INFINITE SERIES - Elsevier.com

1 Chapter 1 INFINITE SERIESChap1 This on-line chapter contains the material on INFINITE SERIES , extracted from the printedversion of the Seventh Edition and presented in much the same organization in whichit appeared in the Sixth Edition. It is collected here for the convenience of instructorswho wish to use it as introductory material in place of that in the printed book. It hasbeen lightly edited to remove detailed discussions involving complex variable theorythat would not be appropriate until later in a course of instruction. For AdditionalReadings, see the printed INTRODUCTION TO INFINITE the most widely used technique in the physicist s toolbox is the use ofinfiniteseries( sums consisting formally of an INFINITE number of terms) to representfunctions, to bring them to forms facilitating further analysis, or even as a preludeto numerical evaluation.

2 The acquisition of skill in creating and manipulating seriesexpansions is therefore an absolutely essential part of the training of one who seekscompetence in the mathematical methods of physics, and it is therefore the first topicin this text. An important part of this skill set is the ability to recognize the functionsrepresented by commonly encountered expansions, and it is also of importance tounderstand issues related to the convergence of INFINITE CONCEPTSThe usual way of assigning a meaning to the sum of an INFINITE number of terms isby introducing the notion of partial sums. If we have an INFINITE sequence of termsu1,u2,u3,u4,u5,.. , we define thei-th partial sum assi=i n=1un.( ) is a finite summation and offers no difficulties.

3 If the partial sumssiconvergeto a finite limit asi ,limi si=S ,( ) INFINITE SERIES n=1unis said to beconvergentand to have the valueS. Notethat wedefinethe INFINITE SERIES as equal toSand that a necessary condition for con-12 CHAPTER 1. INFINITE SERIES vergence to a limit is that limn un= 0. This condition, however, is not sufficientto guarantee it is convenient to apply the condition in Eq. ( ) in a form called theCauchy criterion, namely that for each >0 there is a fixed numberNsuch that|sj si|< for alliandjgreater thanN. This means that the partial sums mustcluster together as we move far out in the seriesdiverge, meaning that the sequence of partial sums approaches ;others may have partial sums that oscillate between two values, as for example n=1un= 1 1 + 1 1 + 1 ( 1)n+.

4 This SERIES does not converge to a limit, and can be calledoscillatory. Often thetermdivergentis extended to include oscillatory SERIES as well. It is important to beable to determine whether, or under what conditions, a SERIES we would like to use The Geometric SeriesThe geometric SERIES , starting withu0= 1 and with a ratio of successive termsr=un+1/un, has the form1 +r+r2+r3+ +rn 1+ .Itsn-th partial sumsn(that of the firstnterms) is1sn=1 rn1 r.( ) attention to|r|<1, so that for largen,rnapproaches zero,snpossessesthe limitlimn sn=11 r,( ) that for|r|<1, the geometric SERIES converges. It clearly diverges (or isoscillatory) for|r| 1, as the individual terms do not then approach zero at largen. Example The Harmonic SeriesAs a second and more involved example, we consider the harmonic SERIES n=11n= 1 +12+13+14+ +1n+.

5 ( ) terms approach zero for largen, limn 1/n= 0, but this is not sufficientto guarantee convergence. If we group the terms (without changing their order) as1 +12+(13+14)+(15+16+17+18)+(19+ +116)+ ,1 Multiply and dividesn= n 1m=0rmby 1 INFINITE SERIES3each pair of parentheses enclosespterms of the form1p+ 1+1p+ 2+ +1p+p>p2p= partial sums by adding the parenthetical groups one by one, we obtains1= 1, s2=32, s3>42, s4>52, .. , sn>n+ 12,and we are forced to the conclusion that the harmonic SERIES the harmonic SERIES diverges, its partial sums have relevance amongother places in number theory, whereHn= nm=1m 1are sometimes referred to asharmonic numbers. We now turn to a more detailed study of the convergence and divergence of SERIES ,considering here SERIES of positive terms.

