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# Foundations of Real Analysis MATH 6101

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This 131 page Class Notes was uploaded by Felicita Lockman on Sunday October 25, 2015. The Class Notes belongs to MATH 6101 at University of North Carolina - Charlotte taught by David Royster in Fall. Since its upload, it has received 19 views. For similar materials see /class/228905/math-6101-university-of-north-carolina-charlotte in Mathematics (M) at University of North Carolina - Charlotte.

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Fall 2008 Given anv semmnce an we associate a new sequence Sn of partial sums s a1a2a3a4an We define the series Zan to be the limit Van lim 3 n gtoo Tl If the sequence of partial sums converges we say th t th infinit eri converges th rwi W say that the series is divergent r3 oar2008 MATH 6101 Eyelid alt1 n0 1 a 1 5n Zak 1aa2a3a kZA 1 a ples 00 1 2 1 n nn 1 In this case we have seen that 1 1 1 1 1 Sn 12 23 34 45 n 1n 11 ii 1 n 1 1 n 1 n Do these converge or diverge 12345mnmZn 111111 1 1 1 1 001 1 2 3 4 n nzln 1 00 1 1 1 1 1 o 2 2 4 9 16 n nzln k12345mkm k1 sn1234441 Sn nn1 2 lim sn 2 limM 00 Tl gtOO Tl gtOO 2 oes not exist as a real number and the series diverges OO 2111111m 11 SH 111n 1 n fimoc limsn 1imn 00 n gtOO Tl gtOO gain the sequence of partial sums does not exist as a real number and the series diverges 05 Nov2008 MATH 6101 1 1 1 1 00 1 1Z 2 3 4 n nzln roves the followin 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 2 3 4 5 6 7 8 9 1o 11 12 13 14 15 16 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 21 2 3 4 5 6 7 8 9 1o 11 12 13 14 15 16 11 gtii gt gt 1 1 1 1 So this one does not add up to a finite number 2 2i nfn n2 i1ltij 2 n n nn1 Therefore it does converges We noted earlier that Euler proved in 1735 that f 1 1ili 2 1an 4 9 16 6 00 2k 1 2k 1 2 lBZklir 2 2k n1 71 here Bn is the nth Bernoulli number Euler only went through the exponent 26 The Bernoulli numbers Bn were discovered by akob Bernoulli in conjunction with computing the sums of powers anOn1n2n3n4nnmln ko or example Zkzin2ln ko 2 2 n 1 1 1 218 n3 l n2 n k 3 2 6 Tl 1 1 1 2k3 2 n4 l n3 n2 ko 4 2 n 1 1 1 1 216 n5 n4 n3 n ko 5 2 30 n 1 1 1 2k5 n6 n5 in4 n2 kzo 6 2 12 12 n 1 1 1 1 1 Zk n7 n6 n5 n3 n H 7 2 2 6 42 n 1 1 1 EH in8 n7ln6 ln4 n2 H 8 2 12 24 12 2k8 2 in9 in8 3n7 ln5 3n3 in quot 1 2kg 2 n10 n k 10 2 klo i ko quot 1 k12 I 13 n kll 1 Z 12 1 1 1 9 l n8 ln6 l n4 n2 4 10 2 12 1 1 n n1 l n9 n7 n5 n3 in 1 11 11 11 11 n12n11n10n8n6n4 n2 2 12 8 6 8 12 1 11 22 6 1 n13 n12n n9 n7 n5 n4 9 n 2 10 3 2730 Bernoulli then states Tl ka 1 np1lnp Anp1WBnp 3 ko p1 2 2 pp 1p 2p 3p 4 Cnp5 pp 1p 2p 3p 4p 5p 6Dn m LAA Azi Bi Czi Di The pth Bernoulli number is the coefficient of n n the polynomial describing ka Other techniques for generating the Bernoulli Ders come Irom p1 k an ko p 1 ko 1p 1 p p1 k zk OO Tl xx 2an e 1 k m 11 En 11 Bn 0 1 12 6912730 1 12 14 76 2 16 16 3617510 4 130 18 43867798 6 142 20 174611330 8 130 22 854513138 10 566 24 2363640912730 BnJr 0 ngtU We will be able to show later that the series 2001 P 7121 n quotIi12 um cm 5 5er map M w me 5amquot minim If the terms an of an infinite series Zan are all ive then the partial sums Sn form a non decreasing sequence m erefore Zan either converges or diverges to co 39 anl is non decreasing for any sequence m e series Zan 1s sa1c1 to converge absomtely if Elanl converges A series converges conditionally if it but not absolutely 1es Theorem t Zan be an absolutely convergent series Then any rearrangement of terms in that ies results in a new series that is also absolutely convergent to the same limit quot39quot Let Zan be a conditionally convergent series T hen for any real number c there is a renrrnnnement nfthe series sneh that the new resulting series will converge to c To be L roven later Let Zan and 21 be two absolutely convergent series Th 91quot i The sum of the two series is again absolutely convergent 2an bn Zan an f39 39 39T I 4 JJ AfL1 JA 1 1A1JA1 convergent Its limit is the product of the limit of the two series 05Nov2008 MATH 6101 23 The Cauchy product of two series Zan and an is defined as The Cauchy product is 00 n nm th Term Test Theorem If Zan converges then an gt0 artial sums conver es sn gtL Note that the sequence Sn1 also converges to L Thus liman limsn sn71limsn limsni1 L L O HHOO HHOO HHOO HHOO Corollagy If a2 1 then 2a diverges nth Term Test If lim an 0 then Zan diverges 05N ov2008 MATH 6 101 Theorem If Zan and an are series so that quot s an 3 bn Then if an converges so does Zan if Zan diverges so does an Proof Set nb0b1b2mbn dn and en are increasing sequences sdnsen 05Nov2008 MATH 6101 Each converges or diverges depending on ether it is bounded or not an converges 2 en converges 2 en ounded 2 dn bounded 2 dn converges 2 Zan converges an diverges 2 dn diverges 2 dn unbounded 2 en unbounded 2 en diverges 2 an diverges 1 r d1ver es n n lt i and we know that the latter converges 2n Theorem t Zan and an be two series Suppose also r 1im lanbnl exists and O lt r ltoo Proof r lim Ianbnl and r is a positive real ber There are constants c and C OltCltCltm N gt 1 if n gt N clt anbnl ltC ssume Zan converges absolutely For n gt N c bnl lt lanl Therefore an converges absolutely by the ompar1son 39139est Assume that an converges absolutely an converes absolutelv b Com arisen Test Theorem uppose an is a decreasing sequence ofpositive terms Then the series Zan converges if and only if m MW39M Z2ka2k converges series Test Corolla or a positive number p Z 1 np converges if and onlyifp gt 1 05Nov2008 MATH 6101 series Test Proof quot p lt 0 then the sequence 1 n19 cuverges to 1nnn1ty Hence the series diverges by the nth Term Test T1 p gt 0 then consider the series Z2na2n Z2n2 p 22139P RV H19 apnmp rrin Qpripq if 0 lt p s 1 2139 2 1 so righthand series diverges if p gt 1 then 211 lt 1 so righthand series converges Now the result follows from the Cauchy Condensation Test Theorem et Zan be a series and let a 1im suplanll Proof Suppose that a lt 1 Then choose an e gt 0 so that a e lt 1 By definition of lim sup there exists N M mt a e lt luban1quot n gt N a 3 111 particular lanlln lt a efor n gt N so lanlta equot for n gt N Since 0 lt a e lt 1 the ermetric series 21 e converges By the Comparison Test Elanl converges This means that Zan also Proof If a gt 1 then there is a subsequence of lanlln that has limit a gt 1 That means that lanl gt 1 for i 39 quotquotquot quot mmquot n The sequence an cannot converge to 0 so Zan cannot converge inFor the series Zln and 21n2 a 1 inc harmonic series diverges and the other conver es so a 1 can not guarantee either convergence or divergence of the series Theorem 39 e series Zan i converges absolutely if lim supan1an lt 1 quotm infan1an s 1 3 lim suplan1an and the test gives no information 05Nov2008 MATH 6101 Theorem quotal 2 a2 2 2 an 2 2 O and an converges to zero then the alternating series Z 1nan converges 0 13 Nov2008 MATH 6101 40 lim na b maxab for a 2 0 and b 2 0 1imn nanb ab Moo 2 0 ifGZbiO hm an bn 1 ifagtb naooanbn 1 if altb undefined if a b 0 lim n gtoo n 1 2n2 2nn1gtn2 2wnn 1m21 2 nn1 2 22 2 2 gtlt2 1gtltgtlt22gtlt2 2 12 2 gtlt2 1gtltgtlt22gtlt2 2 gt00 X X 39gtlt X 00 sinn 9 T121 n Since sin n9 3 1 for all n we have that quot e series 0 1 22n2 1 This diverges by the Comparison Test with the curiae oo 1 hich diverges by the p series 39l est Q139Wmlim 10g nn O and a lt an by the n1 Alternating Series Test this ser1es converges 00 2 n Z Use the Ratlo Test n n12 n n1 O a 0 11m quot1 11m 2 11m n gtoo an n gt00 n n gt00 n O 2 o 13 Nov2008 MATH 6101 00 logn Z Tl1 n log n gt i So the series diverges by the bomparlson l ESL 05mmgooa MATH 101 3 ZOO 1 n logn 1 1 logn lt n lt n logn O the SBI lBS alverges by the Companson 39l est 05 NOV2008 MATH 6101 48 00 1 Z n n n TTQP H19 Rnn r quf 1 11m Tlan 11m O lt 1 n gt00 n gtOO n So the series converges by the Root Test 0 13 Nov2008 MATH 01 49 3 nzn1 This series diverges by the Limit Comparison with 21 n o5Nm2008 MATH 6101 5a D5Nm2008 MATH 1101 