View source for Absolute convergence

{{Short description|Mode of convergence of an infinite series}}
In [[mathematics]], an [[Series (mathematics)|infinite series]] of numbers is said to '''converge absolutely''' (or to be '''absolutely convergent''') if the sum of the [[absolute value]]s of the summands is finite<!-- don't link to [[finite set]], please -->.  More precisely, a [[Real number|real]] or [[Complex number|complex]] series <math>\textstyle\sum_{n=0}^\infty a_n</math> is said to '''converge absolutely''' if <math>\textstyle\sum_{n=0}^\infty \left|a_n\right| = L</math> for some real number <math>\textstyle L.</math> Similarly, an [[improper integral]] of a [[function (mathematics)|function]], <math>\textstyle\int_0^\infty f(x)\,dx,</math> is said to converge absolutely if the integral of the absolute value of the integrand is finite—that is, if <math>\textstyle\int_0^\infty |f(x)|dx = L.</math>  A convergent series that is not absolutely convergent is called [[Conditional convergence|conditionally convergent]].

Absolute convergence is important for the study of infinite series, because its definition guarantees that a series will have some "nice" behaviors of finite sums that not all convergent series possess. For instance, rearrangements do not change the value of the sum, which is not necessarily true for conditionally convergent series.

==Background==
When adding a finite number of terms, [[addition]] is both [[Associative property|associative]] and [[Commutative property|commutative]], meaning that grouping and rearrangement do not alter the final sum. For instance, <math>(1+2)+3</math> is equal to both <math>1+(2+3)</math> and <math>(3+2)+1</math>. However, associativity and commutativity do not necessarily hold for infinite sums. One example is the [[alternating harmonic series]]

<math>S = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots </math>

whose terms are fractions that alternate in sign. This series is [[Convergent series|convergent]] and can be evaluated using the [[Maclaurin series]] for the function <math>\ln(1+x) </math>, which converges for all <math>x </math> satisfying <math>-1<x\leq1 </math>:

<math>\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots </math>

Substituting <math>x=1 </math> reveals that the original sum is equal to <math>\ln2 </math>. The sum can also be rearranged as follows:

<math>S=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac15-\frac{1}{10}\right)-\frac{1}{12}+\cdots </math>

In this rearrangement, the [[Multiplicative inverse|reciprocal]] of each [[Parity (mathematics)|odd number]] is grouped with the reciprocal of twice its value, while the reciprocals of every multiple of 4 are evaluated separately. However, evaluating the terms inside the parentheses yields 

<math>S=\frac12-\frac14+\frac16-\frac18+\frac{1}{10}-\frac{1}{12}+\cdots </math> 

or half the original series. The violation of the associativity and commutativity of addition reveals that the alternating harmonic series is [[Conditional convergence|conditionally convergent]]. Indeed, the sum of the absolute values of each term is <math display="inline">1+\frac12+\frac13+\frac14+\cdots </math>, or the divergent [[Harmonic series (mathematics)|harmonic series]]. According to the [[Riemann series theorem]], any conditionally convergent series can be permuted so that its sum is any finite real number or so that it diverges. When an absolutely convergent series is rearranged, its sum is always preserved.

== Definition for real and complex numbers ==
A sum of real numbers or complex numbers <math display=inline>\sum_{n=0}^{\infty} a_n</math> is absolutely convergent if the sum of the absolute values of the terms <math display=inline>\sum_{n=0}^{\infty} |a_n|</math> [[Convergent series|converges]].

