Every complex Cauchy sequence is convergent Proof Put zn = x + iy Then xn is Cauchy: xx − xm ⩽ zn − zm (as
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A sequence xn ∈ R is said to converge to a limit x if • ∀ϵ > 0, ∃N s t n > N ⇒ xn − x < ϵ A sequence xn ∈ R is called Cauchy sequence if • ∀ϵ, ∃N s t n > N m > N ⇒ xn − xm < ϵ Every convergent sequence is a Cauchy sequence Proof
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Every complex Cauchy sequence is convergent Proof Put zn = x + iy Then xn is Cauchy: xx − xm ⩽ zn − zm (as
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(c) Show that the sequence xn is bounded below by 1 and above by 2 (d) Use (e) Use (d) in a proof to show that Sn is Cauchy and thus converges In the first case, if k = 0, then we wish to prove that k · f(x) = 0, the zero function, has limit 0
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ANALYSIS I
9 The Cauchy Criterion
9.1 Cauchy"s insight
Our difficulty in proving "an→?" is this: What is?? Cauchy saw that it was enough to show that if the terms of the sequence got sufficiently close to each other. then completeness will guarantee convergence. Remark.In fact Cauchy"s insight would let usconstructRout ofQif we had time.9.2 Definition
Let (an) be a sequence [RorC]. We say that (an) is aCauchy sequenceif, for allε >0 there existsN?Nsuch that m,n?N=? |am-an|< ε. [Is that all? Yes, it is!]9.3 Cauchy=?Bounded
Theorem.Every Cauchy sequence is bounded[RorC].
Proof.1>0 so there existsNsuch thatm,n?N=? |am-an|<1. So form?N, |am|?1 +|aN|by the Δ law. So for allm |am|?1 +|a1|+|a2|+···+|aN|.9.4 Convergent=?Cauchy[RorC]Theorem.Every convergent sequence is Cauchy.
Proof.Letan→land letε >0. Then there existsNsuch that k?N=? |ak-l|< ε/2Form,n?Nwe have
|am-l|< ε/2 |an-l|< ε/2 So |am-an|?|am-l|+|an-l|by the Δ law < ε/2 +ε/2 =ε19.5 Cauchy=?Convergent[R]
Theorem.Every real Cauchy sequence is convergent.
Proof.Let the sequence be (an). By the above, (an) is bounded. By Bolzano-Weierstrass (an) has a convergent subsequence (ank)→l, say. So letε >0. Then ?N1such thatr?N1=? |anr-l|< ε/2 ?N2such thatm,n?N2=? |am-an|< ε/2Puts:= min{r|nr?N2}and putN=ns. Then
m,n?N=? |am-an| ?|am-ans|+|ans-l| < ε/2 +ε/2 =ε9.6 Cauchy=?Convergent[C] Theorem.Every complex Cauchy sequence is convergent. Proof.Putzn=x+ iy. Thenxnis Cauchy:|xx-xm|?|zn-zm|(as|?w|?|w|). So x n→x,yn→yand sozn→x+ iy.9.7 Example Let a n= 1-12 +13 +···+(-1)n+1nThen withm?n, andm-nodd we have
|am-an|=?1n+ 1-1n+ 2????
+1n+ 3-1n+ 4+...???? +1m-1-1m1n+ 1????
-1n+ 2+1n+ 3???? -...-1m-2+1m-1???? -1m1n+1?1n
Ifm-nis even, we write
|am-an|=|am-am-1+am|?|am-am-1|+|am|?1N+ 1+1m ?2n Letε >0. ChoseN >12εand convergence follows. Question:What is limn→∞an? Later we"ll see it is log2. 210 Series
10.1 Definition
Note.So far we have always used sequences defined by functionsa:N→R. It is convenient from now on to start off ata0, that is to work with functionsa:N? {0} →R. So let (an)∞n=0be a sequence of real (complex) numbers. Definesm:=?m n=0an, a well defined real (complex) number. Consider the sequence (sn)∞n=0. We call this the series defined by the sequence (an), and we denote it by?∞ n=0an. Note.This is a very odd name indeed! Don"t let it mislead you:?∞ n=0anis just the sequence a0,a0+a1,a0+a1+a2,....
10.2 Examples
(i)The Geometric SeriesLetan:=xnThen?xnis (1,1 +x,1 +x+x2, ... ,1 +x+x2+···+xn, ...) (ii)The Harmonic SeriesLetan:=1n+1. Then?1n+1is1,1 +12
,1 +12 +13 (iii)The Exponential SeriesLetan:=xn/n!. Then?xn/n! is1,1 +x,1 +x,1 +x+x22!
