if, for every ǫ > 0, no matter how small, there exists a number N for which

Transcription

1 Chapter 2 Ifiite Series 2. Sequeces A sequece of complex umbers {z } is a coutably ifiite set of umbers, z, z 2, z 3,...,z,... (2.) That is, for every positive iteger k, there is a umber, the kth term of the sequece, z k, i the set {z }. Mathematically, a sequece is a complexvalued fuctio defied o the positive itegers. We say that the sequece possesses a limit l, lim z = l, or z l as, (2.2) if, for every ǫ > 0, o matter how small, there exists a umber N for which z l < ǫ for all > N. (2.3) (The umber N will deped o ǫ.) That is, {z } =N+ all lie withi a circle of radius ǫ cetered o the poit l i the complex plae. A ecessary ad sufficiet coditio for a sequece {z } to coverge to a limit is Cauchy s criterio: A sequece {z } possesses a limit if ad oly if for every ǫ > 0, o matter how small, it is possible to fid a umber N such that z z m < ǫ for all,m > N. (2.4) (Note that the differece m may be arbitrarily large.) Thus, all elemets of the sequece {z } =N+ lie withi a disk of radius ǫ. Briefly, we say that the Cauchy coditio is z z m 0 for all, m sufficietly large. (2.5) Sequeces havig this property are called Cauchy sequeces. Every Cauchy sequece of complex umbers possesses a limit (which is, of course, a complex umber) this property meas that the complex umbers form a complete space. 7 Versio of August 24, 206

2 8 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES 2.2 Series Suppose we have a sequece {a k } k= from which we costruct the fiite sums s = a k, =, 2, 3,... (2.6) k= Thesetofallthesesums, {s }, itselfformsasequece. Ifthislattersequece has a limit S, s S as, (2.7) the we say that the ifiite series a k = lim k= k= a k (2.8) possesses the limit S (or coverges to S), a k = S. (2.9) k= By the Cauchy criterio, this will be true if ad oly if a k < ǫ (2.0) k=m for ay fixed ǫ > 0, wheever m > N, N a umber depedig o ǫ. Obviously, a ecessary coditio for a k (2.) k= to coverge is for a k 0 as k. However, this is ot sufficiet, as the followig example shows. 2.3 Examples 2.3. Harmoic series Cosider the sum of the reciprocals of the itegers, =. (2.2) Note that if the th term of the series is deoted a = /, we have for the sum of adjacet terms a a 2 = ( ) 2 > = 2 2, (2.3) o matter how large is. This violates Cauchy s criterio, so the harmoic series diverges.

3 2.4. ABSOLUTE AND CONDITIONAL CONVERGENCE9 Versio of August 24, Geometric series Cosider the series ar m, (2.4) m=0 where a is a costat ad r 0. For r, the th partial sum is so s = while the series diverges if r. m=0 S = lim s = a r ar m = a r+, (2.5) r if r <, (2.6) 2.4 Absolute ad Coditioal Covergece Suppose we have a coverget series a. If also a coverges, we say that the origial series coverges absolutely. Otherwise, the origial series is coditioally coverget. (That is, it coverges because of sig alteratios.) A sufficiet coditio for (at least) coditioal covergece is provided by the followig theorem due to Leibitz: If the terms of a series are of alteratig sig ad i additio their absolute values ted to zero, a 0, mootoically, i.e., a > a + for sufficietly large, the a coverges. (2.7) I absolutely coverget series oe ca rearrage the terms without affectig the value of the sum. With coditioally coverget series, oe caot rearrage terms; i fact, such rearragemets ca make a coditioally coverget series coverge to ay desired value, or to diverge! 2.4. Example Cosider the coditioally coverget series formed from the diverget harmoic series by alteratig every other sig: = l2, (2.8) 6 which coverges to the atural logarithm of 2. Multiply this equatio term by term by /2: = l2. (2.9) 2

4 0 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES Add these two series: = 3 l2. (2.20) 2 Sice the reciprocal of each iteger occurs exactly oce i the last series, we would be tempted to rearrage the series to obtai = l2, (2.2) 6 which is idetical to the origial series. There is a obvious cotradictio here! I order to obtai the rearragemet (2.2), we have to go further ad further out i the series (2.20), which apparetly is ot permissible A Theorem About Absolutely Coverget Series Not oly ca absolutely coverget series be rearraged without chagig their value, but they ca be multiplied together term by term: If two series S = u i, (2.22a) T = i= i= v i (2.22b) are both absolutely coverget, the series P = u i v j (2.23) i= j= formed from the product of their terms writte i ay order, is absolutely coverget, ad has a value equal to the product of of the idividual series, 2.5 Covergece Tests P = ST. (2.24) The followig tests ca determie whether a give series is absolutely coverget or ot Compariso test If b > 0 for all ad b is coverget, ad if a b for all, the a is absolutely coverget. (2.25a) Also, if a b > 0 for all, ad b diverges, the a is ot absolutely coverget. (2.25b)