== Sums of more general elements ==

The same definition can be used for series <math display=inline>\sum_{n=0}^{\infty} a_n</math> whose terms <math>a_n</math> are not numbers but rather elements of an arbitrary [[Topological abelian group|abelian topological group]]. In that case, instead of using the [[absolute value]], the definition requires the group to have a [[Norm (mathematics)|norm]], which is a positive real-valued function <math display=inline>\|\cdot\|: G \to \R_+</math> on an abelian group <math>G</math> (written [[Abelian group#Notation|additively]], with [[identity element]] 0) such that:
# The norm of the identity element of <math>G</math> is zero: <math>\|0\| = 0.</math>
# For every <math>x \in G,</math> <math>\|x\| = 0</math> implies <math>x = 0.</math>
# For every <math>x \in G,</math> <math>\|-x\| = \|x\|.</math>
# For every <math>x, y \in G,</math> <math>\|x+y\| \leq \|x\| + \|y\|.</math>

In this case, the function <math>d(x,y) = \|x-y\|</math> induces the structure of a [[metric space]] (a type of [[topology]]) on <math>G.</math> 

Then, a <math>G</math>-valued series is absolutely convergent if <math display=inline>\sum_{n=0}^{\infty} \|a_n\| < \infty.</math>

In particular, these statements apply using the norm <math>|x|</math> ([[absolute value]]) in the space of real numbers or complex numbers.

=== In topological vector spaces ===

If <math>X</math> is a [[topological vector space]] (TVS) and <math display=inline>\left(x_\alpha\right)_{\alpha \in A}</math> is a (possibly [[uncountable]]) family in <math>X</math> then this family is '''absolutely summable''' if<ref>{{Schaefer Wolff Topological Vector Spaces|edition=2|pp=179–180}}</ref> 
# <math display=inline>\left(x_\alpha\right)_{\alpha \in A}</math> is '''summable''' in <math>X</math> (that is, if the limit <math display=inline>\lim_{H \in \mathcal{F}(A)} x_H</math> of the [[Net (mathematics)|net]] <math>\left(x_H\right)_{H \in \mathcal{F}(A)}</math> converges in <math>X,</math> where <math>\mathcal{F}(A)</math> is the [[directed set]] of all finite subsets of <math>A</math> directed by inclusion <math>\subseteq</math> and <math display=inline>x_H := \sum_{i \in H} x_i</math>), and 
# for every continuous [[seminorm]] <math>p</math> on <math>X,</math> the family <math display="inline">\left(p \left(x_\alpha\right)\right)_{\alpha \in A}</math> is summable in <math>\R.</math>
If <math>X</math> is a normable space and if <math display=inline>\left(x_\alpha\right)_{\alpha \in A}</math> is an absolutely summable family in <math>X,</math> then necessarily all but a countable collection of <math>x_\alpha</math>'s are 0.

Absolutely summable families play an important role in the theory of [[nuclear space]]s.

==Relation to convergence==

If <math>G</math> is [[Complete (topology)|complete]] with respect to the metric <math>d,</math> 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.<!-- too HOWTOish? -->

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.<!--Banach spaces are specific here.-->

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.{{Efn|Here, the disk of convergence is used to refer to all points whose distance from the center of the series is less than the radius of convergence. That is, the disk of convergence is made up of all points for which the power series converges.}}

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

Suppose that <math display=inline>\sum \left|a_k\right|, a_k \in \Complex</math> is convergent. Then equivalently, <math display=inline>\sum \left[ \operatorname{Re}\left(a_k\right)^2 + \operatorname{Im}\left(a_k\right)^2 \right]^{1/2}</math> is convergent, which implies that <math display=inline>\sum \left|\operatorname{Re}\left(a_k\right)\right|</math> and <math display=inline>\sum\left|\operatorname{Im}\left(a_k\right)\right|</math> converge by termwise comparison of non-negative terms. It suffices to show that the convergence of these series implies the convergence of <math display=inline>\sum \operatorname{Re}\left(a_k\right)</math> and <math display=inline>\sum \operatorname{Im}\left(a_k\right),</math> for then, the convergence of <math display=inline>\sum a_k=\sum \operatorname{Re}\left(a_k\right) + i \sum \operatorname{Im}\left(a_k\right)</math> would follow, by the definition of the convergence of complex-valued series.

