Absolute convergence

In finite sums, the order in which terms are added is both associative and commutative, meaning that grouping and rearranging do not matter. (1 + 2) + 3 is the same as 1 + (2 + 3), and both are the same as (3 + 2) + 1. However, this is not true when adding infinitely many numbers, and wrongly assuming that it is true can lead to apparent paradoxes. One classic example is the alternating sum

${\displaystyle S=1-1+1-1+1-1...}$

whose terms alternate between +1 and −1. What is the value of S? One way to evaluate S is to group the first and second term, the third and fourth, and so on:

${\displaystyle S_{1}=(1-1)+(1-1)+(1-1)....=0+0+0...=0}$

But another way to evaluate S is to leave the first term alone and group the second and third term, then the fourth and fifth term, and so on:

${\displaystyle S_{2}=1+(-1+1)+(-1+1)+(-1+1)....=1+0+0+0...=1}$

This leads to an apparent paradox: does ${\displaystyle S=0}$ or ${\displaystyle S=1}$?

The answer is that because S is not absolutely convergent, grouping or rearranging its terms changes the value of the sum. This means ${\displaystyle S_{1}}$ and ${\displaystyle S_{2}}$ are not equal. In fact, the series ${\displaystyle 1-1+1-1+...}$ does not converge, so S does not have a value to find in the first place. A series that is absolutely convergent does not have this problem: grouping or rearranging its terms does not change the value of the sum.

A sum of real numbers or complex numbers ${\textstyle \sum _{n=0}^{\infty }a_{n}}$ is absolutely convergent if the sum of the absolute values of the terms ${\textstyle \sum _{n=0}^{\infty }|a_{n}|}$ converges.

The same definition can be used for series ${\textstyle \sum _{n=0}^{\infty }a_{n}}$ whose terms ${\displaystyle a_{n}}$ are not numbers but rather elements of an arbitrary abelian topological group. In that case, instead of using the absolute value, the definition requires the group to have a norm, which is a positive real-valued function ${\textstyle \|\cdot \|:G\to \mathbb {R} _{+}}$ on an abelian group ${\displaystyle G}$ (written additively, with identity element 0) such that:

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

In this case, the function ${\displaystyle d(x,y)=\|x-y\|}$ induces the structure of a metric space (a type of topology) on ${\displaystyle G.}$

Then, a ${\displaystyle G}$-valued series is absolutely convergent if ${\textstyle \sum _{n=0}^{\infty }\|a_{n}\|<\infty .}$

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

If ${\displaystyle G}$ is complete with respect to the metric ${\displaystyle d,}$ 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.[lower-alpha 1]

Proof that any absolutely convergent series of complex numbers is convergent

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

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

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

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

Since ${\textstyle \sum 2\left|a_{k}\right|}$ is convergent, ${\textstyle s_{n}=\sum _{k=1}^{n}\left(a_{k}+\left|a_{k}\right|\right)}$ is a bounded monotonic sequence of partial sums, and ${\textstyle \sum \left(a_{k}+\left|a_{k}\right|\right)}$ must also converge. Noting that ${\textstyle \sum a_{k}=\sum \left(a_{k}+\left|a_{k}\right|\right)-\sum \left|a_{k}\right|}$ is the difference of convergent series, we conclude that it too is a convergent series, 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 ${\displaystyle (X,\|\,\cdot \,\|).}$ Let ${\textstyle \sum x_{n}}$ be an absolutely convergent series in ${\displaystyle X.}$ As ${\textstyle \sum _{k=1}^{n}\|x_{k}\|}$ is a Cauchy sequence of real numbers, for any ${\displaystyle \varepsilon >0}$ and large enough natural numbers ${\displaystyle m>n}$ 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 ${\textstyle \sum _{k=1}^{n}x_{k}}$ is a Cauchy sequence in ${\displaystyle X,}$ hence the series is convergent in ${\displaystyle X.}$[3]

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

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

The Cauchy product is defined as the sum of terms ${\displaystyle c_{n}}$ where:

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

If either the ${\displaystyle a_{n}}$ or ${\displaystyle b_{n}}$ sum converges absolutely then

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

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 ${\displaystyle X}$ and a function ${\displaystyle f:X\to \mathbb {R} .}$ We will give a definition below of the sum of ${\displaystyle f}$ over ${\displaystyle X,}$ written as ${\textstyle \sum _{x\in X}f(x).}$