5 2.5. CONVERGENCE TESTS Versio of August 24, Root test The series a coverges absolutely if from a certai term oward a q <, (2.26) where q 0 is idepedet of. Proof: If the iequality holds, a q. But q coverges for q <, it beig the geometric series, so by 2.5., a coverges Ratio test The series a coverges absolutely if from a certai term oward a + a q <, (2.27) where q 0 is idepedet of. Proof: Without loss of geerality, we may assume the iequality holds for all ; otherwise, we reumber the {a } sequece so that labels the first term for which the iequality (2.27) holds. The a = a a a 2 a 2 q. (2.28) a a a 2 a 3 Covergece is agai assured by compariso with the geometric series. (Whether these tests are satisfied by the first few terms of a series is immaterial, sice a fiite umber of terms of a ifiite seris has o effect o the covergece.) Example Whe does q coverge? If we use the root test, we examie a = q lim = q, lim a (2.29a) while if we use the ratio test, we look at lim a + + a = q lim = q. (2.29b) I either case, we see that the series is absolutely coverget if q <, ad diverget otherwise. The followig are refiemets of the ratio test, which fails (that is, fails to reveal whether the tested series coverges or ot) whe lim a + a =. (2.30) For example, this idetermiate limit results for the case a = /, which yields a diverget series, but also for a = /(l 2 ), which correspods to a coverget sum (see Sec ). Because l = l, which teds to zero as,.

6 2 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES Kummer s test Choose a sequece of positive costats b. If a b b + C > 0, (2.3) a + for all N, where N ad C are fixed umbers, the O the other had, if ad the a coverges absolutely. (2.32) b a a + b + 0, (2.33) b diverges, (2.34) a diverges. (2.35) Proof: If the iequality (2.3) holds, take l N, so that So we have the iequality l=n+ C a l+ b l a l b l+ a l+. (2.36) a l b N a N C Hece, the th partial sum, for > N, is s = i= b a C b N a N C. (2.37) N a i a i + b N a N C. (2.38) i= The right-had side of this iequality is a costat, idepedet of. Therefore, the positive sequece of icreasig terms {s } is bouded above, ad cosequetly possesses a limit. The series is absolutely coverget. If the iequality (2.33) holds, so sice b diverges, so does a. a a N b N b, > N, (2.39)

7 2.5. CONVERGENCE TESTS 3 Versio of August 24, Raabe s test Raabe s criterio for absolute covergece is ( ) a K >, (2.40) a + for all N, where N ad K are fixed. Ad if ( ) a, (2.4) the a + a diverges. (2.42) Proof: I Kummer s test put b = Gauss test If a a + = + h + B() 2, (2.43) where h is a costat ad the fuctio B() is bouded as, the a coverges for h > ad diverges for h. Proof: For h we ca use Raabe s test: ( h lim + B() ) 2 = h. (2.44) For h =, Raabe s test is idetermiate. I that case use Kummer s test with b = l: for large, ( l + h + B() ) (+)l(+) Because l 2 ( + h + B() 2 ) (+) ( h+ B() ) l l ( l+ (h )l < 0, if h. (2.45) =2 l (see homework), the series a diverges. diverges (2.46) )

8 4 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES f Figure 2.: Bouds o a mootoe series provided by a itegral Itegral test If f(x) is a cotiuous, mootoically decreasig real fuctio of x such that the a coverges if ad diverges otherwise. Proof: It is geometrically obvious that dxf(x) < f() = a, (2.47) f() < dxf(x) <, (2.48) dxf(x)+f(), (2.49) for this follows merely from the geometrical meaig of the itegral as the area uder the curve of the fuctio. See Fig Examples The Riema zeta fuctio is defied by the series = ζ(α). (2.50) α We ca test for covergece usig Gauss test, by examiig ( + ) α + α for large. (2.5) Thus the series coverges if α >, ad diverges if α. Cosider the series (l) α. (2.52)