The preceding discussion shows that we need only prove that convergence of <math display=inline>\sum \left|a_k\right|, a_k\in\R</math> implies the convergence of <math display=inline>\sum a_k.</math>

Let <math display=inline>\sum \left|a_k\right|, a_k\in\R</math> be convergent.  Since <math>0 \leq a_k + \left|a_k\right| \leq 2\left|a_k\right|,</math> we have
<math display=block>0 \leq \sum_{k = 1}^n (a_k + \left|a_k\right|) \leq \sum_{k = 1}^n 2\left|a_k\right|.</math>
Since <math display=inline>\sum 2\left|a_k\right|</math> is convergent, <math display=inline>s_n=\sum_{k = 1}^n \left(a_k + \left|a_k\right|\right)</math> is a [[Sequence#Bounded|bounded]] [[Sequence#Increasing and decreasing|monotonic]] [[sequence]] of partial sums, and <math display=inline>\sum \left(a_k + \left|a_k\right|\right)</math> must also converge. Noting that <math display=inline>\sum a_k = \sum \left(a_k + \left|a_k\right|\right) - \sum \left|a_k\right|</math> 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]].<ref>{{Cite book|url=https://archive.org/details/1979RudinW|title=Principles of Mathematical Analysis|last=Rudin|first=Walter|publisher=McGraw-Hill|year=1976|isbn=0-07-054235-X|location=New York|pages=71–72}}</ref>  By the [[Cauchy's convergence test|Cauchy criterion]], <math display=inline>\sum |a_i|</math> converges if and only if for any <math>\varepsilon > 0,</math> there exists <math>N</math> such that <math display=inline>\left|\sum_{i=m}^n \left|a_i\right| \right| = \sum_{i=m}^n |a_i| < \varepsilon</math> for any <math>n > m \geq N.</math> But the triangle inequality implies that <math display=inline>\big|\sum_{i=m}^n a_i\big| \leq \sum_{i=m}^n |a_i|,</math> so that <math display=inline>\left|\sum_{i=m}^n a_i\right| < \varepsilon</math> for any <math>n > m \geq N,</math> which is exactly the Cauchy criterion for <math display=inline>\sum a_i.</math>

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

The above result can be easily generalized to every [[Banach space]] <math>(X, \|\,\cdot\,\|).</math>  Let <math display="inline">\sum x_n</math> be an absolutely convergent series in <math>X.</math> As <math display=inline>\sum_{k=1}^n\|x_k\|</math> is a [[Cauchy sequence]] of real numbers, for any <math>\varepsilon > 0</math> and large enough [[natural number]]s <math>m > n</math> it holds:
<math display=block>\left| \sum_{k=1}^m \|x_k\| - \sum_{k=1}^n \|x_k\| \right| = \sum_{k=n+1}^m \|x_k\| < \varepsilon.</math>

By the triangle inequality for the norm {{math|ǁ⋅ǁ}}, one immediately gets:
<math display=block>\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,</math>
which means that <math display=inline>\sum_{k=1}^n x_k</math> is a Cauchy sequence in <math>X,</math> hence the series is convergent in <math>X.</math><ref>{{citation
|last = Megginson|first = Robert E.|author-link = Robert Megginson
|title = An introduction to Banach space theory
|series = Graduate Texts in Mathematics
|volume = 183 
|publisher = Springer-Verlag
|location = New York 
|year = 1998
|isbn = 0-387-98431-3
|page = 20
}} (Theorem 1.3.9)</ref>

==Rearrangements and unconditional convergence==
===Real and complex numbers===

When a series of real or complex numbers is absolutely convergent, any rearrangement or reordering of that series' terms will still converge to the same value. This fact is one reason absolutely convergent series are useful: showing a series is absolutely convergent allows terms to be paired or rearranged in convenient ways without changing the sum's value.

The [[Riemann rearrangement theorem]] shows that the converse is also true: every real or complex-valued series whose terms cannot be reordered to give a different value is absolutely convergent. 