31 00 1 T121 n Use the Limit Comparison Test with 21 n 1 11n n n 1 11m 11m 11m 11m 1ltOO n gtoo 1 n gtoo n11n n gtoo n nln n gtoo nln n quot39 us the series diverges Chapter 6 Sequences and Series of Real Numbers We often use sequences and series of numbers without thinking about it A decimal representation of a number is an example of a series7 the bracketing of a real number by closer and closer rational numbers gives us an example of a sequence We want to study these objects more closely because this conceptual framework will be used later when we look at functions and sequences and series of functions First7 we will take on numbers Sequences have an ancient history dating back at least as far as Archimedes who used sequences and series in his Method of Exhaustion77 to compute better values of 7139 and areas of geometric gures 61 The Symbols 00 and 00 We often use the symbols 00 and 00 in mathematics7 including courses in high school We need to come to some agreement about these symbols We will often write 00 for 00 when it should not be confusing First of all7 they are not real numbers and do not necessarily adhere to the rules of arithmetic for real numbers There are times that we act as if they do7 so we need to be careful We adjoin 00 and 00 to R and extend the usual ordering to the set R U 00 00 Explicitly7 we will agree that 00 lt a lt 00 for every real number a E R U 00 00 This gives the extended set with an ordering that satis es our usual properties 1 If ab E RU00007 then a S b or b S a 2 lfagbandb a7thenab 3 lfagbandbgc henago We will not extend the usual algebraic structure of the reals to R U 00 00 73 74 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS lnstead when we have to we will discuss the algebra that might involve 00 andor foo Do not apply any theorem that is stated for the real numbers to the symbols 00 or foo The symbols make it convenient to extend our notation about intervals to the real line aoox Rla z aoox Rlaltz foobx lRlz b OObZERlltb Occasionally you will see R foo 00 62 Sequences Sequences are basically countably many numbers arranged in an ordered set that may or may not exhibit certain patterns De nition 61 A sequence of real numbers is a function whose domain is a set of the form n E Z l n 2 m where m is usually 0 or 1 Thus a sequence is a function f N a R Thus a sequence can be denoted by fm fm 1 fm 2 Usually we will denote such a sequence by aih m or amam1am2 where a fi Ifm 1 we may use the notation ann6N Example 61 The sequence 112 131415 is written as 1if1 Keep in mind that this sequence can be thought of as an ordinary function In this case fn1n Example 62 Consider the sequence given by an 71 for n 2 0 This time we have started the sequence with 1 and the terms look like 1 71 1 71 1 71 Note that this time the function has domain N but the range is 711 Example 63 Consider the sequence an cos n E N The rst term in the sequence is cosg cos 60 and the sequence looks like 1 11 7i 7 17i 7 71 e 2 2 2 1 1 7 771 2 2 DlH 1 1 1 27 27 7 27 27 Note that likes its predecessor the function takes on only a nite number of values but the sequence has an in nite number of elements ln n E N the sequence is 1 Viol341 Example 64 If an n MATH 6101 090 Fall 2006 62 SEQUENCES 75 We might use an approximation for each of these and arbitrarily choosing 5 decimal places the sequence would look like 1141421144225141421137973134801132047129684127652125893 We would nd that 1100 104713 and 110000 100092 Example 65 Consider the sequence bn 1 i n E N This is the sequence 3 2 4 3 5 4 2 7 7 7 2 225 237037 244141 248832 252163 254650 256578 258117 259374 or by approximation Again 1100 and 110000 In looking at these examples we might think that some of them are giving us a pattern of numbers that are getting close77 to some other real number Others may not give us that indication We are interested in what the long term behavior of the sequence is What happens for larger and larger values of n Does the sequence approach a real number Could it approach more than one real number De nition 62 A sequence of real numbers is said to converge to a real number L iffor every 6 gt 0 there is an integer N gt 0 such that ifk gt N then lak 7 Ll lt 6 The number L is called the limit of the sequence lf ak converges to L we will write klim ak L or simply 400 ak 7 L If a sequence does not converge then we say that it n n2 diverges 1 05 Note that the N in the de nition depends on the E that we 2 05 were given If you change the value of 6 then you may have to 3 0375 recalculate N n 4 025 Consider the sequence an 7n n E N Now if we look at 5 0 15625 the values that the sequence takes 6 009375 7 00546875 3 3 3 i 8 003125 27 227 237 247 39 H 9 00175781 10 000976562 we might think that the terms are getting smaller and smaller so maybe the limit of this sequence would be 0 Lets take a look and compare how N would vary as E varies Lets start with some simple small numbers and let 6 be 01 001 0001 and 00001 and 000001 MATH 6101 090 Fall 2006 76 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS For E 017 we need to nd an integer N so that N 271V 7 0 lt 01 Look in the table of values here and we see that for N 6 we have satis ed the above condition Following this we get the following by using a calculator or a computer algebra system i i N Ngt0 implies 2N70 lt1 i i N Ngt5 implies 27N70 lt01 i i N N gt 9 implies 7 0 lt 001 i i N N gt 14 implies 271V 7 0 lt 0001 i i N N gt 18 implies 7 7 0 lt 00001 2N i i N N gt 22 implies 271V 7 0 lt 000001 We are going to establish several properties of convergent sequences Many proofs will use a proof much like this next result While this type of argument may not easy to get used to7 it will appear again and again7 so you should try to get as familiar with it as you can Theorem 61 Convergent sequences are bounded Let an n E N be a con uergent sequence Then the sequence is bounded and the limit is unique PROOF The easier property to show is that the limit is unique7 so lets do that rst Suppose the sequence has two limits7 L and K Take any 6 gt 0 Then there is an integer N such that lak7Ll ltgifkgtN Also7 there is another integer N such that lak7Kl ltgifkgtN Then7 by the triangle inequality lL7Kl lt lak7Lllak7Kl ltggeifk gtmaXNN MATH 6101 090 Fall 2006 63 THE ALGEBRA OF CONVERGENT SEQUENCES 77 Therefore lL 7 Kl lt E for any 6 gt 0 But the only way that that can happen is for L K so that the limit is indeed unique Next we need to prove boundedness Since the sequence converges we can take any 6 we wish and tradition shows us to take 6 1 Then there is an integer N so that lak7Ll lt1ifkgtN Fix that integer N Then we have that lanl S lan7LllLl lt1lLlPfor allngtN Now de ne M maxlaklh 1NP Then lanl lt M for all n which makes the sequence bounded I 63 The Algebra of Convergent Sequences This section proves some basic results that do not come as a surprise to the student Theorem 62 If the sequence on converges to L and c E R then the sequence can converges to cL 239e lim can c lim an Hoe Hoe PROOF Let7s assume that c 31 0 since the result is trivial if c 0 Let E gt 0 Since an converges to L we know that there is an N E N so that if n gt N 6 loan 7 Ll lt 7 lCl Thus for n gt N we then have that e lcan7cLl chan7Ll lt lcl 6 M which is what we needed to prove I Theorem 63 If the sequence on converges to L and bn converges to M then the sequence an bn converges to L M ie bn on bn PROOF Let E gt 0 We need to nd an N E N so that if n gt N Kan bn 7 L lt 6 Since an and bn are convergent for the given 6 there are integers N1 N2 6 N so that 6 6 lfngtN1then lan7Ll lt and ifngtN2then lbn7Ml lt Thus if n gt maXN1N2 then 6 e lab7LMl lan7Lllbn7Ml lt e I MATH 6101 090 Fall 2006 78 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS Theorem 64 If the sequence can converges to L and bn converges to M then the sequence an bn converges to L M 239e lim 1 b hm an hm bn n7too n7too n7too The trick with the inequalities here is to look at the inequality lanbn 7 LMl lanbn 7 anM anM 7 LMl lanbn 7 1an lanM 7 LMl lanllbn 7 Ml 7L lMllan 7 Ll39 