DOCSLIB.ORG

61. Rearrangements Here the Difference Between Conditionally and Absolutely Convergent Series Is Further Refined Through The

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
61. Rearrangements Here the Difference Between Conditionally and Absolutely Convergent Series Is Further Refined Through The 61. REARRANGEMENTS 119 61. Rearrangements Here the difference between conditionally and absolutely convergent series is further refined through the concept of rearrangement. Definition 15. (Rearrangement) Let k , n = 1, 2,..., be an integer-valued positive sequence in which { n} every positive integer appears only once (i.e. kn = kn0 if and only if n = n0). Given a series an, put an0 = akn . The series an0 is called a rearrangement of an. P P For a finite sum, aP rearrangement of its terms does not change the value of the sum. This is not generally so for convergent series. Consider an alternating p series − n 1 ∞ ∞ ( 1) − 1 1 1 1 1 (57) a = − = 1 + + + n n − 2 3 − 4 5 − 6 ··· n=1 n=1 X X The series is convergent but not absolutely convergent (its sum is s = ln 2; see Problem 18 in Section 1.6). One of its rearrangements reads ∞ 1 1 1 1 1 1 1 1 (58) a0 =1+ + + + + + n 3 − 2 5 7 − 4 9 11 − 6 ··· n=1 X in which two positive terms are always followed by one negative. Let sn and sn0 be partial sums of (57) and (58), respectively. Put hn = 1 + 1/2 + + 1/n (a partial sum of the harmonic series). Then s = h ···h . Furthermore, 2n 2n − n 1 1 1 1 1 1 1 s0 = 1+ + + + + 3n 3 5 7 ··· 4n 1 − 2 − 4 −···− 2n 1 1 1 −1 1 1 = h h = h h h 4n − 2 − 4 −···− 4n − 2 n 4n − 2 2n − 2 n 1 1 = (h h )+ (h h )= s + s 4n − 2n 2 2n − n 4n 2 2n Taking the limit n in this equality, one finds s0 = s + s/2 = 3s/2 →∞ where s and s0 are the sums of (57) and (58), respectively. Thus, a rearrangement of the series has changed its sum! This fact is not specific for the example considered but inherent to all conditionally convergent series. Terms of a conditionally convergent series occur with different signs (positive and negative). By regrouping positive and negative terms, it will be proved that the sum of a conditionally convergent series can be made any number or . The analysis begins with studying the properties of sums of positive±∞ and negative terms of a conditionally convergent series. 120 9. SEQUENCES AND SERIES + Given a number x, put x± = (x x )/2). The number x = x if + ±| | x > 0 and x = 0 otherwise. Similarly, x− = x if x < 0 and x− = 0 otherwise. Lemma . + 1 Given a series an, consider two series an and an− where an± = (an an )/2 (the series of positive and negative terms). Then ±| | P P P + (i) if an converges absolutely, then an and an− converge. + (ii) if an is conditionally convergent, then a and a− diverge. P P Pn n Proof: Let a = s < and a = t where t < if a P n ∞ | n| P P ∞ n converges absolutely and t = if it is conditionally convergent. Let s± P ∞ P P n be partial sums of an±, sn be partial sums of an and tn be partial sums of a . Since s s and t t as n , one infers that | n| n → n → →∞ + P + P + an an− = an sn sn− = sn s s− = s + −P = + − = + − a + a− = an ⇒ s + s− = tn ⇒ s + s− = t n n | | n n where s± are the limits of s±. If a converges absolutely, then t< n n ∞ and, hence, s± = (t s)/2, that is, both the series a± converge. If ± P n an is conditionally convergent, then t = , and both the sequences ∞ P s± diverge. Pn Theorem 43. (Riemann’s rearrangement theorem) Let an be a series which converges, but not absolutely. Then for any c which is a real number or there exists a rearrangement a0 P ±∞ n whose sequence of partial sums s0 converges to c. { n} P Proof: Let p1, p2,.... denote non-negative terms of an in the order in which they occur, and let q1,q2,... denote negative terms of P an in the order in which they occur. In notations of Lemma 1, the + series pn and a as well as qn and a− may only differ by P n n zero terms (if some an = 0). So, the series pn and qn diverge. ConsiderP the followingP rearrangement.P GivenP a number c, take first P P k1 terms pn so that the number c lies between the partial sums sk1 = p1 + p2 + + pk1 and sk1 1, that is, k1 is defined by the condition s p <c<s··· or c s− < p . If c< 0, then skip this first step. k1 − k1 k1 | − k1 | k1 