===Series with coefficients in more general space===
The term [[unconditional convergence]] is used to refer to a series where any rearrangement of its terms still converges to the same value. For any series with values in a normed abelian group <math>G</math>, as long as <math>G</math> is complete, every series which converges absolutely also converges unconditionally.

Stated more formally:
{{math theorem| Let <math>G</math> be a normed abelian group. Suppose 
<math display="block">\sum_{i=1}^\infty a_i = A \in G, \quad \sum_{i=1}^\infty \|a_i\|<\infty.</math>
If <math>\sigma : \N \to \N</math> is any permutation, then 
<math display="block">\sum_{i=1}^\infty a_{\sigma(i)}=A.</math>}}

For series with more general coefficients, the converse is more complicated. As stated in the previous section, for real-valued and complex-valued series, unconditional convergence always implies absolute convergence. However, in the more general case of a series with values in any normed abelian group <math>G</math>, the converse does not always hold: there can exist series which are not absolutely convergent, yet unconditionally convergent.

For example, in the [[Banach space]] ℓ<sup>∞</sup>, one series which is unconditionally convergent but not absolutely convergent is:
<math display=block>\sum_{n=1}^\infty \tfrac{1}{n} e_n,</math>

where <math>\{e_n\}_{n=1}^{\infty}</math> is an orthonormal basis.  A theorem of [[Aryeh Dvoretzky|A.&nbsp;Dvoretzky]] and [[Claude Ambrose Rogers|C.&nbsp;A.&nbsp;Rogers]] asserts that every infinite-dimensional Banach space has an unconditionally convergent series that is not absolutely convergent.<ref>Dvoretzky, A.; Rogers, C. A. (1950), "Absolute and unconditional convergence in normed linear spaces", Proc. Natl. Acad. Sci. U.S.A. '''36''':192–197.</ref>

===Proof of the theorem===

For any <math>\varepsilon > 0,</math> we can choose some <math>\kappa_\varepsilon, \lambda_\varepsilon \in \N,</math> such that:
<math display=block>\begin{align}
\text{ for all } N > \kappa_\varepsilon  &\quad \sum_{n=N}^\infty \|a_n\| < \tfrac{\varepsilon}{2} \\ 
\text{ for all } N > \lambda_\varepsilon &\quad \left\|\sum_{n=1}^N a_n - A\right\| < \tfrac{\varepsilon}{2}
\end{align}</math>

Let
<math display=block>\begin{align}
N_\varepsilon &=\max \left\{\kappa_\varepsilon, \lambda_\varepsilon \right\} \\
M_{\sigma,\varepsilon} &= \max \left\{\sigma^{-1}\left(\left\{ 1, \ldots, N_\varepsilon \right\}\right)\right\}
\end{align}</math>
where <math>\sigma^{-1}\left(\left\{1, \ldots, N_\varepsilon\right\}\right) = \left\{\sigma^{-1}(1), \ldots, \sigma^{-1}\left(N_\varepsilon\right)\right\}</math> so that <math>M_{\sigma,\varepsilon}</math> is the smallest natural number such that the list <math>a_{\sigma(1)}, \ldots, a_{\sigma\left(M_{\sigma,\varepsilon}\right)}</math> includes all of the terms <math>a_1, \ldots, a_{N_\varepsilon}</math> (and possibly others).