l So for large values of n lb 7 Ml and lan 7 Ll are small and is constant Now by Theorem 61 shows that lanl is bounded so that we will be able to show that lanbn 7 LMl is small PROOF Let E gt 0 By Theorem 61 there is a constant K gt 0 such that lanl S K for all n Since bn is convergent for the given 6 there is an integer N1 6 N so that e lfn gtN1 then lbn7Ml lt Also since an is convergent there is an integer N2 6 N so that e lfngtN then an7L lt 2 l l 2lMl1 Thus if N maXN1N2 then if n gt N lanbn 7 LMl S lanllbn 7 Ml 7L lMllan 7 Ll E E E E K7 M 7 7 2Kl l2lMl1lt22 6 l Lemma 61 If the sequence can converges to L and bn converges to M and if an ltbnfor alln 2m thenL S M Lemma 62 If the sequence can converges to L if an 31 0 for all n E N and if L 31 0 then glblal l n E N gt 0 PROOF Let E gt 0 Since an converges to L there is an N E N so that if n gt Nthen lan 7 Ll lt lLl2 Now if n gt N we must have that lanl 2 lLl2 If not then the triangle inequality would imply NW NW lLllL7ananlSlL7anllanlltTTlMl39 Nowweset m min Elall 7 27 1702l7laNl Thenclearlymgt0andlanlmeor allnEN I MATH 6101 090 Fall 2006 63 THE ALGEBRA OF CONVERGENT SEQUENCES 79 Theorem 65 If the sequence can converges to L if an 31 0 for all n E N and if L 31 0 then the sequence 1an converges to 1L 239e lim 1an 1 lim an PROOF Let E gt 0 By Lemma 62 there is an m gt 0 such that lanl 2 m for all n Since an is convergent there is an integer N E N so that if n gt N lLianl lt emlLl Then for n gt N 7 lt lt6 an L lanLl mm 1 1 ilaniLl laniLl Theorem 66 Suppose that the sequence bn converges to M and if an converges to L If bn 31 0 for all n E N and lfM 31 0 then the sequence anbn converges to an lim an W 31330 a hm b PROOF We use two of the previous theorems to prove this By Theorem 65 1bn converges to 1M7 so le w n 1 liulj1noangL39 1 Example 66 Let p gt 0 then lim 7 0 naoo 71 Let E gt 0 and let N 2117 Then n gt N implies that n gt i and hence E gt Since 7 gt 07 this shows that n gt N implies 7 0 lt 6 Example 67 Let lal lt 17 then lim 1 0 Hoe Suppose a 31 07 because lim 1 0 is clear for a 0 Since lal lt 17 we can write lal 7 where b gt 0 By the binomial theorem 1 lbl 1 1br1nb b2b 21nbgtnb l 011Ll1lt1 a a 1bn nb39 Now consider6gt0and letN6 Then ifngtN7 we havengtand hence la 70llt lte MATH 6101 090 Fall 2006 80 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS Example 68 1 Let an n1 7 1 and note that an 2 0 for all 71 By Theorem 63 it is suf cient for us to show that limiH00 an 0 Since 1 an 711 we have n 1 an For n 2 2 the binomial theorem gives us 71 1 an 21 nan 7 Mai gt 71a Thus7 n gt 7 Dag7 so ai lt Thus7 we have shown that an lt A for n 2 2 Thus7 liman 0 Example 69 lima1 1 for a gt 0 Suppose a 2 1 Then for n 2 a we have 1 3 a1 3 711 Since limn1 17 it easily follows that limaV 1 Suppose that 0 lt a lt 1 Then i gt 17 so that limi1 1 Thus7 llri limm 1 1 7 1 lima1 lima1 1 De nition 63 For a sequerice an we write lim an oo provided for each M gt 0 there is a number N such that n gt N implies that an gt M In this case we will say that an diverges to 00 We can make a similar de nition for lim an 700 Of course7 we cannot use the previous theorems when dealing with in nite limits Theorem 67 Let an arid bn be sequerices such that lim an oo aridlim bn gt 0 Theri lim anbn oo PROOF Let M gt 0 Choose a real number in so that 0 lt m lt limb Whether lim bn 00 or not7 there exists N1 so that if n gt N1 then bn gt m Since liman oo there is an N2 so that if n gt N2 then an gt m Setting N maXN1N2 means that for n gt N anbn gt m M I MATH 6101 090 Fall 2006 64 MONOTONlClTY AND CAUCHY SEQUENCES 81 Theorem 68 For a sequence an ofposz39tz39ue real numbers lirn an 00 if and only if lirn a 0 PROOF Let an be a sequence of positive numbers We need to show i q 1 If lirn an 00 then lirn 7 0 61 an and 1 If lirn 7 0 then lirn an 00 62 an To prove 61 we will suppose that lirn an 00 Let 6 gt 0 and let M 16 Since an diverges to 007 there is an N so that if n gt N then an gt M 16 Therefore7 ifNgtnthen6gt1angt0so ifn gt Nthen 1 l 770 lt6 an Thus7 lirn1an 0 This proves 61 To prove 62 suppose that lirn1an 0 and let M gt 0 Let 6 1M Then since 6 gt 0 there is an N so that ifn gt Nthen 7 lt 6 Since an gt 0 we then know that 1 1 f gt N th 0 lt 7 lt 7 1 n en an M and hence ifngtNthenangtM This means that lirn an 00 and 62 holds I 64 Monotonicity and Cauchy Sequences The previous section showed us how to work with convergent sequences7 but does not tell us how to determine quickly if a sequence does converge We need such a tool De nition 64 A sequence an of real numbers is called monotonically increas 39ing 1 if an 3 an for all n and an is called a monotonically decreasingz if an 2 an for all n If a sequence is monotonically increasing or monotonically decreasing7 we will call it a monotonic sequence or a monotone sequence Theorem 69 All bounded monotone sequences converge 1Sometimes called nondecreasing 2Sometimes called nonincreasingi MATH 6101 090 Fall 2006 82 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS PROOF Let an be a bounded monotonically increasing sequence and let S an mod n E N Since the sequence is bounded on lt M for some real number M and for all n E N This means that the set S is bounded and thus it has a least upper bound Let u lub S Let E gt 0 Since u lubS and E gt 0 u 7 E is not an upper bound for S This means that there must be some N so that IN gt u 7 6 Since an is monotonically increasing we have that for all n gt N on 2 IN and hence for all n gt N it follows that u 7 E lt an S u Thus loan 7 ul lt E for all n gt N Thus liman u lub S The proof for bounded monotonically decreasing sequences is the same with the greatest lower bound playing the role of the least upper bound Now we can also handle unbounded monotone sequences Theorem 610 Let an be a sequence of real numbers i If an is an L J J t 39 quot2 tncycwmy sequence then liman 00 ii If an is an unbounded monotonically decreasing sequence then lim an 700 Let an be a bounded sequence of real numbers While it may converge or may not converge the limiting behavior of an depends only on the tails77 of the sequence or sets of the form an l n gt N This leads us to a concept that we can discuss without knowing a priori if a given sequence converges or diverges Let uN glban l n gt N infan l n gt N and let on luban l n gt N supan l n gt N We have seen that if lim an exists then it must lie in the interval uN N As N increases the sets an l n gt N get smaller so we have u1 UZSU3 ando12o22o32 By the above theorem the limits u limNH00 uN and o limNH00 UN both exist and u S 1 since UN 3 UN for all N If the limit exists then UN 3 lim an 3 UN so u 3 lim an S 1 These numbers u and 1 turn out to be useful whether lim an exists or not De nition 65 Let an be a sequence of real numbers We de ne limsup an Alim lubsn l n gt N lim inf an Alim glbsn l n gt N Note that we do not require that an be bounded We will take some precautions and adopt the following conventions If an is not bounded above lubsn l n gt N 00 for all N and we de ne lim sup an 00 Likewise if an is not bounded below glbsn l n gt N 700 and we de ne lim inf an 7oo in this case MATH 6101 090 Fall 2006 64 MONOTONlClTY AND CAUCHY SEQUENCES 83 Now is it true that lirn sup an luban l n gt N Not necessarily because while it is true that lirn sup an S luban l n gt N some of the values an may be much larger than lirn sup an Note that lirn sup an is the largest value that in nitely many an7s can get close to Theorem 611 Let an be a sequence of real numbers i If lirn an is de ned as a real number 00 or 00 then lirn inf an lirn an lirn supan ii If lirninfan lirnsupan then lirnan is de ned and lirnan lirninfan lirn supan PROOF Let us use the notation from above ieLet uN glban l n gt N m luban l n gt N u lirnuN lirninfan and i lirnizN lirnsup an i Suppose lirn an 00 Let M gt 0 be a positive number Then there is N E N so that ifn gt N then an gt M Then uN glban l n gt N 2 M It follows that m gt N implies that um 2 M Thus the sequence UN satis es the condition that lirnuN 00 or lirn inf an 00 Likewise we can