Lecture Notes On Advanced Calculus II

1 Lecture Notes On Advanced Calculus IIJie WuDepartment of MathematicsNational University of SingaporeContentsChapter 1. Sequences of Real Numbers51. Sequences52. Limits of Sequences53. Sequences which tend to 84. Techniques For Computing Limits95. The Least Upper Bounds and the Completeness Property ofR126. Monotone Sequences157. Subsequences168. The Limit Superior and Inferior of a Sequence179. Cauchy Sequences and the Completeness ofR25 Chapter 2. Series of Real Numbers311. Series312. Tests for Positive Series343. The Dirichlet Test and Alternating Series434.

2 Absolute and Conditional Convergence495. Remarks on the various tests for convergence /divergence of series50 Chapter 3. Sequences and Series of Functions531. Pointwise Convergence532. Uniform Convergence573. Uniform convergence of{Fn}and Continuity634. Uniform convergence and Integration655. Uniform convergence and Differentiation706. Power Series757. Differentiation of Power Series838. Taylor Series87 Bibliography973 CHAPTER 1 Sequences of Real Numbers1. SequencesAsequenceis an ordered list of numbers. For example,1,2,3,4,5,6 The order of the sequence is important.

3 For example,2,1,4,3,6,5is different from above sequence. Aninfinitesequence is a list which does not example,1,1/2,1/3,1/4,1/5, We are going to study infinite sequences. We denote by{an}the sequencea1, a2, a3, , an, are some examples of infinite sequences.(1). 1,12,13, ,1n, (2).13,132,133,134, (3). 1, 2,3, 4,5, Can you find a formula for each of the above sequences?Answer:(1).an= 1/n. (2).an= 1/3n. (3). ( 1)n Limits of {an}isA, and is written aslimn an=A,if for any >0, there is a natural numberNsuch that for everyn > N, we have|an A|<.

4 Some sequences do not satisfy the above. We call such Sequences which satisfy the above definition, and is finite, are the following limits by using Ndefinition1) limn 1n= SEQUENCES OF REAL NUMBERS2) limn n2n2+ 1= ) limn (34)n= (1). Given any >0, we want to findNsuch that 1n 0 < forn > N, ,n >1 forn > N. ChooseNto be the smallest integer such thatN 1 .(Nis found now!) Whenn > N, thenn > N 1 or 1n 0 < . Thus limn 1n= 0.(2). Given any >0, we want to findNsuch that n2n2+ 1 1 < forn > n2n2+ 1 1 < n n2+ 1 1 < n n2+ 1 n2+ 1 < n2 (n2+ 1) n2+ 1(n+ n2+ 1) < 1 n2+ 1(n+ n2+ 1)< n2+ 1(n+ n2+ 1)>1 Observe that n2+ 1(n+ n2+ 1)> nforn 1.

5 ChooseNto be the smallest integer such thatN 1 . Then, forn > N, n2+ 1(n+ n2+ 1)> N2+ 1(N+ N2+ 1)> N 1 or n2n2+ 1 1 < . ThusNis found and hence the result.(3). Given any >0, we want to findNsuch that (34)n 0 < forn > that(34)n< nln(34)<ln( ) n >ln( )ln(3/4)( (3/4)<0!!) ChooseNto be the smallest positive integer such thatN ln( )ln(3/4). Whenn > N, thenn > N ln( )ln(3/4)or (34)n 0 < . The proof is finished. {an}has a limit, then the limit is LIMITS OF limits of{an}. Suppose thatA6=B. Choose =|A B|2. Then >0 becauseA6=B.

6 By definition, there existsN1andN2suchthat|an A|< forn > N1and|an B|< forn > N2. Forn >max{N1, N2}, wehave|A B|=|(A an) + (an B)| |A an|+|an B|<2 = 2|A B|2=|A B|,which is a contradiction. ThusA=B. (Squeeze or Sandwich Theorem).Given3sequences{an},{bn},{cn}such that(i)an bn cnfor everynand(ii) limn an=A= limn cn,thenlimn bn= any >0, there existsN1andN2such that|cn A|< forn > N1and|an A|< forn > N2. LetN= max{N1, N2}. Forn > N, we have < cn A < and < an A < A < cn< A+ andA < an< A+ .ThusA < an bn cn< A+ or|bn A|< .By definition, we have limn bn=Aand hence the result.

7 Above theorem is still applicable if the inequalityan bn cnis limits1) limn 1 + )(3n 14n+ 1) (1). Since0 1 + sinnn 2nand limn 2n= limn 0 = 0, we havelimn 1 + sinnn= 0.(2). Since0 (3n 14n+ 1)n (34)n81. SEQUENCES OF REAL NUMBERSand limn (34)n= limn 0 = 0, we havelimn (3n 14n+ 1)n= 0. 3. Sequences which tend to {an}tends to+ if for each positive numberk, there is anNsuch thatan> kfor alln > such sequences, we write asan + asn orlimn an= + . following sequences tend to + 1)an= )an= (3/2) sequences lnn, n2and etc then tend to . (Reciprocal Rule).

8 Consider a sequence{an}.(i)Ifan>0for allnandlimn 1an= 0, thenlimn an= + .(ii)Iflimn an= , thenlimn 1an= only prove (i). For each positive integerk, there existsNsuch that 1an 0 <1kforn > Nbecause limn 1an= 0. Then, forn > N,an> kbecausean>0. Thisfinishes the proof. limn 1 n= 0, we have limn n= . Similarly, sincelimn n= + , we have limn 1 n= TECHNIQUES FOR COMPUTING LIMITS94. Techniques For Computing a continuous function. Thenlimn f(an) =f( limn an).Idea of the definition of continuity, whenx x0,f(x) f(x0).Now limn an=Ameans thatan Awhenn.

9 Thusf(an) f(A) whenn , that is,limn f(an) =f(A) =f( limn an). sin(n 2n+ 1)= limn ( 2 + 1/n)= sin( 2)= (L Hopital s Rule).Supposean=f(n),bn=g(n). Iflimn f(n)g(n)isof the form or00, thenlimn f(n)g(n)= limn f (n)g (n).History the theorem is named after Marquis de l Hospital(1661-1704), it should be called Bernoulli s rule. The story is that in 1691, l Hopitalasked Johann Bernoulli (1667-1748) to provide, for a fee, lectures on the new sub-ject of Calculus . L Hopital subsequently incorporated these lectures into the firstcalculus text,L Analyse des infiniment petis (Analysis of infinitely small quantities),published in 1696.

10 The initial version of what is now known as l Hopital s rule firstappeared in this thatlimn (1 +xn)n= ln[(1 +xn)n]= limn nln(1 +xn)= limn ln(1 +xn)1/n= limn 11+xn ( xn2) 1n2= limn x1 +xn= limn (1 +xn)n=ex. anandlimn bnexist, then(1). limn (an+bn) = limn an+ limn bn,(2). limn kan=klimn an,(3). limn anbn= limn anlimn bn,(4). limn anbn=limn anlimn bn,providedbn6= 0 and limn bn6= SEQUENCES OF REAL the limit ofln(n2+ 3n+ 22 + 4n+ 2n2)+ cos(1 n). [ln(n2+ 3n+ 22 + 4n+ 2n2)+ cos(1 n)]= limn [ln((n2+ 3n+ 2)/n2(2 + 4n+ 2n2)/n2)+ cos(1 n)]= limn [ln(1 + 3/n+ 2/n22/n2+ 4/n+ 2)+ cos(1 n)]= ln(1 + 0 + 00 + 0 + 2)+ cos 0 = ln(12)+ 1 = 1 ln 2.