Finally for any [[integer]] <math> N > M_{\sigma,\varepsilon}</math> let
<math display=block>\begin{align}
I_{\sigma,\varepsilon} &= \left\{ 1,\ldots,N \right\}\setminus \sigma^{-1}\left(\left \{ 1, \ldots, N_\varepsilon \right \}\right) \\
S_{\sigma,\varepsilon} &= \min \sigma\left(I_{\sigma,\varepsilon}\right) = \min \left\{\sigma(k) \ : \ k \in I_{\sigma,\varepsilon}\right\} \\
L_{\sigma,\varepsilon} &= \max \sigma\left(I_{\sigma,\varepsilon}\right) = \max \left\{\sigma(k) \ : \ k \in I_{\sigma,\varepsilon}\right\} \\
\end{align}</math>
so that 
<math display="block">\begin{align}
\left\|\sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)}\right\| 
&\leq \sum_{i \in I_{\sigma,\varepsilon}} \left\|a_{\sigma(i)}\right\| \\
&\leq \sum_{j = S_{\sigma,\varepsilon}}^{L_{\sigma,\varepsilon}} \left\|a_j\right\| && \text{ since } \sigma(I_{\sigma,\varepsilon}) \subseteq \left\{S_{\sigma,\varepsilon}, S_{\sigma,\varepsilon} + 1, \ldots, L_{\sigma,\varepsilon}\right\}  \\
&\leq \sum_{j = N_\varepsilon + 1}^{\infty} \left\|a_j\right\| && \text{ since } S_{\sigma,\varepsilon} \geq N_{\varepsilon} + 1 \\
&< \frac{\varepsilon}{2}
\end{align}</math>
and thus
<math display=block>\begin{align}
\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\| \\
&< \left\|\sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \frac{\varepsilon}{2}\\
&< \varepsilon
\end{align}</math>

This shows that
<math display=block>\text{ for all } \varepsilon > 0, \text{ there exists } M_{\sigma,\varepsilon}, \text{ for all } N > M_{\sigma,\varepsilon} \quad \left\|\sum_{i=1}^N a_{\sigma(i)} - A\right\| < \varepsilon,</math>
that is:
<math display=block>\sum_{i=1}^\infty a_{\sigma(i)} = A.</math>

[[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
<math display=block>\sum_{n=0}^\infty a_n = A \quad \text{ and } \quad \sum_{n=0}^\infty b_n = B.</math>

The Cauchy product is defined as the sum of terms <math>c_n</math> where:
<math display=block>c_n = \sum_{k=0}^n a_k b_{n-k}.</math>

If {{em|either}} the <math>a_n</math> or <math>b_n</math> sum converges absolutely then
<math display=block>\sum_{n=0}^\infty c_n = A B.</math>

==Absolute convergence over sets==

A generalization of the absolute convergence of a series, is the absolute convergence of a sum of a function over a set.  We can first consider a countable set <math>X</math> and a function <math>f : X \to \R.</math>  We will give a definition below of the sum of <math>f</math> over <math>X,</math> written as <math display=inline>\sum_{x \in X} f(x).</math>  

First note that because no particular enumeration (or "indexing") of <math>X</math> has yet been specified, the series <math display=inline>\sum_{x \in X}f(x)</math> cannot be understood by the more basic definition of a series.  In fact, for certain examples of <math>X</math> and <math>f,</math> the sum of <math>f</math> over <math>X</math> may not be defined at all, since some indexing may produce a conditionally convergent series.  

Therefore we define <math display=inline>\sum_{x \in X} f(x)</math> only in the case where there exists some bijection <math>g : \Z^+ \to X</math> such that <math display=inline>\sum_{n=1}^\infty f(g(n))</math> is absolutely convergent.  Note that here, "absolutely convergent" uses the more basic definition, applied to an indexed series.  In this case, the value of the '''sum of <math>f</math> over <math>X</math>'''<ref>{{Cite book|title=Analysis I|last=Tao|first=Terence|publisher=Hindustan Book Agency|year=2016|isbn=978-9380250649|location=New Delhi|pages=188–191}}</ref> is defined by 
<math display=block>\sum_{x \in X}f(x) := \sum_{n=1}^\infty f(g(n))</math>

Note that because the series is absolutely convergent, then every rearrangement is identical to a different choice of bijection <math>g.</math>  Since all of these sums have the same value, then the sum of <math>f</math> over <math>X</math> is well-defined.  

Even more generally we may define the sum of <math>f</math> over <math>X</math> when <math>X</math> is uncountable.  But first we define what it means for the sum to be convergent.