show that lirn sup an 00 We do the case that lirn an 00 sirnilarly Suppose that lirn an L E R Let E gt 0 There exists an N E N so that lan Ll lt E for n gt N Thus an lt L E for n gt N This means that UNluban l ngtN Le Also if m gt N then 11m 3 Le for all 6 gt 0 no matter how small This means that lirn sup an S L lirn an Similarly we can show that lirnan S lirn inf an Since lirn inf an S lirn sup an these inequalities give us that lirn inf an lirn an lirn sup an ii If lirn inf an lirn sup an i00 it is easy to show that lirn an i00 Suppose that lirn inf an lirn sup an L E R We need to show that lirn an L Let E gt 0 Since L lirnuN there is an N0 6 N so that lL luban l n gt N0l lt 6 Thus luban l n gt No lt L 6 so an lt L6f0r allngt N0 Sirnilarly since L lirn uN there is an N1 6 N so that lLglban l n gt N1l lt 6 MATH 6101 090 Fall 2006 84 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS Thus glban l n gt N1 gt L 7 6 so angtL76for allngtN1 These two conditions tell us that L 7 E lt an lt L E for n gt maXN0N1 or equivalently lan 7 Ll lt E for n gt maXN0 N1 This proves that lim an L as needed This tells us that if an converges then lim inf an lim sup an so for large N the numbers luban l n gt N and glban l n gt N must be close together This means that all of the numbers in the set an l n gt N must be close together This leads to the following de nition De nition 66 A sequence an of real numbers is called a Cauchy sequence if for each 6 gt 0 there is a number N E N so that if mn gt N then lan 7 am lt 6 Lemma 63 Convergent sequences are Cauchy sequences PROOF Suppose that lim an L Note that lan7aml lan7LL7aMl S lan7Lllam7Ll Thus given any 6 gt 0 there is an N E N so that if h gt N then lak 7 Ll lt Thus if mn gt N we have 6 2 lan7aml lan7Lllam7Ll ltg 6 Thus an is a Cauchy sequence I Lemma 64 Cauchy sequences are bounded This leads us to the following theorem Theorem 612 A sequence is a convergent sequence if and only if it is a Cauchy sequence MATH 6101 090 Fall 2006 65 SUBSEQUENCES 85 PROOF We have just proven half of this above That means that we are left to show that any Cauchy sequence must converge To see this let an be a Cauchy sequence From the above lemma we know that it is bounded That means then that we only need to show that lim inf an lim sup an Let E gt 0 Since an is a Cauchy sequence7 there is an N E N so that if mn gt N then lan 7 aml lt 6 ln particular7 an lt am E for all mn gt N This shows that am E is an upper bound for an l n gt N Thus vN luban l n gt N 3 am E for m gt N Now7 this shows that vN 7 E is a lower bound for am l m gt N7 so that vN 7 E S glbam l m gt N uN Therefore limsupan S vN 3 UN E S liminfan 6 Since this holds for all 6 gt 07 we have that lim sup an 3 lim inf an and this is enough to give us that the two quantities are equal 65 Subsequences So far we have learned the basic de nitions of a sequence a function from the nat ural numbers to the reals7 the concept of convergence7 and we have extended that concept to one which does not pre suppose the unknown limit of a sequence Cauchy sequence Unfortunately7 however7 not all sequences converge We will now intro duce some techniques for dealing with those sequences The rst is to change the sequence into a convergent one extract subsequences and the second is to modify our concept of limit as we did with lim sup and lim inf De nition 67 Let an be a sequence When we ecctract from this sequence only certain elements and drop the remaining ones we obtain a new sequences consisting of an in nite subset of the original sequence That sequence is called a subsequence and denoted by am One can extract in nitely many subsequences from any given sequence Example 610 Take the sequence 71 7 which we know does not converge Ex tract every other member7 starting with the rst Does this sequence converge What if we extract every other member7 starting with the second What do you get in this case Example 611 Take the sequence Extract three different subsequences of your choice Do these subsequences converge ls so7 to what limit The last example is an indication of a general result Theorem 613 If an is a convergent sequence then every subsequence of that sequence converges to the same limit MATH 6101 090 Fall 2006 86 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS ii Ifis a sequence such that every possible subsequence emtracted from that sequences converges to the same limit then the original sequence also converges to that limit The next statement is probably one on the most fundamental results of basic real analysis and generalizes the above proposition It also explains why subsequences can be useful even if the original sequence does not converge Theorem 614 Bolzano Weierstrass Let an be a sequence of real numbers that is bounded Then there eists a subsequence am that converges PROOF Since the sequence is bounded there exists a number M such that an lt M for all n Then either 7M0 or 0 M contains in nitely many elements of the sequence Say that 0M does Choose one of them and call it am Now either 0 M2 or M2 M contains in nitely many elements of the original sequence Say it is 0 Choose one of those elements and call it am Again either 0 M4 or M4M2 contains in nitely many elements of the original sequence This time say it is M4 Pick one of those elements and call it am Keep on going in this way halving each interval from the previous step at the next step and choosing one element from that new interval Here is what we get 0 an1 7 an2 lt M because both are in 0M o an2 7 am lt M2 because both are in 0M2 o an3 7 and lt M4 because both are in M2M4 and in general we see that ant 7 ank1 lt M2k717 because both are in an interval of length M2k 1 So this proves that consecutive elements of this subsequence are close together That is not enough however to say that the sequence is Cauchy since for that not only consecutive elements must be close together but all elements must get close to each other eventually So take any 6 gt 0 and pick an integer N such that for any hm gt N with m gt k MATH 6101 090 Fall 2006 66 SERIES 87 wehave lank 7 anml 1ank 7 ank1 717 ank1 7 ank2 717 39 39 39 717anm717 anm1 S law 7 ank111ank17 ank2l quot 39 1anm717 anml 1 1 1 1 Mlt2kj27w 2m72 M 1 1 1 F1 392mikil M 1 1 S w 12 M 1quot 2M M 71 i 72k71 2k72 j1 In order for this to be less than 6 for m7 k gt N we would need to take N so that M 721V lt E 2N gt M e N72ln2 gt lnMi lne l M fl N gt 21 ln2 This proves what we wanted I 66 Series Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an in nite series The old Greeks already won dered about this7 and actually did not have the tools to quite understand it This is illustrated by the old tale of Achilles and the Tortoise Zeno7s Paradox Achilles and the Tortoise Achilles7 a fast runner7 was asked to race against a tortoise Achilles can run 10 yards per second7 the tortoise only 5 yards per second The track is 100 yards long Achilles7 being a fair sportsrnan7 gives the tortoise 10 yard advantage Who will win Both start running7 with the tortoise being 10 yards ahead After one second7 Achilles has reached the spot where the tortoise started The tortoise7 in turn7 has run 5 yards Achilles runs again and reaches the spot the tortoise has just been The tortoise7 in turn7 has run 25 yards Achilles runs again to the spot where the tortoise has just been The tortoise7 in turn7 has run another 125 yards ahead MATH 6101 090 Fall 2006 88 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS This continuous for a while but whenever Achilles manages to reach the spot where the tortoise has just been a second ago the tortoise has again covered a little bit of distance and is still ahead of Achilles Hence as hard as he tries Achilles only manages to cut the remaining distance in half each time implying of course that Achilles can actually never reach the tortoise So the tortoise wins the race which does not make Achilles very happy at all What is wrong with this line of thinking Let us look at the difference between Achilles and the tortoise and so on In general we have Time