Next, take first m1 terms qn where m1 is the smallest integer such that sk1+m1 = sk1 + q1 + + qm1 <c, that is, m1 is defined by the conition s + q >c>s··· or c s < q . This can always be k1+m1 m1 k1+m1 | − k1+m1 | | m1 | done because of partial sums of pn can be larger than any number, while partial sums of qn can be smaller than any number thanks to the divergence of these series. So,P P s s s , 1 n k where c s < p 1 ≤ n ≤ k1 ≤ ≤ 1 | − k1 | k1 s s s , k n k + m where c s < q k1 ≥ n ≥ k1+m1 1 ≤ ≤ 1 1 | − k1+m1 | | k1 | 61. REARRANGEMENTS 121 Next, take k2 next terms of pn where k2 is the smallest integer such that sk1+m1+k2 > c, and take m2 next term of qn where m2 is the smallest integer for which sk1+m1+k1+m2 < c and so on. At the nth step of the procedure, let n1 be the integer for which the last term in sn1 is pkn and let n2 be the integer for which the last term in sn2 is qmn , i.e. n2 = n1 + mn. The partial sums of the constructed rearrangement oscillate about c reaching local minima sn1 and local maxima sn1 : s s s , n n n n1 ≤ n ≤ n2 1 ≤ ≤ 2 (59) c s < p , c s < q | − n1 | kn | − n2 | | mn | By convergence of the series a , a 0 as n 0. Hence, p and q n n → → n n also converge to zero and so do the subsequences pkn 0 and qmn 0 Thus, all local maxima and minimaP of the sequence of→ partial sum →s { n} converge to c by (59) which shows that sn c. Finally, if c = , one can take any divergent sequence c → (or ) and construct±∞ n → ∞ −∞ a rearrangement such that sk1 overshoots c1, sk1+m1 undershoots c1, then sk1+m1+k2 overshoots c2 and sk1+m1+k2+m2 undershoots c2 and so on. Obviously, this sequence of partial sum diverges. Absolutely convergent series have a drastically different property. Theorem 44. (Rearrangement and absolute convergence) If a series an converges absolutely, then every rearrangement of an converges, and they all converge to the same sum P P Proof: Let t = a + a + + a is a partial sum of the n | 1| | 2| ··· | n| series of absolute values. The sequence tn converges to a number t by the hypothesis, that is, for any ε > {0 there} is an integer N such that t t <ε for all n > N. Therefore | − n| n a = t t = t t + t t | k| | n − N+1| | n − − N+1| k=XN+1 t t + t t < 2ε ≤ | n − | | − N+1| So, by taking N large enough the sum of any number of terms ak , k > N, can be made smaller than any preassigned positive number.| | Let sn and sn0 be partial sums of an and its rearrangement an0 . One can take n > N large enough so that s0 contains a1, a2,...,aN (i.e. the P n P integers 1, 2,...,N are in the set of integers k1, k2,...,kn in notations of Definition 15). Then the difference sn0 sn contains only terms a with k > N (the terms a , a ,..., |a −are cancelled).| Therefore | k| 1 2 N s0 s < 2ε for all n > N. If s0 s0 and s s, then s0 s < 2ε | n − n| n → n → | − | which shows that s0 = s because ε> 0 is arbitrary. 122 9. SEQUENCES AND SERIES Thus, an absolutely convergent series is much like a finite sum. The sum does not depend on the order in which the summation is carried out. In contrast, the sum of a conditionally convergent series depends on the summation order. This is the characteristic difference between these two classes of convergent series. 61.1. Strategy for testing series. It would not be wise to apply test for convergence in a specific order to find one that finally works. Instead, a proper strategy, as with integration, is to classify the series according to its form. One should also keep in mind that a conclusion about the convergence of a series can be reached in different ways. 1. Special series. A series an coincide with (or is a combination of or is equivalent to) special series such as a p series, alternating p series, geometric series, telescopicP series, etc.− Their convergence properties− are known. 2. Series similar to special ones. A series an has a form that is similar to one of the special series, then one of the comparison P tests should be considered. For example, if an is a rational or alge- braic function (contains roots of polynomials), then the series should be compared with a p series. − 3. Necessary condition for convergence. The condition an 0 as n is always easier to check than to investigate the series→ → ∞ an for convergence.