Let <math>X</math> be any set, countable or uncountable, and <math>f : X \to \R</math> a function.  We say that '''the sum of <math>f</math> over <math>X</math> converges absolutely''' if 
<math display=block>\sup\left\{\sum_{x \in A} |f(x)|: A\subseteq X, A \text{ is finite }\right\} < \infty.</math>

There is a theorem which states that, if the sum of <math>f</math> over <math>X</math> is absolutely convergent, then <math>f</math> takes non-zero values on a set that is at most countable.  Therefore, the following is a consistent definition of the sum of <math>f</math> over <math>X</math> when the sum is absolutely convergent.  
<math display=block>\sum_{x \in X} f(x) := \sum_{x \in X : f(x) \neq 0} f(x).</math>

Note that the final series uses the definition of a series over a [[countable set]].

Some authors define an iterated sum <math display=inline>\sum_{m=1}^\infty \sum_{n=1}^\infty a_{m,n}</math> to be absolutely convergent if the iterated series <math display=inline>\sum_{m=1}^\infty \sum_{n=1}^\infty |a_{m,n}| < \infty.</math><ref>{{Cite book|title=The Way of Analysis|last=Strichartz|first=Robert|publisher=Jones & Bartlett Learning|year=2000|isbn=978-0763714970|pages=259–260}}</ref>  This is in fact equivalent to the absolute convergence of <math display=inline>\sum_{(m,n) \in \N \times \N} a_{m,n}.</math>  That is to say, if the sum of <math>f</math> over <math>X,</math> <math display=inline>\sum_{(m,n) \in \N \times \N} a_{m,n},</math> converges absolutely, as defined above, then the iterated sum <math display=inline>\sum_{m=1}^\infty \sum_{n=1}^\infty a_{m,n}</math> converges absolutely, and vice versa.

==Absolute convergence of integrals==

The [[integral]] <math display=inline>\int_A f(x)\,dx</math> of a real or complex-valued function is said to '''converge absolutely''' if <math display=inline>\int_A \left|f(x)\right|\,dx < \infty.</math>  One also says that <math>f</math> is '''absolutely integrable'''.  The issue of absolute integrability is intricate and depends on whether the [[Riemann integral|Riemann]], [[Lebesgue integral|Lebesgue]], or [[Henstock–Kurzweil integral|Kurzweil-Henstock]] (gauge) integral is considered; for the Riemann integral, it also depends on whether we only consider integrability in its proper sense (<math>f</math> and <math>A</math> both [[Bounded function|bounded]]), or permit the more general case of improper integrals.

As a standard property of the Riemann integral, when <math>A=[a,b]</math> is a bounded [[Interval (mathematics)|interval]], every [[continuous function]] is bounded and (Riemann) integrable, and since <math>f</math> continuous implies <math>|f|</math> continuous, every continuous function is absolutely integrable.  In fact, since <math>g\circ f</math> is Riemann integrable on <math>[a,b]</math> if <math>f</math> is (properly) integrable and <math>g</math> is continuous, it follows that <math>|f|=|\cdot|\circ f</math> is properly Riemann integrable if <math>f</math> is.  However, this implication does not hold in the case of improper integrals.  For instance, the function <math display=inline>f:[1,\infty) \to \R : x \mapsto \frac{\sin x}{x}</math> is improperly Riemann integrable on its unbounded domain, but it is not absolutely integrable:
<math display=block>\int_1^\infty \frac{\sin x}{x}\,dx = \frac{1}{2}\bigl[\pi - 2\,\mathrm{Si}(1)\bigr] \approx 0.62, \text{ but } \int_1^\infty \left|\frac{\sin x}{x}\right| dx = \infty.</math>
Indeed, more generally, given any series <math display=inline>\sum_{n=0}^\infty a_n</math> one can consider the associated [[step function]] <math>f_a: [0,\infty) \to \R</math> defined by <math>f_a([n,n+1)) = a_n.</math>  Then <math display=inline>\int_0^\infty f_a \, dx</math> converges absolutely, converges conditionally or diverges according to the corresponding behavior of <math display=inline>\sum_{n=0}^\infty a_n.</math>

