}$ is a bounded monotonic sequence of partial sums, and ${\textstyle \sum (a_{k}+|a_{k}|)}$ must also converge. Noting that ${\textstyle \sum a_{k}=\sum (a_{k}+|a_{k}|)-\sum |a_{k}|}$ is the difference of convergent series, we conclude that it too is a convergent series, as desired.

#### Alternative proof using the Cauchy criterion and triangle inequality

By applying the Cauchy criterion for the convergence of a complex series, we can also prove this fact as a simple implication of the triangle inequality.[1] By the Cauchy criterion, ${\textstyle \sum |a_{i}|}$ converges if and only if for any ${\displaystyle \epsilon >0}$, there exists ${\displaystyle N}$ such that ${\textstyle {\big |}\sum _{i=m}^{n}|a_{i}|{\big |}=\sum _{i=m}^{n}|a_{i}|<\epsilon }$ for any ${\displaystyle m,n\geq N}$. But the triangle inequality implies that ${\textstyle {\big |}\sum _{i=m}^{n}a_{i}{\big |}\leq \sum _{i=m}^{n}|a_{i}|}$, so that ${\textstyle {\big |}\sum _{i=m}^{n}a_{i}{\big |}<\epsilon }$ for any ${\displaystyle m,n\geq N}$, which is exactly the Cauchy criterion for ${\textstyle \sum a_{i}}$.

### 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 ${\displaystyle \scriptstyle \sum _{k=1}^{n}\|x_{k}\|}$ is a Cauchy sequence of real numbers, for any ε > 0 and large enough natural numbers m > n it holds:

${\displaystyle \left|\sum _{k=1}^{m}\|x_{k}\|-\sum _{k=1}^{n}\|x_{k}\|\right|=\sum _{k=n+1}^{m}\|x_{k}\|<\varepsilon .}$

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

${\displaystyle \left\|\sum _{k=1}^{m}x_{k}-\sum _{k=1}^{n}x_{k}\right\|=\left\|\sum _{k=n+1}^{m}x_{k}\right\|\leq \sum _{k=n+1}^{m}\|x_{k}\|<\varepsilon ,}$

which means that ${\displaystyle \scriptstyle \sum _{k=1}^{n}x_{k}}$ is a Cauchy sequence in X, hence the series is convergent in X.[2]

## 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 ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ with values in a normed abelian group G and a permutation σ of the natural numbers, one builds a new series ${\displaystyle \sum _{n=0}^{\infty }a_{\sigma (n)}}$, 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
${\displaystyle \sum _{i=1}^{\infty }a_{i}=A\in G,\quad \sum _{i=1}^{\infty }\|a_{i}\|<\infty }$
and let σ : NN be a permutation. Then:
${\displaystyle \sum _{i=1}^{\infty }a_{\sigma (i)}=A.}$

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 Banach space, for example:

${\displaystyle a_{n}={\tfrac {1}{n}}e_{n},}$

where ${\displaystyle \{e_{n}\}_{n=1}^{\infty }}$ is an orthonormal basis. A theorem of A. Dvoretzky and C. A. Rogers asserts that every infinite-dimensional Banach space admits an unconditionally convergent series that is not absolutely convergent.[3]

### Proof of the theorem

For any ε > 0, we can choose some ${\displaystyle \kappa _{\varepsilon },\lambda _{\varepsilon }\in \mathbf {N} }$, such that:

{\displaystyle {\begin{aligned}\forall N>\kappa _{\varepsilon }&\quad \sum _{n=N}^{\infty }\|a_{n}\|<{\tfrac {\varepsilon }{2}}\\\forall N>\lambda _{\varepsilon }&\quad \left\|\sum _{n=1}^{N}a_{n}-A\right\|<{\tfrac {\varepsilon }{2}}\end{aligned}}}

Let

{\displaystyle {\begin{aligned}N_{\varepsilon }&=\max \left\{\kappa _{\varepsilon },\lambda _{\varepsilon }\right\}\\M_{\sigma ,\varepsilon }&=\max \left\{\sigma ^{-1}\left(\left\{1,\dots ,N_{\varepsilon }\right\}\right)\right\}\end{aligned}}}

Finally for any integer ${\displaystyle N>M_{\sigma ,\varepsilon }}$ let

{\displaystyle {\begin{aligned}I_{\sigma ,\varepsilon }&=\left\{1,\ldots ,N\right\}\setminus \sigma ^{-1}\left(\left\{1,\dots ,N_{\varepsilon }\right\}\right)\\S_{\sigma ,\varepsilon }&=\min \left\{\sigma (k)\ :\ k\in I_{\sigma ,\varepsilon }\right\}\\L_{\sigma ,\varepsilon }&=\max \left\{\sigma (k)\ :\ k\in I_{\sigma ,\varepsilon }\right\}\end{aligned}}}