Load more

Recommended publications
  • Ch. 15 Power Series, Taylor Series
    Ch. 15 Power Series, Taylor Series 서울대학교 조선해양공학과 서유택 2017.12 ※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다. Seoul National 1 Univ. 15.1 Sequences (수열), Series (급수), Convergence Tests (수렴판정) Sequences: Obtained by assigning to each positive integer n a number zn z . Term: zn z1, z 2, or z 1, z 2 , or briefly zn N . Real sequence (실수열): Sequence whose terms are real Convergence . Convergent sequence (수렴수열): Sequence that has a limit c limznn c or simply z c n . For every ε > 0, we can find N such that Convergent complex sequence |zn c | for all n N → all terms zn with n > N lie in the open disk of radius ε and center c. Divergent sequence (발산수열): Sequence that does not converge. Seoul National 2 Univ. 15.1 Sequences, Series, Convergence Tests Convergence . Convergent sequence: Sequence that has a limit c Ex. 1 Convergent and Divergent Sequences iin 11 Sequence i , , , , is convergent with limit 0. n 2 3 4 limznn c or simply z c n Sequence i n i , 1, i, 1, is divergent. n Sequence {zn} with zn = (1 + i ) is divergent. Seoul National 3 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 1 Sequences of the Real and the Imaginary Parts . A sequence z1, z2, z3, … of complex numbers zn = xn + iyn converges to c = a + ib . if and only if the sequence of the real parts x1, x2, … converges to a . and the sequence of the imaginary parts y1, y2, … converges to b. Ex.
    [Show full text]
  • Arxiv:1207.1472V2 [Math.CV]
    SOME SIMPLIFICATIONS IN THE PRESENTATIONS OF COMPLEX POWER SERIES AND UNORDERED SUMS OSWALDO RIO BRANCO DE OLIVEIRA Abstract. This text provides very easy and short proofs of some basic prop- erties of complex power series (addition, subtraction, multiplication, division, rearrangement, composition, differentiation, uniqueness, Taylor’s series, Prin- ciple of Identity, Principle of Isolated Zeros, and Binomial Series). This is done by simplifying the usual presentation of unordered sums of a (countable) family of complex numbers. All the proofs avoid formal power series, double series, iterated series, partial series, asymptotic arguments, complex integra- tion theory, and uniform continuity. The use of function continuity as well as epsilons and deltas is kept to a mininum. Mathematics Subject Classification: 30B10, 40B05, 40C15, 40-01, 97I30, 97I80 Key words and phrases: Power Series, Multiple Sequences, Series, Summability, Complex Analysis, Functions of a Complex Variable. Contents 1. Introduction 1 2. Preliminaries 2 3. Absolutely Convergent Series and Commutativity 3 4. Unordered Countable Sums and Commutativity 5 5. Unordered Countable Sums and Associativity. 9 6. Sum of a Double Sequence and The Cauchy Product 10 7. Power Series - Algebraic Properties 11 8. Power Series - Analytic Properties 14 References 17 arXiv:1207.1472v2 [math.CV] 27 Jul 2012 1. Introduction The objective of this work is to provide a simplification of the theory of un- ordered sums of a family of complex numbers (in particular, for a countable family of complex numbers) as well as very easy proofs of basic operations and properties concerning complex power series, such as addition, scalar multiplication, multipli- cation, division, rearrangement, composition, differentiation (see Apostol [2] and Vyborny [21]), Taylor’s formula, principle of isolated zeros, uniqueness, principle of identity, and binomial series.