6 SERIES with terms of both signs are TESTIf term by term a SERIES of termsunsatisfies 0 un an, where theanform aconvergent SERIES , then the SERIES nunis also convergent. Lettingsiandsjbepartial sums of theuseries, withj > i, the differencesj siis jn=i+1un, andthis is smaller than the corresponding quantity for theaseries, thereby provingconvergence. A similar argument shows that if term by term a SERIES of termsvnsatisfies 0 bn vn, where thebnform a divergent SERIES , then nvnis the convergent seriesanwe already have the geometric SERIES , whereas theharmonic SERIES will serve as the divergent comparison seriesbn. As other SERIES areidentified as either convergent or divergent, they may also be used as the known seriesfor comparison A Divergent SeriesTest n=1n p,p= , for convergence.

7 Sincen > n 1andbn=n 1forms the divergent harmonic SERIES , the comparison test shows that nn Generalizing, nn pis seen to be divergent for allp 1. CAUCHY ROOT TESTIf (an)1/n r <1 for all sufficiently largen, withrindependent ofn, then nanisconvergent. If (an)1/n 1 for all sufficiently largen, then nanis language of this test emphasizes an important point: the convergence ordivergence of a SERIES depends entirely upon what happens for largen. Relative toconvergence, it is the behavior in the large-nlimit that first part of this test is verified easily by raising (an)1/nto thenth getan rn< 1. INFINITE SERIESS incernis just thenth term in a convergent geometric SERIES , nanis convergentby the comparison test. Conversely, if (an)1/n 1, thenan 1 and the SERIES mustdiverge.

8 This root test is particularly useful in establishing the properties of powerseries (Section ).D ALEMBERT (OR CAUCHY) RATIO TESTIfan+1/an r <1 for all sufficiently largenandris independent ofn, then nanis convergent. Ifan+1/an 1 for all sufficiently largen, then nanis test is established by direct comparison with the geometric SERIES (1+r+r2+ ). In the second part,an+1 anand divergence should be reasonably not quite as sensitive as the Cauchy root test, this D Alembert ratio testis one of the easiest to apply and is widely used. An alternate statement of the ratiotest is in the form of a limit: Iflimn an+1an <1,convergence,>1,divergence,= 1,indeterminate.( ) of this final indeterminate possibility, the ratio test is likely to fail at crucialpoints, and more delicate, sensitive tests then become necessary.

9 The alert readermay wonder how this indeterminacy arose. Actually it was concealed in the firststatement,an+1/an r <1. We might encounteran+1/an<1 for allfinitenbutbe unable to choose anr <1and independent of nsuch thatan+1/an rfor allsufficiently largen. An example is provided by the harmonic SERIES , for whichan+1an=nn+ 1< an+1an= 1,no fixed ratior <1 exists and the test D Alembert Ratio TestTest nn/2nfor convergence. Applying the ratio test,an+1an=(n+ 1)/2n+1n/2n=12n+ +1an 34forn 2,we have convergence. CAUCHY (OR MACLAURIN) INTEGRAL TESTThis is another sort of comparison test, in which we compare a SERIES with an , we compare the area of a SERIES of unit-width rectangles with the areaunder a INFINITE SERIES5 Figure : (a) Comparison of integral and sum-blocks leading.

10 (b) Comparison ofintegral and sum-blocks (x) be a continuous,monotonic decreasing functionin whichf(n) = nanconverges if 1f(x)dxis finite and diverges if the integral is partial sum issi=i n=1an=i n=1f(n).But, becausef(x) is monotonic decreasing, see Fig. (a),si i+11f(x)dx .On the other hand, as shown in Fig. (b),si a1 i1f(x)dx .Taking the limit asi , we have 1f(x)dx n=1an 1f(x)dx+a1.( ) the INFINITE SERIES converges or diverges as the corresponding integral convergesor integral test is particularly useful in setting upper and lower bounds on theremainder of a SERIES after some number of initial terms have been summed. That is, n=1an=N n=1an+ n=N+1an,( ) N+1f(x)dx n=N+1an N+1f(x)dx+aN+1.( ) 1. INFINITE SERIESTo free the integral test from the quite restrictive requirement that the interpo-lating functionf(x) be positive and monotonic, we shall show that for any functionf(x) with a continuous derivative, the INFINITE SERIES is exactly represented as a sumof two integrals:N2 n=N1+1f(n) = N2N1f(x)dx+ N2N1(x [x])f (x)dx.