Then

{\displaystyle {\begin{aligned}\left\|\sum _{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 }}\left\|a_{\sigma (i)}\right\|\\&\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{j=S_{\sigma ,\varepsilon }}^{L_{\sigma ,\varepsilon }}\left\|a_{j}\right\|\\&\leq \left\|\sum _{j=1}^{N_{\varepsilon }}a_{j}-A\right\|+\sum _{j=N_{\varepsilon }+1}^{\infty }\left\|a_{j}\right\|&&S_{\sigma ,\varepsilon }\geq N_{\varepsilon }+1\\&<\varepsilon \end{aligned}}}

This shows that

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

that is:

${\displaystyle \sum _{i=1}^{\infty }a_{\sigma (i)}=A}$

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

${\displaystyle \sum _{n=0}^{\infty }a_{n}=A}$ and ${\displaystyle \sum _{n=0}^{\infty }b_{n}=B}$.

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

${\displaystyle c_{n}=\sum _{k=0}^{n}a_{k}b_{n-k}.}$

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

${\displaystyle \sum _{n=0}^{\infty }c_{n}=AB.}$

## Absolute convergence of integrals

The integral ${\displaystyle \int _{A}f(x)\,dx}$ of a real or complex-valued function is said to converge absolutely if ${\displaystyle \int _{A}\left|f(x)\right|\,dx<\infty .}$ One also says that ${\displaystyle f}$ is absolutely integrable. The issue of absolute integrability is intricate and depends on whether the Riemann, Lebesgue, or Kurzweil-Henstock (gauge) integral is considered; for the Riemann integral, it also depends on whether we only consider integrability in its proper sense (${\displaystyle f}$ and ${\displaystyle A}$ both bounded), or permit the more general case of improper integrals.

As a standard property of the Riemann integral, when ${\displaystyle A=[a,b]}$ is a bounded interval, every continuous function is bounded and (Riemann) integrable, and since ${\displaystyle f}$ continuous implies ${\displaystyle |f|}$ continuous, every continuous function is absolutely integrable. In fact, since ${\displaystyle g\circ f}$ is Riemann integrable on ${\displaystyle [a,b]}$ if ${\displaystyle f}$ is (properly) integrable and ${\displaystyle g}$ is continuous, it follows that ${\displaystyle |f|=|\cdot |\circ f}$ is properly Riemann integrable if ${\displaystyle f}$ is. However, this implication does not hold in the case of improper integrals. For instance, the function ${\displaystyle f:[1,\infty )\to \mathbb {R} ,\ \ x\mapsto x^{-1}\sin x}$ is improperly Riemann integrable on its unbounded domain, but it is not absolutely integrable:

${\displaystyle \int _{1}^{\infty }{\frac {\sin x}{x}}dx={\frac {1}{2}}{\big [}\pi -2\mathrm {Si} (1){\big ]}\approx 0.62}$, but ${\displaystyle \int _{1}^{\infty }{\Big |}{\frac {\sin x}{x}}{\Big |}dx=\infty }$.

Indeed, more generally, given any series ${\displaystyle \sum _{n=0}^{\infty }a_{n}}$ one can consider the associated step function ${\displaystyle f_{a}:[0,\infty )\rightarrow \mathbb {R} }$ defined by ${\displaystyle f_{a}([n,n+1))=a_{n}}$. Then ${\displaystyle \int _{0}^{\infty }f_{a}\,dx}$ converges absolutely, converges conditionally or diverges according to the corresponding behavior of ${\displaystyle \sum _{n=0}^{\infty }a_{n}.}$

The situation is different for the Lebesgue integral, which does not handle bounded and unbounded domains of integration separately (see below). The fact that the integral of ${\displaystyle |f|}$ is unbounded in the examples above implies that ${\displaystyle f}$ is also not integrable in the Lebesgue sense. In fact, in the Lebesgue theory of integration, given that ${\displaystyle f}$ is measurable, ${\displaystyle f}$ is (Lebesgue) integrable if and only if ${\displaystyle |f|}$ is (Lebesgue) integrable. However, the hypothesis that ${\displaystyle f}$ is measurable is crucial; it is not generally true that absolutely integrable functions on ${\displaystyle [a,b]}$ are integrable. Let ${\displaystyle S\subset [a,b]}$ be a nonmeasurable subset and consider ${\displaystyle f=\chi _{S}-1/2,}$ where ${\displaystyle \chi _{S}}$ is the characteristic function of ${\displaystyle S}$. Then ${\displaystyle f}$ is not Lebesgue measurable and thus not integrable, but ${\displaystyle |f|\equiv 1/2}$ is a constant function and clearly integrable.

On the other hand, a function ${\displaystyle f}$ may be Kurzweil-Henstock integrable (or "gauge integrable") while ${\displaystyle |f|}$ is not. This includes the case of improperly Riemann integrable functions.

In a general sense, on any measure space ${\displaystyle A}$, 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.