lDi erence 10 t12172i32 27LyardS Now we want to take the limit as 71 goes to in nity to nd out when the distance between Achilles and the tortoise is zero But that involves adding in nitely many numbers in the above expression for the time and we the Greeks and Zeno dont know how to do that However if we de ne 1 1 1 1 1 7 i i i s 22223 2 then dividing by 2 and subtracting the two expressions 171 1 sniisni 7 1 sn2172n Now 5 is a simple sequence for which we know how to take limits In fact from the last expression it is clear that lim 5 2 as n approaches in nity Hence we have mathematically correctly computed that Achilles reaches the tortoise after exactly 2 seconds and then of course passes it and wins the race A much simpler calculation not involving in nitely many numbers gives the same result or equivalently solving for 5 o Achilles runs 10 yards per second so he covers 20 yards in 2 seconds MATH 6101 090 Fall 2006 66 SERIES 89 o The tortoise runs 5 yards per second and has an advantage of 10 yards So it also reaches the 20 yard mark after 2 seconds 0 Therefore both are even after 2 seconds Of course Achilles will nish the race after 10 seconds while the tortoise needs 18 seconds to nish and Achilles will clearly win The problem with Zeno7s paradox is that Zeno was uncomfortable with adding in nitely many numbers In fact his basic argument was that if you add in nitely many numbers then 7 no matter what those numbers are 7 you must get in nity If that was true it would take Achilles in nitely long to reach the tortoise and he would loose the race However reducing the in nite addition to the limit of a sequence we have seen that this argument is false One reason for looking so carefully at sequences is that it allows us to to quickly obtain the properties of in nite series We know at least theoretically how to deal with nite sums of real numbers TL E akamam1an km More interest in mathematics though tends to lie in the area of in nite series 00 Zakamam1am2 km What do we mean by this in nite series 6 De ne the nth partial sum Sn by V L Snamam1anZak km This now gives us a sequence the sequence of partial sums The in nite series ak is said to converge provided the sequence of partial sums converges to a real number S In this case we de ne 227 an S Thus 2 an S means lim Sn S or lim 6 S 7 km If a series does not converge we say that it diverges We can then say that a series diverges to 00 if lim 5 00 or that it diverges to 00 if lim 5 00 Some texts will indicate that the symbol 23 an has no meaning unless the series converges or diverges to 00 or 00 Thus Zf01 will have no meaning MATH 6101 090 Fall 2006 90 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS 1 1 1 1 1 Example 612 g 27 1 E Z g E is an in nite series The sequence of partial sums looks like 3 7 15 8157S7S7 0 7 1 27 2 47 3 8 7 We saw above that this sequence converges to 27 so i limSn 2 2n M3 3 H 0 Example 613 The harmonic series is i771133 n7 2 3 4 n1 The rst few terms in the sequence of partial sums are 3 11 25 137 51SSisi 7 1 7 2 27 3 6 7 4 127 5 7 49 363 761 7129 7381 6 757 7 8 789 7510 20 140 280 2520 2520 This series diverges to 00 To prove this we need to estimate the nth term in the sequence of partial sums The 71 partial sum for this series is S i 1 1 1 1 1 N 7 2 3 4 Nquot Now consider the following subsequence extracted from the sequence of partial sums 81 1 1 1 1 S 17 7 7 4 234 gt11117111712 2 4 4 7 2 27 2 S 711111111 87 2 3 4 5 6 7 3 gt1l ll llll 71333717 2 4 4 3 3 3 3 7 2 2 27 MATH 6101 090 Fall 2006 66 SERIES 91 In general by induction we have that that k S gt 1 7 2k 7 2 for all k Hence the subsequence 52k extracted from the sequence of partial sums SN is unbounded But then the sequence SN cannot converge either and must in fact diverge to in nity If the terms can of an in nite series 2a are all nonnegative then the partial sums Sn form a nondecreasing sequence so by Theorems 69 and 610 2a either converges or diverges to 00 In particular 2 lanl is meaningful for any sequences can whatsoever The series 2a is said to converge absolutely if E lanl converges Example 614 A series of the form 20 or for constants a and r is called a geo metric sem39es For 7 31 1 the partial sums are given by 1 7 Tn1 1 7 7 Taking the limit as N goes to in nity gives us that a 7 if M lt 1 Z 1 0 00 if a 31 0 and lr gt21 Example 615 p Serieleor a positive number p 1 Z 7 converges if and only if p gt 1 n n1 The exact value of this series for p gt 1 is extremely dif cult to determine A few are known The rst of these below is due to Euler 1 2 Z 4 1 7139 2m If p gt 1 then the sum of the series is 10 ie the Riemann zeta function evaluated at p There If p is an even integer then there are formulas like the above but there are no elegant formulas for p an odd integer A series converges conditionally if it converges but not absolutely MATH 6101 090 Fall 2006 92 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS Example 616 1 Does the series Zf071 converge absolutely7 conditionally or not at all 2 Does the series 2206 converge absolutely7 conditionally7 or not at all 00 71 n1 3 Does the series E converge absolutely7 conditionally7 or not at all this n n1 series is called alternating harmonic series Conditionally convergent sequences are rather dif cult with which to work Several operations that one would expect to be true do not hold for such series The perhaps most striking example is the associative law Since a b b a for any two real numbers a and b7 positive or negative7 one would expect also that changing the order of summation in a series should have little effect on the outcome Not true Theorem 615 Order of Summation Let Zan be an absolutely convergent series Then any rearrangement of terms in that series results in a new series that is also absolutely convergent to the same limit ii Let be a conditionally convergent series Then for any real number c there is a rearrangement of the series such that the new resulting series will converge to c This will be proved later One sees7 however7 that conditionally convergent series probably contain a few surprises Absolutely convergent series7 however7 behave just as one would expect Theorem 616 Algebra on Series Let Zan and Eb be two absolutely conver gent seri39es Then i The sum of the two series is again absolutely convergent Its limit is the sum of the limit of the two series ii The di erence of the two series is again absolutely convergent Its limit is the di erence of the limit of the two series 111 e pro uc o e wo series is again a so u e y convergen s imi is e Th d t th t 39 39 39 b l tl t It l39 39t 39 th product of the limit of the two series The Cauchy product of two series 2 an and 2 bn of real is de ned as follows The Cauchy product is 0 an b on where on ia wk nm nm nm h0 for n 0 1 2 777 MATH 6101 090 Fall 2006 67 CONVERGENCE TESTS 93 67 Convergence Tests De nition 68 We say that a series Zan satis es the Cauchy criterion if its se quence of partial sums is a Cauchy sequence This means that for each 6 gt 0 there exists a number N such that if mn gt N then 5 7 SM lt 6 Nothing is lost in this de nition if we impose the restriction n 2 m Moreover7 it is only a notational matter to work with m 7 1 where m S n instead of m where m lt ii That means that the de nition is equivalent to for each 6 gt 0 there exists a number N such that if n 2 m gt N then 5 7 mill lt 6 The reason for doing this is that Sn 7 Sm1 ELM ak Then this condition can be rewritten as for each 6 gt 0 there exists a number N such that if n 2 m gt N then ELM lt 6 Theorem 617 A series converges if and only if it satis es the Cauchy criterion Corollary 61 If a series 2 an converges then lim an 0 It is often easier to prove that a limit exists or that a series converges than it is to determine its exact value As an example consider the following Theorem 618 Comparison Test Let Zan be a series where an 2 0 for all n i IfZan converges and lbnl 3 an for all n then Eb converges ii IfZan 00 and bn 2 an for all n then Eb 00 PROOF For n 2 m we have bk E 1k M M M lbkl S k k