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 <math>|f|</math> is unbounded in the examples above implies that <math>f</math> is also not integrable in the Lebesgue sense.  In fact, in the Lebesgue theory of integration, given that <math>f</math> is [[Measurable function|measurable]], <math>f</math> is (Lebesgue) integrable if and only if <math>|f|</math> is (Lebesgue) integrable.  However, the hypothesis that <math>f</math> is measurable is crucial; it is not generally true that absolutely integrable functions on <math>[a,b]</math> are integrable (simply because they may fail to be measurable): let <math>S \subset [a,b]</math> be a nonmeasurable [[subset]] and consider <math>f = \chi_S - 1/2,</math> where <math>\chi_S</math> is the [[Indicator function|characteristic function]] of <math>S.</math>  Then <math>f</math> is not Lebesgue measurable and thus not integrable, but <math>|f| \equiv 1/2</math> is a constant function and clearly integrable.

On the other hand, a function <math>f</math> may be Kurzweil-Henstock integrable (gauge integrable) while <math>|f|</math> is not.  This includes the case of improperly Riemann integrable functions.

In a general sense, on any [[measure space]] <math>A,</math> the Lebesgue integral of a real-valued function is defined in terms of its positive and negative parts, so the facts:

# <math>f</math> integrable implies <math>|f|</math> integrable
# <math>f</math> measurable, <math>|f|</math> integrable implies <math>f</math> integrable

are essentially built into the definition of the Lebesgue integral.  In particular, applying the theory to the [[counting measure]] on a [[Set (mathematics)|set]] <math>S,</math> one recovers the notion of unordered summation of series developed by Moore–Smith using (what are now called) nets.  When <math>S = \N</math> 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 analysis|functional analytic]] approach, obtaining the [[Bochner integral]].

==See also==

* {{annotated link|Cauchy principal value}}
* {{annotated link|Conditional convergence}}
* {{annotated link|Convergence of Fourier series}}
* {{annotated link|Fubini's theorem}}
* {{annotated link|Modes of convergence (annotated index)}}
* {{annotated link|Radius of convergence}}
* {{annotated link|Riemann series theorem}}
* {{annotated link|Unconditional convergence}}
* {{annotated link|1/2 − 1/4 + 1/8 − 1/16 + · · ·}}
* {{annotated link|1/2 + 1/4 + 1/8 + 1/16 + · · ·}}

== Notes ==

{{notelist}}
{{reflist|group=note}}

==References==

{{reflist}}

===General references===

* {{Narici Beckenstein Topological Vector Spaces|edition=2}} <!-- {{sfn|Narici|Beckenstein|2011|p=}} -->
* Walter Rudin, ''Principles of Mathematical Analysis'' (McGraw-Hill: New York, 1964).
* {{Pietsch Nuclear Locally Convex Spaces|edition=2}} <!-- {{sfn|Pietsch|1979|p=}} -->
* {{cite book|last=Robertson|first=A. P.|title=Topological vector spaces|publisher=University Press|publication-place=Cambridge England|year=1973|isbn=0-521-29882-2|oclc=589250 }} <!-- {{sfn|Robertson|1973|p=}} -->
* {{Ryan Introduction to Tensor Products of Banach Spaces|edition=1}} <!-- {{sfn|Ryan|2002|p=}} -->
* {{Trèves François Topological vector spaces, distributions and kernels}} <!-- {{sfn|Trèves|2006|p=}} -->
* {{Wong Schwartz Spaces, Nuclear Spaces, and Tensor Products}} <!-- {{sfn|Wong|1979|p=}} -->

{{Series (mathematics)}}
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[[Category:Series (mathematics)]]
[[Category:Integral calculus]]
[[Category:Summability theory]]
[[Category:Convergence (mathematics)]]