    [Show full text]
  • INFINITE SERIES in P-ADIC FIELDS
    INFINITE SERIES IN p-ADIC FIELDS KEITH CONRAD 1. Introduction One of the major topics in a course on real analysis is the representation of functions as power series X n anx ; n≥0 where the coefficients an are real numbers and the variable x belongs to R. In these notes we will develop the theory of power series over complete nonarchimedean fields. Let K be a field that is complete with respect to a nontrivial nonarchimedean absolute value j · j, such as Qp with its absolute value j · jp. We will look at power series over K (that is, with coefficients in K) and a variable x coming from K. Such series share some properties in common with real power series: (1) There is a radius of convergence R (a nonnegative real number, or 1), for which there is convergence at x 2 K when jxj < R and not when jxj > R. (2) A power series is uniformly continuous on each closed disc (of positive radius) where it converges. (3) A power series can be differentiated termwise in its disc of convergence. There are also some contrasts: (1) The convergence of a power series at its radius of convergence R does not exhibit mixed behavior: it is either convergent at all x with jxj = R or not convergent at P n all x with jxj = R, unlike n≥1 x =n on R, where R = 1 and there is convergence at x = −1 but not at x = 1. (2) A power series can be expanded around a new point in its disc of convergence and the new series has exactly the same disc of convergence as the original series.
    [Show full text]
  • Series: Convergence and Divergence Comparison Tests
    Series: Convergence and Divergence Here is a compilation of what we have done so far (up to the end of October) in terms of convergence and divergence. • Series that we know about: P∞ n Geometric Series: A geometric series is a series of the form n=0 ar . The series converges if |r| < 1 and 1 a1 diverges otherwise . If |r| < 1, the sum of the entire series is 1−r where a is the first term of the series and r is the common ratio. P∞ 1 2 p-Series Test: The series n=1 np converges if p1 and diverges otherwise . P∞ • Nth Term Test for Divergence: If limn→∞ an 6= 0, then the series n=1 an diverges. Note: If limn→∞ an = 0 we know nothing. It is possible that the series converges but it is possible that the series diverges. Comparison Tests: P∞ • Direct Comparison Test: If a series n=1 an has all positive terms, and all of its terms are eventually bigger than those in a series that is known to be divergent, then it is also divergent. The reverse is also true–if all the terms are eventually smaller than those of some convergent series, then the series is convergent. P P P That is, if an, bn and cn are all series with positive terms and an ≤ bn ≤ cn for all n sufficiently large, then P P if cn converges, then bn does as well P P if an diverges, then bn does as well. (This is a good test to use with rational functions.
    [Show full text]
  • Divergent and Conditionally Convergent Series Whose Product Is Absolutely Convergent
    DIVERGENTAND CONDITIONALLY CONVERGENT SERIES WHOSEPRODUCT IS ABSOLUTELYCONVERGENT* BY FLORIAN CAJOKI § 1. Introduction. It has been shown by Abel that, if the product : 71—« I(¥» + V«-i + ■•• + M„«o)> 71=0 of two conditionally convergent series : 71—0 71— 0 is convergent, it converges to the product of their sums. Tests of the conver- gence of the product of conditionally convergent series have been worked out by A. Pringsheim,| A. Voss,J and myself.§ There exist certain conditionally- convergent series which yield a convergent result when they are raised to a cer- tain positive integral power, but which yield a divergent result when they are raised to a higher power. Thus, 71 = «3 Z(-l)"+13, 7>=i n where r = 7/9, is a conditionally convergent series whose fourth power is con- vergent, but whose fifth power is divergent. || These instances of conditionally * Presented to the Society April 28, 1900. Received for publication April 28, 1900. fMathematische Annalen, vol. 21 (1883), p. 327 ; vol. 2(5 (1886), p. 157. %Mathematische Annalen, vol. 24 (1884), p. 42. § American Journal of Mathematics, vol. 15 (1893), p. 339 ; vol. 18 (1896), p. 195 ; Bulletin of the American Mathematical Society, (2) vol. 1 (1895), p. 180. || This may be a convenient place to point out a slight and obvious extension of the results which I have published in the American Journal of Mathematics, vol. 18, p. 201. It was proved there that the conditionally convergent series : V(_l)M-lI (0<r5|>). Sí n when raised by Cauchy's multiplication rule to a positive integral power g , is convergent whenever (î — l)/ï <C r ! out the power of the series is divergent, if (q— 1 )¡q > r.
