Absolute convergence

Absolute convergence   (Redirected from Absolute convergence theorem) Jump to navigation Jump to search This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (February 2013) (Learn how and when to remove this template message) 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 }|f(x)|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 - a convergent series that is not absolutely convergent is called conditionally convergent, while 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 ln 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}}ln 2.} Contents 1 Background 1.1 Explanation 2 Definition for real and complex numbers 3 Sums of more general elements 3.1 In topological vector spaces 4 Relation to convergence 4.1 Proof that any absolutely convergent series of complex numbers is convergent 4.1.1 Alternative proof using the Cauchy criterion and triangle inequality 4.2 Proof that any absolutely convergent series in a Banach space is convergent 5 Rearrangements and unconditional convergence 5.1 Real and complex numbers 5.2 Series with coefficients in more general space 5.3 Proof of the theorem 6 Products of series 7 Absolute convergence over sets 8 Absolute convergence of integrals 9 See also 10 Notes 11 References 11.1 Works cited 11.2 General references Background In finite sums, the order in which terms are added does not matter. 1 + 2 + 3 is 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, 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: rearranging its terms does not change the value of the sum.

Explanation This is an example of a mathematical sleight of hand. If the terms of S are rearranged in such a way that every term remains in its original position, one finds that S is either the infinite series {displaystyle S=1-1+1-1+...+1-1+1-1} or with equal possibility, that {displaystyle S=1-1+1-1+...+1-1+1} Evaluating S as before, by grouping every -1 with the +1 preceding it or by grouping every +1 except the first with the -1 preceding it, gives in the first case: {displaystyle S_{1}=(1-1)+....+(1-1)=0+....+0=0} {displaystyle S_{2}=1+(-1+1)+....+(-1+1)-1=1+0+....+0-1=1-1=0} and in the second case: {displaystyle S_{1}=(1-1)+....+(1-1)+1=0+....+0+1=1} {displaystyle S_{2}=1+(-1+1)+....+(-1+1)=1+0+...+0=1} This reveals the trick: the definition of S was interpreted as defining its last term as negative when evaluating {displaystyle S_{1}=0} but positive when evaluating {displaystyle S_{2}=1} when in fact the definition of S didn't define (and the rearrangement was independent of) either option.

Definition for real and complex numbers 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.

Sums of more general elements 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 |:Gto mathbb {R} _{+}} on an abelian group {displaystyle G} (written additively, with identity element 0) such that: The norm of the identity element of {displaystyle G} is zero: {displaystyle |0|=0.} For every {displaystyle xin G,} {displaystyle |x|=0} implies {displaystyle x=0.} For every {displaystyle xin G,} {displaystyle |-x|=|x|.} For every {displaystyle x,yin 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}|0,} there exists {displaystyle N} such that {textstyle left|sum _{i=m}^{n}left|a_{i}right|right|=sum _{i=m}^{n}|a_{i}|mgeq 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 left|sum _{i=m}^{n}a_{i}right|mgeq 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 {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}|0,} we can choose some {displaystyle kappa _{varepsilon },lambda _{varepsilon }in mathbb {N} ,} such that: {displaystyle {begin{aligned}{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}-Aright|<{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,ldots ,N_{varepsilon }right}right)right}end{aligned}}} where {displaystyle sigma ^{-1}left(left{1,ldots ,N_{varepsilon }right}right)=left{sigma ^{-1}(1),ldots ,sigma ^{-1}left(N_{varepsilon }right)right}} so that {displaystyle M_{sigma ,varepsilon }} is the smallest natural number such that the list {displaystyle a_{sigma (0)},ldots ,a_{sigma left(M_{sigma ,varepsilon }right)}} includes all of the terms {displaystyle a_{0},ldots ,a_{N_{varepsilon }}} (and possibly others). Finally for any integer {displaystyle N>M_{sigma ,varepsilon }} let {displaystyle {begin{aligned}I_{sigma ,varepsilon }&=left{1,ldots ,Nright}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) : kin I_{sigma ,varepsilon }right}\L_{sigma ,varepsilon }&=max sigma left(I_{sigma ,varepsilon }right)=max left{sigma (k) : kin I_{sigma ,varepsilon }right}\end{aligned}}} so that {displaystyle {begin{aligned}left|sum _{iin I_{sigma ,varepsilon }}a_{sigma (i)}right|&leq sum _{iin I_{sigma ,varepsilon }}left|a_{sigma (i)}right|\&leq sum _{j=S_{sigma ,varepsilon }}^{L_{sigma ,varepsilon }}left|a_{j}right|&&{text{ since }}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{aligned}}} and thus {displaystyle {begin{aligned}left|sum _{i=1}^{N}a_{sigma (i)}-Aright|&=left|sum _{iin sigma ^{-1}left({1,dots ,N_{varepsilon }}right)}a_{sigma (i)}-A+sum _{iin I_{sigma ,varepsilon }}a_{sigma (i)}right|\&leq left|sum _{j=1}^{N_{varepsilon }}a_{j}-Aright|+left|sum _{iin I_{sigma ,varepsilon }}a_{sigma (i)}right|\&0,{text{ there exists }}M_{sigma ,varepsilon },{text{ for all }}N>M_{sigma ,varepsilon }quad left|sum _{i=1}^{N}a_{sigma (i)}-Aright|

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