k H 3 H 3 H 3 where the rst inequality follows from the Triangle lnequality Since Zan converges7 it satis es the Cauchy criterion It then follows from the above that 2 bn also satis es the Cauchy criterion and hence it also converges ii Let Sn and Tn be the sequences of partial sums for Zan and 2b respec tively Since bn 2 an for all n we then clearly have that Tn 2 Sn for all n Since lim Sn 007 we conclude that lian 007 and El 00 I Corollary 62 Absolutely convergent series are convergent PROOF Suppose that Zan is absolutely convergent This means that Eb con verges where bn lanl for all n Then lanl S bn7 so that Zan converges by the Comparison Test MATH 6101 090 Fall 2006 94 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS Theorem 619 Limit Comparison Test Suppose Zan and El are two in nlte series Suppose also that r lim lanbnl em39sts and 0 lt r lt 00 Then Zan conuerges absolutely if and only lebn conuerges absolutely PROOF Since r limlanbnl exists7 and r is between 0 and 007 there exist con stants c and C7 0 lt c lt C lt 00 such that for some N gt1 we have that if n gt N an lt0 n Assume that Zan converges absolutely For n gt N we have that clbnl lt lanl Therefore7 2 bn converges absolutely by the Comparison Test Now assume that 2 bn converges absolutely From the above inequality we have that lanl lt Clbnl for n gt N But since the series 0 El also converges absolutely7 we can use again the Comparison Test to see that 2 an must converge absolutely I Theorem 620 Cauchy Condensation Test Suppose an is a decreasing se quence of positive terms Then the series Zan conuerges if and only if the series 2 2ka2k conuerges 00 PROOF Assume that 2 an converges Since an is a decreasing sequence7 we have n1 that 21671 7 12k 7 12k 12k 12k 12k 2k S azkil1 02k712 02k712k72 12k E am m2k 11 Therefore7 we have that N N 2k 2 kil 2 12k 3 am i am h1 h1 m2k11 m2 Now the partial sums on the right are bounded7 by assumption Hence the partial sums on the left are also bounded Since all terms are positive7 the partial sums now form an increasing sequence that is bounded above7 hence it must converge Multiplying the left sequence by 2 will not change convergence7 and hence the series 221 2ka2k converges Now7 assume that 221 2ka2k converges We have 2N am azkil1 02k712 02k712k72 12k m2k 11 l 02k71 02k71 azka azka 2k7102k71 MATH 6101 090 Fall 2006 67 CONVERGENCE TESTS 95 Therefore similar to above we get 2 N 2k N kil am i am 3 2 12k m2 k1 m2k11 k1 Now the sequence of partial sums on the right is bounded by assumption There fore the left side forms an increasing sequence that is bounded above and therefore must converge Corollary 63 For a positive number p 00 1 E 7 converges if and only ifp gt 1 n1 n 1 PROOF If p lt 0 then the sequence 7 diverges to in nity Hence the series n diverges by the Divergence Test lfp gt 0 then consider the series 00 TL 00 TL 1 00 7L 22 lt2 321p 39 This right hand side is a geometric series Thus we know that o if 0 lt p S 1 then 21 2 1 hence the right hand series diverges o ifp gt 1 then 21 lt 1 and the right hand series converges Now the result follows from the Cauchy Condensation Test I Theorem 621 Root Test Let 2a be a series and let oz limsup lanlln The series 2a i converges absolutely ifa lt 1 ii diverges ifa gt 1 iii Otherwise oz 1 and the test gives no information Although this Root Test is more dif cult to apply it is better than the Ratio Test in the following sense There are series for which the Ratio Test give no information yet the Root Test will be conclusive We will use the Root Test to prove the Ratio Test but you cannot use the Ratio Test to prove the Root Test It is important to remember that when the Root Test gives 1 as the answer for the lim sup then no conclusion at all is possible The use of the lim sup rather than the regular limit has the advantage that we do not have to be concerned with the existence of a limit On the other hand if the regular limit exists it is the same as the lim sup so that we are not giving up anything using the lim sup MATH 6101 090 Fall 2006 96 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS PROOF i Suppose that 04 lt 1 Then choose an E gt 0 so that 04 E lt 1 Then by the de nition of the limit superior there is a natural number N such that 04 7 E lt lublanll l n gt Noz 6 ln particular7 we have lanlln lt 04 E for n gt N7 so lanl lt 04 6 for n gt N Since 0 lt 04 E lt 17 the geometric series ZN1oz 6 converges Thus7 the Comparison Test shows that 2NH an converges This means that Z on also converges ii If 04 gt 17 then there is a subsequence of 1 that has limit 04 gt 1 That means that lanl gt 1 for in nitely many choices of n In particular7 the sequence on cannot converge to 07 so the series Eon cannot converge iii For the series and for the series 2 04 turns out to be 1 Since the harmonic series diverges and the series converges7 the equality oz 1 cannot guarantee either convergence or divergence of the series I Theorem 622 Ratio Test A series Zan of nonzero series i converges absolutely if lim sup lanHanl lt 1 ii diverges is lim inf lanHanl gt 1 iii Otherwise lim inf lanHanl S 1 3 lim sup lanHanl and the test gives no infor mation PROOF i Suppose that lim sup lanHanl lt 1 Then choose an E gt 0 so that lan anHe lt 1 Then by the de nition of the limit superior there is a natural number N such that for n gt N m an lt176 Multiplying both sides by lanl we get lan l lt 17 6lanl for n gt N Therefore7 we also have lan l lt17 elan1l lt 17 6Zlanl for n gt N MATH 6101 090 Fall 2006 67 CONVERGENCE TESTS 97 Repeating this procedure we get that lakl lt 17 ek NlaNl for h gt N These terms form a convergent geometric series Thus the Comparison Test shows that 2 an also converges The other two parts are proven in much the same fashion as the previous theorem The next lemma is used to prove Abel7s Convergence Test It is computational in nature Lemma 65 Summation by Parts Consider the two sequences an and Let Sn 2a be the n th partial sum Then for any 0 S m S n we have k1 n n1 2 ab Snbn i swam 2 59a 7 om jm jm PROOF Just be careful with the subscripts Z ajbj 05 Sjil j jm jm Z Sij i 2 5911 m jm n n71 Z Sjb i 5le jm jm71 n71 Theorem 623 Abel7s Test Consider the series Zanbn Suppose that i the partial sums SN ZN n 1 an form a bounded sequence ii the sequence bn is decreasing iii limb 0 Then the series 2 anbn converges MATH 6101 090 Fall 2006 98 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS This test is rather sophisticated Its main application is to prove the Alternating Series test but one can sometimes use it for other series as well if the more obvious tests do not work PROOF First let7s assume that the partial sums SN are bounded by K Next since the sequence bn converges to zero we can choose an integer N such that lbnl lt 62K Using the Summation by Parts lemma we then have n n1 Z ajbj lSnbn Smilbml Z 510 bi jm 739m n71 Klbnl Klbml K lbj 43ml Fm But the sequence bn is decreasing to zero so in particular all terms must be positive and all absolute values inside the summation above are super uous But then the sum is a telescoping sum Therefore all that remains is the rst and last term and we have 2 ajbj Kbm bn bm i b 2mm 739m But by our choice of N this is less than 6 if we choose n and m larger than the predetermined N Theorem 624 Alternating Series Test faL 2 a2 2 2 an 2 0 and an converges to zero then the alternating series 271 an converges This test does not prove absolute convergence In fact when checking for absolute convergence the term 7alternating series7 is meaningless It is important that the series truly alternates that is each positive term is followed by a negative one and visa versa If that is not the case the alternating series test does not apply while Abel7s Test may still work PROOF Let an 71 Then the formal sum 221 can has bounded partial sums although the sum does not converge Then with the given choice of bn Abel7s test applies directly showing that the series converges I 671 Riemann Series Theorem This is Theorem 615 We will prove it here Let me restate it in two parts