    [Show full text]
  • Dirichlet’S Test
    4. CONDITIONALLY CONVERGENT SERIES 23 4. Conditionally convergent series Here basic tests capable of detecting conditional convergence are studied. 4.1. Dirichlet’s test. Theorem 4.1. (Dirichlet’s test) n Let a complex sequence {An}, An = k=1 ak, be bounded, and the real sequence {bn} is decreasing monotonically, b1 ≥ b2 ≥ b3 ≥··· , so that P limn→∞ bn = 0. Then the series anbn converges. A proof is based on the identity:P m m m m−1 anbn = (An − An−1)bn = Anbn − Anbn+1 n=k n=k n=k n=k−1 X X X X m−1 = An(bn − bn+1)+ Akbk − Am−1bm n=k X Since {An} is bounded, there is a number M such that |An|≤ M for all n. Therefore it follows from the above identity and non-negativity and monotonicity of bn that m m−1 anbn ≤ An(bn − bn+1) + |Akbk| + |Am−1bm| n=k n=k X X m−1 ≤ M (bn − bn+1)+ bk + bm n=k ! X ≤ 2Mbk By the hypothesis, bk → 0 as k → ∞. Therefore the right side of the above inequality can be made arbitrary small for all sufficiently large k and m ≥ k. By the Cauchy criterion the series in question converges. This completes the proof. Example. By the root test, the series ∞ zn n n=1 X converges absolutely in the disk |z| < 1 in the complex plane. Let us investigate the convergence on the boundary of the disk. Put z = eiθ 24 1. THE THEORY OF CONVERGENCE ikθ so that ak = e and bn = 1/n → 0 monotonically as n →∞.
    [Show full text]
  • Chapter 3: Infinite Series
    Chapter 3: Infinite series Introduction Definition: Let (an : n ∈ N) be a sequence. For each n ∈ N, we define the nth partial sum of (an : n ∈ N) to be: sn := a1 + a2 + . an. By the series generated by (an : n ∈ N) we mean the sequence (sn : n ∈ N) of partial sums of (an : n ∈ N) (ie: a series is a sequence). We say that the series (sn : n ∈ N) generated by a sequence (an : n ∈ N) is convergent if the sequence (sn : n ∈ N) of partial sums converges. If the series (sn : n ∈ N) is convergent then we call its limit the sum of the series and we write: P∞ lim sn = an. In detail: n→∞ n=1 s1 = a1 s2 = a1 + a + 2 = s1 + a2 s3 = a + 1 + a + 2 + a3 = s2 + a3 ... = ... sn = a + 1 + ... + an = sn−1 + an. Definition: A series that is not convergent is called divergent, ie: each series is ei- P∞ ther convergent or divergent. It has become traditional to use the notation n=1 an to represent both the series (sn : n ∈ N) generated by the sequence (an : n ∈ N) and the sum limn→∞ sn. However, this ambiguity in the notation should not lead to any confu- sion, provided that it is always made clear that the convergence of the series must be established. Important: Although traditionally series have been (and still are) represented by expres- P∞ sions such as n=1 an they are, technically, sequences of partial sums. So if somebody comes up to you in a supermarket and asks you what a series is - you answer “it is a P∞ sequence of partial sums” not “it is an expression of the form: n=1 an.” n−1 Example: (1) For 0 < r < ∞, we may define the sequence (an : n ∈ N) by, an := r for each n ∈ N.