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

Therefore we define ${\textstyle \sum _{x\in X}f(x)}$ only in the case where there exists some bijection ${\displaystyle g:\mathbb {Z} ^{+}\to X}$ such that ${\textstyle \sum _{n=1}^{\infty }f(g(n))}$ 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 ${\displaystyle f}$ over ${\displaystyle X}$[5] is defined by

$${\displaystyle \sum _{x\in X}f(x):=\sum _{n=1}^{\infty }f(g(n))}$$

Note that because the series is absolutely convergent, then every rearrangement is identical to a different choice of bijection ${\displaystyle g.}$ Since all of these sums have the same value, then the sum of ${\displaystyle f}$ over ${\displaystyle X}$ is well-defined.

Even more generally we may define the sum of ${\displaystyle f}$ over ${\displaystyle X}$ when ${\displaystyle X}$ is uncountable. But first we define what it means for the sum to be convergent.

Let ${\displaystyle X}$ be any set, countable or uncountable, and ${\displaystyle f:X\to \mathbb {R} }$ a function. We say that the sum of ${\displaystyle f}$ over ${\displaystyle X}$ converges absolutely if

$${\displaystyle \sup \left\{\sum _{x\in A}|f(x)|:A\subseteq X,A{\text{ is finite }}\right\}<\infty .}$$

There is a theorem which states that, if the sum of ${\displaystyle f}$ over ${\displaystyle X}$ is absolutely convergent, then ${\displaystyle f}$ takes non-zero values on a set that is at most countable. Therefore, the following is a consistent definition of the sum of ${\displaystyle f}$ over ${\displaystyle X}$ when the sum is absolutely convergent.

$${\displaystyle \sum _{x\in X}f(x):=\sum _{x\in X:f(x)\neq 0}f(x).}$$

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

Some authors define an iterated sum ${\textstyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{m,n}}$ to be absolutely convergent if the iterated series ${\textstyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }|a_{m,n}|<\infty .}$[6] This is in fact equivalent to the absolute convergence of ${\textstyle \sum _{(m,n)\in \mathbb {N} \times \mathbb {N} }a_{m,n}.}$ That is to say, if the sum of ${\displaystyle f}$ over ${\displaystyle X,}$ ${\textstyle \sum _{(m,n)\in \mathbb {N} \times \mathbb {N} }a_{m,n},}$ converges absolutely, as defined above, then the iterated sum ${\textstyle \sum _{m=1}^{\infty }\sum _{n=1}^{\infty }a_{m,n}}$ converges absolutely, and vice versa.

The integral ${\textstyle \int _{A}f(x)\,dx}$ of a real or complex-valued function is said to converge absolutely if ${\textstyle \int _{A}\left|f(x)\right|\,dx<\infty .}$ 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 ${\textstyle f:[1,\infty )\to \mathbb {R} :x\mapsto {\frac {\sin x}{x}}}$ is improperly Riemann integrable on its unbounded domain, but it is not absolutely integrable:

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

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

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 (simply because they may fail to be measurable): 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 (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. ${\displaystyle f}$ integrable implies ${\displaystyle |f|}$ integrable
2. ${\displaystyle f}$ measurable, ${\displaystyle |f|}$ integrable implies ${\displaystyle f}$ integrable

are essentially built into the definition of the Lebesgue integral. In particular, applying the theory to the counting measure on a set ${\displaystyle S,}$ one recovers the notion of unordered summation of series developed by Moore–Smith using (what are now called) nets. When ${\displaystyle S=\mathbb {N} }$ 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.

• Cauchy principal value – Method for assigning values to certain improper integrals which would otherwise be undefined
• Conditional convergence – a property of infinite seriesPages displaying wikidata descriptions as a fallback
• Convergence of Fourier series
• Fubini's theorem – Conditions for switching order of integration in calculus
• Modes of convergence (annotated index) – Annotated index of various modes of convergence
• Radius of convergence – Domain of convergence of power series
• Riemann series theorem – Unconditional series converge absolutely
• Unconditional convergence – Order-independent convergence of a sequence
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• 1/2 + 1/4 + 1/8 + 1/16 + · · · – Mathematical infinite seriesPages displaying short descriptions of redirect targets