MATH 6101 090 Fall 2006 67 CONVERGENCE TESTS 99 We are used to using the law of commutativity when adding real numbers It does not matter in which order we calculate a sum 3 M k1 we always end up with the same number We can express this in a more formal way by introducing the notion of rearrangement A rearrangement is a one to one onto function 7T 1n a1n As a consequence every number k E 1 n can be written as k 7TZ for some Z 6 1 n Often such a function is called a permutation of the numbers 1 n For example take the set 123 and consider the map This is a rearrangement or permutation of the the three numbers Question ln how many ways can we rearrange the numbers 1 n Returning to the series we can express the fact that it does not matter in which order we add the numbers by the equation 2 am z m k1 k1 What do we mean by a rearrangement of a series If we denote the natural numbers by N we have to be precise what we mean by a rearrangement of N We have to specify the rearrangement 7139 for in nitely many numbers Again we must have that 7139 is one to one and onto This means that every element k E N can be written as t for some Z 6 N With the help of such a map 7139 we can rearrange series simply by writing 2 W k It is natural to expect that for any convergent series it doe not really matter how we sum up this series 239e E a k E 1k k k The Reimann Series Theorem tells us rst that it is not true in general but it is true under certain reasonable conditions Theorem 625 Riemann Series Theorem Let 2k ak be an absolutely conver gent series then for any rearrangement 7139 we have that 2k a k 2k ak MATH 6101 090 Fall 2006 100 CHAPTER 6 SEQUENCES AND SERIES OF REAL NUMBERS Thus absolutely convergent series really do behave like nite sums when it comes to changing the order of summation PROOF We will do this by looking at the partial sums The partial sum for the rearranged series we will call whereas Since we are only interested in absolute convergence we may assume that the numbers 1k gt 0 De ne the number D k 7 k n 315 W l This number describes the range of the rearrangement in particular we have that Mk S k Dn for all 1 S k S 71 Now we have that n nDn Rn 2074103 2 ak k1 161 In the rst expression the n indices 7Tk for k 1 n are distinct numbers each somewhere between 1 and n Here is one place that we use the fact that the function 7139 is one to one These indices are also included on the right side of the equation but there may be more and hence the inequality sign In particular we realize that nDn Rn S Sn Z ak kn1 Since the series 2k ak is absolutely convergent we have that RnSSn f aim kn1 and as n a 00 Sn has a limit L and 211 ak a 0 Hence Rn is a bounded sequence and since it is increasing it has a limit R which satis es R S L Now we have to show that the converse is true Notice that we have not used the assumption that 7139 is onto which means that we do not miss any terms Pick any n MATH 6101 090 Fall 2006 67 CONVERGENCE TESTS 101 and notice that there exists a number so that the numbers a17 an are among the numbers a l a Em Thus Em Sn E 2 am k1 As n a 00 a 00 and since both sides converge7 we have L S R and our theorem is proven The following is one of the most scandalous results in series Theorem 626 Let 2k ak be a conditionally convergent series For any given num ber L there eists a rearrangement 7139 of N so that the series 2k a k converges to Thus we see that our expectations fail in a spectacular way and conditionally conver gent sequences continue to be promiscuous PROOF We assume that L is positive We have to nd the desired rearrangement The procedure goes as follows Make two boxes7 in one we put all the positive numbers ak in an ordered fashion and in the other we put all the negative numbers ak also in an ordered fashion Since the series is conditionally convergent we have that the series consisting of the numbers in the positive box diverges to 00 whereas the series consisting of the numbers in the negative box has to diverge towards foo Now start picking numbers from the positive box until the sum overshoots L for the rst time Notice that this is possible since the series in the positive box diverges Then we start adding the numbers from the negative box until we undershoot the number L for the rst time Notice again7 that this must happen since the series from the negative box diverges Now we keep doing this7 and produce in this fashion a sequence of partial sums of a rearranged series that oscillates about the value L it remains to see that this sequence converges to L7 but this follows immediately from the fact that the numbers ak a 0 as h a 00 MATH 6101 090 Fall 2006 10292008 MATH 6101 Fall 2008 The Cauchy Property 00 and oo 1They are not real numbers and do not numbers 2We often act as ifthey do 3We need guidelines u A A 1 J lt 1 lt for every real number a e R u no ice 00 a 1 Ha e Rthen we de nethe following 1 a no 30 7 no em a 3Ifz1gt0thenaxmmandaxrm no 4Ifalt0thenaxmrmandaxrm 30 We may adopt the following conventions 0 and arm 10202000 lens of Sequences 0011005 311315me Cameslmmue 0mm mew orem Thmnmglullulunlfwy cowarng mmw sbaumied Is the mumSe m7 15mm 0m everyhnnn led sequeuw mnvexges Mdapmnfnmmummmpb Aseqmwmnhsimnming mum m everyn Aseqmwa1sdwmxing nquot we everyn Aeeqmcemmeememmeme eememueasmnmeuem 10202000 Examples g ml an mp0 nfanmuwasmgseqmw mummm 10202000 M Theorem ThmnnuEwy baundedmnomm muamwnuayes Pmnf Lat 12 hman mmasmg sequlmealu l v r hummnm rmemlmnubnMam l nalln hnm lrmdubSaudlmwn Theorem Pmnf SmwulhSaudgt M munsz mm mm themsalumgerkm 0m Hmnm Thus anrukt ua wkaudhuafu lubS 10202000 Consequences g m dermal mplusemam mm number mumps g g g Lemming 4 Themquot we Jameyg 2 Lem nude 00 17 Consequences 2 Leta mudaanaquot 12 0 may 2 W7 Consequences 2 1202 mrlam mm mm anhatdnes mlwergquot Mam m u LV JXZLSHLX SL XZD he downtown2 10202000 Theorem mum L21 12 he asequeuw nfml mlmhus 0 ma 15 annnhnmuderlmnlmtnmm y 00 Mtg 15 annnhnmudedmnlm nmm ly Theorem mum Suppnse 0m 12 15 annulme musing sequeuw and 0 15 mmm demasmg sequins swhtha an bnmu w cf 0 a 102mmquot 090quot ThmnnuEuwy mmcemmzmm mmmmembwqumi Pm Lung be aseqmw Wesaythata Imansaammanngxfanmm ua m gt0 mnuhernn mte umbernfdmummug msll lntz 000 mqu be a Theorem Pmn mmmwl 10202000 Theorem Pmn mmmwl 0 km 12 has 6m numbernf 0mmme bmmmnsmma ythu hmxsawpnasn manSamSmwn w 175 an a a Thus wmsau s Tha mmsthat lnanhw mthmexislsa have gt00 u 1 SO Thmnm39Euaybomiedwqmnampham Bowng 63111va The Cauchy Property mam u Asequeuw 12 15 said u have annudeszn am am A lt 5 nalanKalu lm 0 Hum eqnwalems39atmem r Izmrkacm rganotl na wzl 10202000 The Cauchy Property 0 m 0501mmquot mam have anmdesznthaufnmgtK wu amr nk t Defmmons LA 12 he handed r mumgem m mm 0 dues mam memughehavmrnfa depends nldynnthe I017an seqmu 12 n gt M 100 gMan 0 gt00 Lewflnbimwm rm mm anems 0mm hesmu0 1 Defmmons Astuuases e se39sangtN get mumms 0500 0020203 Ln 0 lJmu3ldu 100M Enthexistrwhy39 Clamuusu 10202000 Defmmons x mnamscsmmsmqnm m ushmansu umumu dwh mma emnrm Mmqu uhusupahuluhmngtml kmmfmglbmnww 11m mf and hm sup Nntz Du 1m qu 0m 12 hmmd mmquot sm hnmuda lahnveJuhmn m and we sup an m 21mgquot 15 m bun012d helm glh an rm and we 000 hm mquot w 11m mf and hm sup Ismr wmamusumduhmm quotm Mw anl because whde ms me am hm s 11 s 100 9 mm mm vslwsanmayhemwhlalgu than hm sup 12 quotmg 0m 0m sup 17 15 the largest vain 0m m mzdy my an smuget dnse n 10202000 11m mf and hm sup Thmlun L0 12 heasequlmenfml mlmhus g I manxsrle lw asamlnmhenmnr mmmnmm 110000 17115 Proof Ncnmhamnmmqnm1m ufglhmn w gtM 2M SmfmgtNtheuum2NL unsuplznmm Dn hemsethatlmgm 4 mm Proof Suppnse 0m m N 0 Sn M an tELef2gtnThm15 lt2 lrngtN anltLotfnrngtN Thusuflnbfan wMSLot w M rquot 0 10202000 Emma 0 sz Smwhunfa slmswamwhaw mmwmn 01mm Proof hm ma a mme u gmmm 00 mpg 1 mm 1mg Lelbu Smld nu k ml t sn 0m Mm mde wm1mangtultmm unwrapquot Proof SmdarlysnwL 1m hmsNcn Sn um Lrglbm wka Thuglbmn ngtNgtL 7 My 1 mm 1025200110er116me ngtmaxNuN Eqmalmly aquot in an gtmaxNuN ThlspmvesfhafhmanL 11m mf and hm sup This mils mm 02 my than hm mquot hm sup 12 10202000 M M Elhmn