    [Show full text]
  • Approximating the Sum of a Convergent Series
    Approximating the Sum of a Convergent Series Larry Riddle Agnes Scott College Decatur, GA 30030 [email protected] The BC Calculus Course Description mentions how technology can be used to explore conver- gence and divergence of series, and lists various tests for convergence and divergence as topics to be covered. But no specific mention is made of actually estimating the sum of a series, and the only discussion of error bounds is for alternating series and the Lagrange error bound for Taylor polynomials. With just a little additional effort, however, students can easily approximate the sum of many common convergent series and determine how precise that approximation will be. Approximating the Sum of a Positive Series Here are two methods for estimating the sum of a positive series whose convergence has been established by the integral test or the ratio test. Some fairly weak additional requirements are made on the terms of the series. Proofs are given in the appendix. ∞ n P P Let S = an and let the nth partial sum be Sn = ak. n=1 k=1 1. Suppose an = f(n) where the graph of f is positive, decreasing, and concave up, and the R ∞ improper integral 1 f(x) dx converges. Then Z ∞ Z ∞ an+1 an+1 Sn + f(x) dx + < S < Sn + f(x) dx − . (1) n+1 2 n 2 (If the conditions for f only hold for x ≥ N, then inequality (1) would be valid for n ≥ N.) an+1 2. Suppose (an) is a positive decreasing sequence and lim = L < 1.
    [Show full text]
  • Calculus Online Textbook Chapter 10
    Contents CHAPTER 9 Polar Coordinates and Complex Numbers 9.1 Polar Coordinates 348 9.2 Polar Equations and Graphs 351 9.3 Slope, Length, and Area for Polar Curves 356 9.4 Complex Numbers 360 CHAPTER 10 Infinite Series 10.1 The Geometric Series 10.2 Convergence Tests: Positive Series 10.3 Convergence Tests: All Series 10.4 The Taylor Series for ex, sin x, and cos x 10.5 Power Series CHAPTER 11 Vectors and Matrices 11.1 Vectors and Dot Products 11.2 Planes and Projections 11.3 Cross Products and Determinants 11.4 Matrices and Linear Equations 11.5 Linear Algebra in Three Dimensions CHAPTER 12 Motion along a Curve 12.1 The Position Vector 446 12.2 Plane Motion: Projectiles and Cycloids 453 12.3 Tangent Vector and Normal Vector 459 12.4 Polar Coordinates and Planetary Motion 464 CHAPTER 13 Partial Derivatives 13.1 Surfaces and Level Curves 472 13.2 Partial Derivatives 475 13.3 Tangent Planes and Linear Approximations 480 13.4 Directional Derivatives and Gradients 490 13.5 The Chain Rule 497 13.6 Maxima, Minima, and Saddle Points 504 13.7 Constraints and Lagrange Multipliers 514 CHAPTER Infinite Series Infinite series can be a pleasure (sometimes). They throw a beautiful light on sin x and cos x. They give famous numbers like n and e. Usually they produce totally unknown functions-which might be good. But on the painful side is the fact that an infinite series has infinitely many terms. It is not easy to know the sum of those terms.