9 2.6. SERIES OF FUNCTIONS 5 Versio of August 24, 206 Let s use Raabe s test: ( ) α ( ) α l(+) l+l(+/) = + α l l l. (2.53) Because ( α ) = α 0 as, (2.54) l l we coclude that the series is diverget. To test for covergece of let us use the itegral test: =2 (l) α, (2.55) 2 dx x(lx) α = l2 = = d(l x) (lx) α α (lx) α x=2, α, l(lx) x=2, α = { fiite α >, α. (2.56) Thus the series coverges if α > ad diverges for other real α. 2.6 Series of Fuctios 2.6. Cotiuity A (complex-valued) fuctio f(z) of a complex variable is cotiuous at z 0 if f(z) f(z 0 ) as z z 0 (2.57) from ay directio. That is, give ǫ > 0 we may fid a δ > 0 such that f(z) f(z 0 ) < ǫ wheever z z 0 < δ. (2.58) I other words, z lies withi a circle of radius δ aroud z Uiform Covergece Cosider the ifiite series f(z) = g i (z) (2.59) i=

10 6 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES f x 2ǫ = f = f Figure 2.2: Uiform covergece of the partial sum f (x) to the limit f(x). For all x, f (x) is withi a bad of width 2ǫ about f(x). costructed from the sequece of fuctios {g i } i=. The coditio that this series coverge is expressed i terms of the partial sums, f (z) = g i (z) (2.60) thusly: give ǫ > 0 we ca fid a iteger N so that for > N i= f +p (z) f (z) < ǫ for all p > 0. (2.6) ThisisCauchy scriterio. IgeeraltheN requiredforthistooccurwilldeped o the poit z. If, however, Eq. (2.6) holds for all z if > N idepedet of z, we say that the series coverges uiformly throughout the regio of iterest. Equivaletly, there exists a fuctio f(z) such that f(z) f (z) < ǫ for all > N, N idepedet of z. (2.62) That is, the partial sum f is everywhere uiformly close to f, the limitig fuctio. This situatio is illustrated i Fig. 2.2 for a real fuctio of a real variable. Cotrast absolute ad uiform covergece through the followig examples. The series ( ) +z 2 (2.63) isolycoditioallycoverget,becauseasymptoticallythetermsbecome( ) /. O the other had, for real z it is uiformly coverget because N+p ( ) +z 2 < N +z 2 N, (2.64) =N+ which is the Cauchy criterio with ǫ = /N. I cotrast, cosider, for real z, the series z 2 S(z) = (+z 2 ) (2.65)

11 2.6. SERIES OF FUNCTIONS 7 Versio of August 24, 206 which coverges absolutely. For z = 0, S(0) = 0; ad for z 0, S(z) = z 2 (+z 2 ) = z 2 = +z 2. (2.66) +z 2 Thus S(z) is discotiuous at z = 0. The followig theorem shows that this series caot be uiformly coverget there. Theorem If a series of cotiuous fuctios of z is uiformly coverget for all values of z i a give closed domai, the sum is cotiuous throughout the domai. Proof: Let f (z) = g i (z). (2.67) Sice i= we ca fid, for ay ǫ > 0, a value of such that throughout the domai. The f (z) f(z) uiformly, (2.68) f (z) f(z) < ǫ for all z (2.69) f(z) f(z ) = f(z) f (z)+f (z) f(z )+f (z ) f (z ) f(z) f (z) + f(z ) f (z ) + f (z) f (z ).(2.70) Sice the f s are cotiuous, we ca fid a δ for ay give ǫ such that Therefore, f (z) f (z ) < ǫ wheever z z < δ. (2.7) f(z) f(z ) < 3ǫ wheever z z < δ. (2.72) QED. Eve if the limit fuctio is cotiuous, covergece to it eed ot be uiform, as the followig example shows: Example Cosider the sequece of cotiuous fuctios, x, 0 x /, f (x) = (2/ x), / x 2/, 0, otherwise. (2.73) This fuctio i sketched i Fig Note that the maximum of the fuctio

12 8 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES f 0 / 2/ x Figure 2.3: Sketch of the fuctio f (x) give by Eq. 2.73). f (x) is. O the other had, for all x, lim f (x) = 0, (2.74) which is certaily a cotiuous limit fuctio. But the covergece to this limit is ot uiform, for there is always a poit, x = /, for which ( ) 0 f = (2.75) o matter how large is. So the covergece is ouiform. Properties of Uiformly Coverget Series Cosider the series of fuctios of a real variable, f(x) = g (x). (2.76). If the g are cotiuous, we ca itegrate term by term if g is uiformly coverget over the domai of itegratio: b a dxf(x) = b a dxg (x). (2.77) 2. If the g ad g = d dx g are cotiuous, ad g is uiformly coverget, the we ca differetiate term by term: f (x) = Coditio for Uiform Covergece g (x). (2.78) The followig coditio is sufficiet, but ot ecessary, to esure that a series is uiformly coverget.