wmmlslbednsetnge wn39fhls means haka n heumhexsm wset anxwmmwmmgm Theorems 4 am rz zrmsmrzwoxa zw Mondaummcgmm mx b nrlltz HmngtN m urumxsxuruoxumru kwm ugmd cm wmm Theorem only 00 hm whme ymemm slmmla r mm m mm mlwex ge 12101 mmcmw wqu smmmn Lumpa Proof 10202000 mmsaner smhaufmwN wu anr v n v rquot u mpgtNThsshnws hammousannpper Thus Proof Them mmpansusu smmno Sm 005 must all 2 gt n we have um sup an hm mquot qmmmesam equal Problems Cnmpme the 1mm 0 2x195 mud Problems Cnmpme m m n 2x195 10252003 Problems Cnmpme m m n 2x195 fwd Problems Cnmpme m m n 2x195 mud Problems Cnmpm the mm M 2x195 an n and ui wk 4 1192008 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 1152003 MATH 6101 Fall 2008 Calculus from Archimedes to Fermat 9182008 A Request Please de ne a relative maximum Please de ne a relative minimum How can you tell them apart The Derivative A Chronology Used od hue to solve particular problems Discovered as a general concept red and developed in applications to mathematics and physics 4 De ned rigorously w til 8 o 9182008 Curves and Tangents Greeks mainly known from work of Archimedes had studied some curves a circle 7 Come sections parabola ellipse hyperbola e Spirals 7 Others defined as loci ofpomts Muslim scholars studied afewmore Many problems studied especially nding their tangents and areas W mna i Move to Medieval Europe 1591 r Frangois Viete Vieta e Isdgege in mm nalyticam introduced symbolic algebra without an equal sign W mm 5 Algebra and Curves n the 1630 s Descartes and Fermat independently discoveredinvented analytic geometry W mna Algebra and Curves with this algebra there was an explosion of curves to study work New method required for nding tangents and areas 9182008 Algebra and Curves Tangents Areas Extrema r from the Greeks came isoperimetric problems 7 quot 39 perimeter which one has the maximal areaquot rmat and Descartes had hopes for these being answered by symbolic algebra W warm de Roberval s Method of Tangents a 5 r T a motion A curve is sketohea by a moving pD39nl The anger is he sam of motion W warm de Roberval s Method of Tangents 9182008 Vectors V1 and V2 at a polnt P V 1S tn the same dlreetlon as parabola s and the pomt p V2 1S perpendlcular to the dlrecmx The tangent to the graph at y VV1V2 eyelotd but could not generallze W new Fennat s Method of Derivatives Fermat s Illustration rm n n in 39 39 39 product ofthe parts will be a maximum Let b length ofthe line a length of the rst part abra abra Pappus ofAlexandrla e a problem Whlch In general has two soluuons wlll haye only one solutlon m the case ot a maxlmum W mam Fermat39s Method Suppose that there is a second solution Then the k a t wouldbe brt1 e be are Multiply the two parts together berazraerearezabrt1272t1ebere2 By Pappus there is only one solution so set these e 31 t one another t1brt12t1brt1zr2t1ebare2 e e W mam Fermat39s Method abrazabrazr2aebere2 e be 21 e b Now Fermat says suppress aquot and we get 1 b2 which is the point at which the maximum occurs W mm 1 9182008 Fermat39s Method Note that Fermat did NOT caueiniinite1ysman say thatevanished use alimit exptain why he could divide by e and then treat it as 0 At this point he did not make the connection tan ent W mm Fermat39s Method 7 Modern Notation 7 11m Finding tangents Draw pointx andw39 l consider a point a distance a away From the gure the following relationship exists Lex se 7 fx e W mm 5 Fermat39s Method 7 Modern Notabon 3 sum m x Kn Xe Thedeumumamnshsdx mmal slaps fmy 5132003 Fermat39s Method 7 Modern Notahon ftxx nerawe x rix ye x y a few 4 He sets 2 n 35 m mk 4 y Fermat and Tangent yam aw zlvmhymm mnmwsv mammme anaxmmpm m mwimhhx ll mam manme mmnm mnlulmz rm WEEK w 9182008 Tangents Descartes Isaac Barrow John W 15 Re e S Christopher Huygens All had methods of nding the tangent By 1660 we had what is nowknown as Fermat s tongentline has slope u Had no connecrion to the process ofcompu ng areas W mm a Early Calculations of Area We say what Archimedes had done with the area between the parabola and a secant line rm i i a r r reometric series 7 referrinr arithmetic series Areas of general curves needed symbolic algebra W mm s Bonaventura Cavalieri 1598 7 1647 o r at i o E inmlpora n Kepler39s hemy of in nitesimally small georrieoic quantities Allowed him to nd simpiy and rapidly area and volume of various geometric gures W mm a 5132003 Hemwd wcum astheslnunf spnm39snr m lmslhls Ldeewsem mdwmb15 thatmmpnsedan Keplerhadrlnlwm hm mhmhmsm Cav xen39s Method Cavalxen39s Method Cavahmmmmlzd u mun rshlgmmhn vahn my Hzm md uahmwmnfmwmdamhhxmnm mm Cavalieri 5 Method For at any distance xalong the rains the height of the parabola would be X2 Therefore the area of the rectangle enclosing the rectangular subdivisions at a point xwas equal to gt00 or s From his earlier result the area underneath the parabola is equal to 13 the area of the ooundingrectangle 1 Area under 12 13 3 W mum as 9182008 J o hn Wallis Wallis showed that the area function for the curve y 1m is A n1 is true not only ior positive integers but for negative and fractional exponents as well Also integrated polynomials W mum i F rmat slntegration Format used the concept of in nite series 1 Kala 23K 21K W mm a Fermat39s Integranon Chanse n lt elt 1 HM mayquot 04 exre xxuyh 2K172Xe quot lreequotquot x quot w e xr xllw21ree xquot xree 1W 172 lw quot e mhexrm 5132003 Fermat39s Integrauon Hym mmermq f u 1 exm m H mnn 3 1 175 an A m k Hm Hg Fermat39s Integranon 1104 1 175 172quotquot E lEXEEx 51 M A f 175X1EE 5 J Mxm 15 E 43quot 5132003 Fermat39s Integranon m s L m A111 m Xm q M 111 mquot MATH 6101 Fall 2008 In nite Series and Convergence 1152008 Definition Given any sequence a we associate a new sequence 5quot of partial sums 5 u aquot We de ne the szrizs 21 to be the limit an limM 5 lfthe sequence ofparrial sums converges we say that the in nite series converges Otherwise we say that the series is divergent Wm mam Examples Z 1 1 quota nn1 Wm mam 1152003 Examples 1 7 mm Inthlsmsewehaveseeu m squot 9 x2 23 34 45 with Other Examples nu these mlwexge m dx mg xz3 s n uaqmmungumppemsaupmsmus wr r rrrr w 4 h l w u dim mmq aas numgxpmmmsm u v s 2 H 47 seuegpnql sammpsms mp pus mqlmul rm 2 um um seueg pumeg saxxmp sumsqu mnmhas aqua mm m Wu 2 H W mm M mang y svszxz seueg 13nd HGGZSU u n Dnes hlsmlwexge my my Wehmwthat 2W nmxm 2 gt Mn1 n quot 1 quot 2 lt Zr 2mm 2 Human dues muvmgs 1152003 Con rmed Wemtedmher m dnpmwdmms m 1 1 1 4 2w TV 7 Wealmhmwmnm z 5 2quot m whmB s a thmmuII number Euler mdywemmmugnmmmm Bernoth Numbers The Eenmulh umbnanwmdxsmvmdby he suns nfpnwexs Elfaquotxquotzquot3quot quot mar 25quot 37 Bernoth Numbers 16 swam z o 1 L m w i l Qi mq 1152003 Bernoth Numbers wz Zng egm quot x x s x s 2 z a z m k nN u39w39r uu m m m n a z a 7 m 3 m Bernoth Numbers Eenmulh mm 2 z 3 3 5b 1707700172 VSXHrmzrsXprblmyu z 3 m 78 Bernoth Numbers Thepth Eenmu mmhens the me wmnfu Dmmduuqus ugemnmg w enm h mlmhus mm m 1152003 Bernoth Numbers 8 Senes We muhe able m Shaw lam am the genes g muverresxfaudnldyxfpgt L Theeasexpmnf mums ahtde cams 1152008 Absolute Convergence If the terms a of an in nite series 2a are all nonnegative then the partial sums 5quot form a nonedecreasing sequence Therefore go either converges or diverges to oo glo l is nonedecreasing for any sequence The series go is saidto conoerge absolutely if Zla l converges Conditional Convergence A series converges conditionally ifit converges but not absolutely Does the series 24quot converge absolutely conditionally or not at all Does the series 202quot converge absolutely conditionally or not at all Does the series armn converge absolutely conditionally or not at all this series is called alternating harmonic series mom min a Order of Summatlon Thmlun xJLelZa bean mm comm m 71m any mungme o wmsm m abmlmlywnwym othemebmh mes M vmy mI mm zhmmz mm mew1 mmgem c Tn he 1me lam 1152003 Algebra of Serles mmmmmmmmxm TM 3 mxmnf amxmlmahnhndy mmmmbkm39zb mmmammgmmxmmmw mmmmnudemdm mmnnf awa Algebra of Serles mummammhasmmzm 6 whch

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