    [Show full text]
  • Divergence and Integral Tests Handout Reference: Brigg’S “Calculus: Early Transcendentals, Second Edition” Topics: Section 8.4: the Divergence and Integral Tests, P
    Lesson 17: Divergence and Integral Tests Handout Reference: Brigg's \Calculus: Early Transcendentals, Second Edition" Topics: Section 8.4: The Divergence and Integral Tests, p. 627 - 640 Theorem 8.8. p. 627 Divergence Test P If the infinite series ak converges, then lim ak = 0. k!1 P Equivalently, if lim ak 6= 0, then the infinite series ak diverges. k!1 Note: The divergence test cannot be used to conclude that a series converges. This test ONLY provides a quick mechanism to check for divergence. Definition. p. 628 Harmonic Series 1 1 1 1 1 1 The famous harmonic series is given by P = 1 + + + + + ··· k=1 k 2 3 4 5 Theorem 8.9. p. 630 Harmonic Series 1 1 The harmonic series P diverges, even though the terms of the series approach zero. k=1 k Theorem 8.10. p. 630 Integral Test Suppose the function f(x) satisfies the following three conditions for x ≥ 1: i. f(x) is continuous ii. f(x) is positive iii. f(x) is decreasing Suppose also that ak = f(k) for all k 2 N. Then 1 1 X Z ak and f(x)dx k=1 1 either both converge or both diverge. In the case of convergence, the value of the integral is NOT equal to the value of the series. Note: The Integral Test is used to determine whether a series converges or diverges. For this reason, adding or subtracting a finite number of terms in the series or changing the lower limit of the integration to another finite point does not change the outcome of the test.
    [Show full text]
  • 1 Convergence Tests
    Lecture 24Section 11.4 Absolute and Conditional Convergence; Alternating Series Jiwen He 1 Convergence Tests Basic Series that Converge or Diverge Basic Series that Converge X Geometric series: xk, if |x| < 1 X 1 p-series: , if p > 1 kp Basic Series that Diverge X Any series ak for which lim ak 6= 0 k→∞ X 1 p-series: , if p ≤ 1 kp Convergence Tests (1) Basic Test for Convergence P Keep in Mind that, if ak 9 0, then the series ak diverges; therefore there is no reason to apply any special convergence test. P k P k k Examples 1. x with |x| ≥ 1 (e.g, (−1) ) diverge since x 9 0. [1ex] k X k X 1 diverges since k → 1 6= 0. [1ex] 1 − diverges since k + 1 k+1 k 1 k −1 ak = 1 − k → e 6= 0. Convergence Tests (2) Comparison Tests Rational terms are most easily handled by basic comparison or limit comparison with p-series P 1/kp Basic Comparison Test 1 X 1 X 1 X k3 converges by comparison with converges 2k3 + 1 k3 k5 + 4k4 + 7 X 1 X 1 X 2 by comparison with converges by comparison with k2 k3 − k2 k3 X 1 X 1 X 1 diverges by comparison with diverges by 3k + 1 3(k + 1) ln(k + 6) X 1 comparison with k + 6 Limit Comparison Test X 1 X 1 X 3k2 + 2k + 1 converges by comparison with . diverges k3 − 1 √ k3 k3 + 1 X 3 X 5 k + 100 by comparison with √ √ converges by comparison with k 2k2 k − 9 k X 5 2k2 Convergence Tests (3) Root Test and Ratio Test The root test is used only if powers are involved.
    [Show full text]
  • Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012
    Series Convergence Tests Math 122 Calculus III D Joyce, Fall 2012 Some series converge, some diverge. Geometric series. We've already looked at these. We know when a geometric series 1 X converges and what it converges to. A geometric series arn converges when its ratio r lies n=0 a in the interval (−1; 1), and, when it does, it converges to the sum . 1 − r 1 X 1 The harmonic series. The standard harmonic series diverges to 1. Even though n n=1 1 1 its terms 1, 2 , 3 , . approach 0, the partial sums Sn approach infinity, so the series diverges. The main questions for a series. Question 1: given a series does it converge or diverge? Question 2: if it converges, what does it converge to? There are several tests that help with the first question and we'll look at those now. The term test. The only series that can converge are those whose terms approach 0. That 1 X is, if ak converges, then ak ! 0. k=1 Here's why. If the series converges, then the limit of the sequence of its partial sums n th X approaches the sum S, that is, Sn ! S where Sn is the n partial sum Sn = ak. Then k=1 lim an = lim (Sn − Sn−1) = lim Sn − lim Sn−1 = S − S = 0: n!1 n!1 n!1 n!1 The contrapositive of that statement gives a test which can tell us that some series diverge. Theorem 1 (The term test). If the terms of the series don't converge to 0, then the series diverges.
    [Show full text]