MAT25 LECTURE 11 NOTES - math.ucdavis.edu

1 MAT25 LECTURE 11 NOTESNATHANIEL convergence TheoremDefinition 1: Increasing, Decreasing, and MonotoneA sequence (an) isincreasingifan an+1for alln Nanddecreasingifan an+1for alln N. Asequence ismonotoneif it is either increasing or 1: Monotone convergence Theorem (Abbott Theorem )If a sequence is monotone and bounded, then it consider two cases. (anis increasing). Since (an) is bounded,??implies that the setA={an|n N}is a bounded subset ofR. Hence the Least Upper Bound Property ofRimplies thatAhas a supremum. Lets= supA. We willshow that (an) s. Given >0, sinces= supA, there exists some elementaN Asuch thats < anyn N, since (an) is increasing, we have thataN an, and therefores < anas well.

2 Butsincean A, it follows thatan s < s+ . Therefores < an< s+ , and hence|an s|< for alln N. Hence we have shown that for all >0, there existsN Nsuch that ifn N, then|an s|< ,which implies that (an) s. Hence indeed (an) converges. (anis decreasing). Homework. Exercise 1: MCT Decreasing CaseProve the Monotone convergence Theorem (??) for decreasing .. SeriesDefinition 2: SeriesLet (bn) be a sequence of real numbers. Aninfinite series(or justseries) is a formal expression of the form n=1bn=b1+b2+b3+b4+b5+..Definition 3: Sequence of Partial SumsGiven a series n=1bnof real numbers, thesequence of partial sumsof this series is the sequence (sm)wheresm=m n=1bn=b1+b2+.

3 + LECTURE 11 NOTES2 Definition 4: convergence of a SeriesA series n=1bnconverges to a real numberBif its sequence of partial sums (sm) converges (in the senseof??) toB. In this case we write n=1bn=B. If a series does not converge to any real number, we 1: convergence of Series (Abbott Example )Consider the series n= of the terms in this series are positive, hence the sequence of partial sums (sm) withsm= 1 +14+19+..+1m2,is an increasing sequence. Furthermore, we computesm= 1 +14+19+..+1m2= 1 +12 12+13 13+..+1m 1m<1 +12 11+13 12+..+1m 1m 1= 1 +(1 12)+(12 13)+..+(1m 1 1m)= 1 + 1 1m< implies that the sequence (sm) of partial sums is bounded.

4 Since we already showed it was increasing,by the Monotone convergence Theorem (??), (sm) converges to somes R, hence the series n=11n2converges tos. However at this point in the course we are not yet able to LECTURE 11 NOTES3 Example 2: Harmonic Series (Abbott Example )Consider theharmonic series n= sequence (sm) of partial sums of the harmonic series is given bysm= 1 +12+13+..+ particular, we list several terms of this sequences2= 1 + 1 +12+(13+14)>1 +12+(14+14)= 1 +12+12s8= 1 +12+(13+14)+(15+16+17+18).>1 +12+(14+14)+(18+18+18+18)= 1 +12+12In general, we computes2k= 1 +12+(13+14)+.

5 +(12k 1+ 1+..+12k)>1 +12+(14+14)+..+ 12k+..+12k 2k 1 = 1 +12+12+..+12 k= 1 +k(12).Therefore the sequence of partial sums (sm) is unbounded. Since convergent sequences are bounded (??),(sm) cannot be .. 5: Subsequence (Abbott Definition )Let (an) be a sequence of real numbers, and letn1< n2< ..be an increasing sequence of natural the sequence(an1,an2,an3,..)is called asubsequenceof (an) and is denoted by (ank) wherek Nindexes the LECTURE 11 NOTES4 Example 3: SubsequencesConsider the sequence (an) = (1n) n=1. One subsequence of this sequence is(ank) = (1,1/3,1/5,1/7.)

6 Wherenk= 2k 2: convergence of Subsequences (Abbott Theorem )Let (an) be a sequence of real numbers which converges toa R. Then any subsequence (ank) of (an) alsoconverges >0 be arbitrary. Because (an) a, there existsN Nsuch that for alln N, we have|an a|< .Note thatnk kfor allk N. Therefore ifk N, we havenk k N, which implies that|ank a|< .Therefore (ank) aas well. Example 4: convergence of Subsequences (Abbott Example )Let 0< b <1. Then multiplying both sides of the inequalityb <1 bybkyieldsbk+1< bkfor allk , we haveb > b2> b3> ..Therefore the sequence (bn) is decreasing, and bounded (by|bk|<1 for allk N), the Monotone Conver-gence Theorem (?)

7 ?) implies that (bn) converges to some limit`. Consider the subsequence (b2n). By??,(b2n) converges to the same limit`as (bn). But we have thatb2n=bn bn. Therefore the algebraic limittheorem implies that`= limn (bn bn) =(limn bn) (limn bn)=` implies that either`= 0 or`= 1. However all terms of this sequence satisfybn b. Therefore by theOrder Limit Theorem (??),` b <1. Hence`= 5: Using Subsequences to show their Parent Sequences Diverge (Abbott Example )Recall the sequence (an) given byan= ( 1)n+1. Explicitly, this sequence is given by(1, 1,1, 1,1, 1,..).We can finally show that (an) diverges.

8 Suppose, for contradiction, that (an) a. Then we have thesubsequence (a1,a3,a5,..) = (1,1,1,..). This is a constant sequence, and therefore converges to 1. By??, it follows thata= 1. Similarly, though, the subsequence (a2,a4,a6,..) = ( 1, 1, 1,..) converges to 1. Thereforea= 1. Since we showed that limits are unique and 16= 1, this contradicts our assumptionthat (an) converges. Hence it .. TheoremTheorem 3: Bolzano-Weierstrass Theorem (Abbott Theorem )Every bounded sequence contains a convergent (an) be a bounded sequence. Then there exists someM >0 such that|an| Mfor alln N.

9 Thisimplies thatan [ M,M] for alln LECTURE 11 NOTES5 We bisect the interval [ M,M] into two closed intervals of equal length. One of these intervals mustcontain infinitely many terms of the sequence (an). LetI1be that interval, and letan1be any point inI1. Next we bisectI1into two closed intervals, and note that one of these intervals must contain infinitelymany terms of the sequence (an). LetI2be that interval, and choosean2inside this interval which satisfiesn2 n1. In general, we bisectIk 1into two closed intervals, one which must contain infinitely many terms of (an).

10 LetIkbe this closed interval, and chooseank Iksuch thatnk> nk we have obtained a subsequence (an1,an2,..) of (an) and a sequence of nested intervalsI1 I2 I3 ..By the Nested Interval Property (??), the intersection n=1 Inis nonempty, and therefore contains some claim that (ank) converges tox. Let >0 be arbitrary. Then the length of the intervalIkisM(12)k , the sequence(12k 1)converges to 0. Therefore, by the Algebraic Limit Theorem (??), the sequence(M(12)k 1)converges to 0 as well. Hence, there exists someN Nsuch that ifk N, then|M(12)k 1|< .But thenank,x Ik, hence|ank x|<.