13 2.7. POWER SERIES 9 Versio of August 24, 206 If g (z) < a, where {a } is a sequece of costats such that a coverges, the g (z) coverges uiformly ad absolutely. Proof: The hypothesis implies N+p N+p g (z) < a, (2.79) =N so that if N is chose so that N+p =N a < ǫ, the N+p g (z) < ǫ z. (2.80) 2.7 Power Series =N =N By a power series, we mea a series of the form, c z = c 0 +c z +c 2 z , (2.8) wherethe c sformasequeceofcomplex costats, adz is acomplexvariable. If a power series coverges for oe poit, z = z 0, it coverges uiformly ad absolutely for all z satisfyig z η, (2.82) where η is ay positive umber less tha z 0. Proof: Sice c z0 coverges, it must be true that the terms are bouded, c z0 < M, (2.83) where M is idepedet of (but ot of z 0 ). Hece if Eq. (2.82) is satisfied, ( ) η c z c η < M <, (2.84) z 0 siceη/ z 0 <. Thisprovesabsolutecovergece. Uiformcovergecefollows from the theorem above Radius of Covergece Use the root test to determie where the power series coverges. That test says if c z (2.85) lim c z <, the series coverges, (2.86a)

14 20 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES z-plae ρ Figure 2.4: Circle of covergece of a power series. The series (2.85) coverges iside the circle, ad diverges outside. The radius of covergece ρ is give by Eq. (2.87). while if lim c z >, the series diverges. (2.86b) Therefore, the power series coverges withi a circle of covergece of radius ρ, the radius of covergece, where ρ = lim c, (2.87) ad diverges outside that circle, as show i Fig More detailed examiatio is required to determie whether or ot the series coverges o the circle of covergece Properties of Power Series Withi the Circle of Covergece. The fuctio defied by the power series is cotiuous. [This follows from the theorem i Sec ] 2. It may be differetiated or itegrated term by term. [This follows from the theorem above, togther with the fact that if c z coverges, so does c z, by the ratio test, (+)c + z c z = + c + c z. (2.88) Now if z lies withi the circle of covergece, lim c + c z <. (2.89) Sice lim ( + )/ =, covergece of the differetiated series is assured.]

15 2.7. POWER SERIES 2 Versio of August 24, Two such power series may be multiplied together term by term, withi the smaller of the two circles of covergece. [This follows from the theorem i Sec ] 4. The power series is uique. [It suffices to show that if Ideed, f(z) = c z = 0 z, c = 0. (2.90) f(0) = c 0 = 0, f (0) = c = 0,..., f () (0) =!c = 0.] (2.9a) (2.9b) (2.9c) Taylor Expasio The Taylor expasio for a real fuctio of a real variable is obtaied from the above argumet. If we write a fuctio as a power series, the f(x) = c x, (2.92) c =! f() (0). (2.93) Hece, the power series is the Taylor series of the fuctio it represets, f(x) =! f() (0)x. (2.94) Hypergeometric Fuctio The hypergeometric fuctio F is defied by the power series F(a,b;c;z) = A z = (a) (b) (c) z!. (2.95) Here the coefficiets are defied i terms of the Pochhammer symbol, (a) = a(a+)(a+2) (a+ ) = Γ(a+). (2.96) Γ(a)

16 22 Versio of August 24, 206 CHAPTER 2. INFINITE SERIES To determie covergece, we examie A z Γ(a+)Γ(b+) = A + z+ Γ(c+) ( )( ) + c + = ( )( + a + b = z Γ(c++) Γ(a++)Γ(b++) (+)!! z z + ) z [+ ( )] (c+ a b)+o 2, (2.97) where O(/ 2 ) meas that the ext term goes to zero as at least as fast as / 2. Accordig to the ratio test, the radius of covergece of this series is z = ; that is, the series diverges for z >, ad coverges uiformly ad absolutely for ay z such that z η <. The remaiig questio is what happes o the circle of covergece, z =. Accordig to Gauss test, the series is the absolutely coverget if c > a +b [if the costats are complex, if Re(c a b) > 0]. For the poit z = the series is certaily diverget if this coditio is ot satisfied; however, if < Re(c a b) 0 the series is coditioally coverget o the uit circle except for the exceptioal poit z =. O the other had, if Re(c a b) the series is diverget o the uit circle because the terms i the series icrease i magitude.