(iv)The Cosine SeriesLet a n=? x2m2m!(-1)mifn= 2m0 otherwise
Then ?anis?1,1,1-x22!
,1-x22! ,1-x22! +x44!10.3 Convergence and Divergence
With the setup above if (sn) converges, then we say "the series?∞ n=0isconvergent".Otherwise we say that the series isdivergent.
310.4 More examples
(i) Letan=xn. (a) Ifx= 1 then?xnis divergent. Proof.s0= 1,s1= 2, ...,sn=n+ 1 by induction.(b) Ifx?= 1, thensn=1-xn+11-x. Proof.Pure algebra.(c) If|x|<1 then?xnis convergent. (d) If|x|>1 then?xnis divergent. Proof.If|x|<0 thenxn→0 asn→ ∞. So by AOLsn→1/(1-x). Ifsn→s thenxn+1→11-(1-x)sby AOL. But if|x|>1,xnis not convergent.(ii) Letan=1n . Then?1n is divergent.Proof.s2n-1= 1 +12
+···+12 n= = 1 + ?12 ?+?13 +14 ?+?15 +16 +17 +18 ?1 +12 +24+48
+...+2n-12 n ?n2 so (sn) has a subsequence which is not convergent.(iii) Letan=1(n+1)2. Then?1n
2is convergent.
Proof.
s n= 1 +14 +19 +...+1n2+1(n+1)2
?1 +11·2+12·3+...+1n(n+1) = 1 +?11 -12 ?+?12 -13 ?+...+?1n -1n+1? = 2-1n+1?2 So (sn) is monotone increasing [sn+1=sn+1(n+1)2?sn] and bounded by 2. So (sn) is convergent by Bolzano-Weierstrass (??).410.5 Notation and its abuse
More notation: if the series
n=0anis convergent then we often denote the limit by?∞ n=0an, and call it thesum. Note.We must take great care, but this double use is traditional. I will try to distinguish the two uses, and say "The series?an" and "The sum?an". I suggest that you do the same for a bit. Note.Just attaching the label "sum" to the symbol?∞ n=0andoes not turn it into a proper mathematical sum. Look at our axioms forR. They only speak of adding pairs of real numbers, which we can extend using axiom A2 to finite sets of real numbers. But given an infinite set of real numbers we can"t simply "add" them and get a "sum". Instead we have to talk about sequences ...10.6 Tails
Let (an) be a sequence and (sn) be the corresponding series. Sometimes we want to look at (ak,ak+1,ak+2,...). We write this series?∞ n=kan. We put S n=ak+ak+1+···+anand note that what we said about Tails of sequences, and adding constants, ensures that (sn) is convergent if and only ifSnis convergent.10.7 Cauchy"s criterion
We rewrite Cauchy Criterion for series.
Theorem.The series?∞
n=0anis convergent if and only if for allε >0there existsN?N such that l?k > N=?? ????l n=ka nA genuine
sum Note.Clearly in practice when we estimate the sum we"ll use theΔlaw when we can.10.8 Absolute Convergence
Letanbe a sequence. Then we say that?anisabsolutely convergentif the series?|an| is convergent.10.9 Absolute Convergence=?Convergence
Theorem.If?anis absolutely convergent then it is convergent.Proof.Letε >0 Then there existsN?Nsuch that
l?k?N=?? ????l k|an|? 5 So l?k?N=?? ????l ka n? ?????l? k|an|By the Δ law=?
????l k|an|? ????< εNote.Why is absolute convergence a good thing to have? Because it makes use of Cauchy criterion easy!10.10 Examples
(i) ?xnabsolutely convergent if|x|<1 (ii) ?(-1)nn2is absolutely convergent
(iii) ?sinnn3is absolutely convergent
(iv) ?(-1)nn+1is convergent, but not absolutely convergent.10.11 Re-arrangements
Letp:N-→None-to-one and onto. We can then putbn=ap(n)and consider?bn, which we call arearrangementof the series?an. Funny this can happen! Later on we will be able to prove thatExample.
1-12 +13 -14 +...→log2 1 + 13 +15 -12 +17 +19 +111-14 +...→log(2/3) Theorem.If(an)is absolutely convergent and(bn)is a rearrangement of(an)then?bnis absolutely convergent too.