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LECTURES 16 AND 17: SUMMARY

In these two lectures, we proved a number of fundamental results about convergence of sequences and series. We started with an easy observation: that if a sequence converges to some number, then the terms of the sequence are eventually all close to one another. More precisely, we proved the following. Proposition 1.If a sequence(an)converges, then for any >0there existsNsuch that janamj< for allm;nN.

Proof.Given >0. Let

A:= limn!1an:

There existsNsuch that for allnN,

janAj< =2:

Thus, for allm;nN,

janamj=j(anA) + (Aam)j jamAj+janAj< : Remarkably, the converse of this proposition is also true. To state it in a cleaner way, we made a definition: Definition.A sequence(an)is said to be aCauchy sequenceiff for any >0there existsNsuch that janamj< for allm;nN. In other words, a Cauchy sequence is one in which the terms eventually cluster together. We will prove (over the course of2 +lectures) the following theorem: Theorem 2(Cauchy Criterion).A sequence is Cauchy iff it converges. Whyisthisuseful? Usuallywhenweexploretheconvergenceofasequence, wefirstguesswhether or not it converges (and what it converges to), and then verify the guess with anproof. This approach works for nice sequences, but often it"s not clear how to guess about the convergence. The Cauchy Criterion allows us to shift from an external point of view - one in which we know not only the sequence, but also the limit of that sequence - to an internal one, where we can decide convergence based purely on the behavior of the sequence itself. One nice example of this is the construction ofR. One way of doing this is to consider all Cauchy

sequences consisting of rational numbers. Every such Cauchy sequence converges tosomething,Date: March 7th and 12th, 2013.

butthissomethingmightbeirrational. ToeveryCauchysequenceofrationalnumbers, weassociate a symbol(intuitively,is the limit of the sequence). We can define addition and multiplication of real numbers in terms of operations on the underlying Cauchy sequences, and thus construct all ofR. Although this is actually a bit painful to carry out rigorously, the underlying idea is elegant and straightforward, and what makes it work is that we know that Cauchy sequences converge.

Before proving Theorem

2 , we prove the following nice consequence:

Corollary 3.If the series1X

n=1janjconverges, then so does1X n=1a n. If 1X n=1janjconverges, we say that the series1X n=1a nconverges absolutely. Thus, the above corollary

asserts that if a series converges absolutely, then it converges. It turns out that the converse to this

is false, as we shall see soon.

Proof of Corollary.Let

A N:=X nNjanjandSN:=X nNa n: We are given that(AN)converges, and wish to prove that(SN)does as well. Given >0. Since(AN)converges, Theorem2 implies that (AN)is Cauchy, whence9Ksuch that jANAMj< for allM;N > K. Pick anyM;N > K; without loss of generality, we haveNM. Then jSNSMj=jaM+1+aM+2++aNj jaM+1j+jaM+2j++jaNj=jANAMj< : Thus,(SN)is Cauchy. The Cauchy Criterion implies that it converges. Thus, the Cauchy Criterion makes this result rather trivial to prove. I challenge the reader to come up with an alternative proof of the Corollary, without relying on the Cauchy Criterion. This can be accomplished (you"ll see one approach on the assignment), but it isn"t easy.

We are now ready to prove the Cauchy Criterion.

Proof of Theorem

2 .We"ve already proved that if a sequence converges, it is Cauchy. It therefore suffices to prove that a Cauchy sequence(an)must converge. The tricky part is that we have no idea,a priori, what it converges to! This is what makes the proof challenging. The proof proceeds in several steps, which we isolate and prove subsequently.

Step 1.Since(an)is Cauchy, it must be bounded.

Step 2.Since(an)is bounded, it has a convergent subsequence(ank). Letx:= limk!1ank.

Step 3.limn!1an=x.

This is the big picture of the proof. We now justify the individual steps. Steps 1 and 3 are fairly straightforward; Step 2 will take some more effort. We proceed in order. Proposition(Step 1).If a sequence(an)is Cauchy, then it is bounded. Proof.There existsNsuch thatjanamj<1for alln;mN. In particular, we have janaNj<1for allnN, whence a n2(aN1;aN+ 1) for allnN. Thus,fan:nNgis bounded. Also,fan:n < Ngis bounded (since it"s finite). We conclude that the entire range of the sequencefangis bounded, as claimed. Our proof of Step 2 will rely on the following result: Theorem(Monotone Subsequence Theorem).Every sequence has a monotone subsequence. We will prove this in the next lecture. For the time being, we take it on faith and prove Step 2. Theorem(Step 2, aka Bolzano-Weierstrass).Every bounded sequence has a convergent subse- quence. Proof.Suppose(an)is a bounded sequence. By the Monotone Subsequence Theorem, it has a monotone subsequence(ank). But then this subsequence is both bounded and monotone, whence (by the MCT) it is convergent.

Finally, we prove Step 3:

Proposition(Step 3).If a subsequence of a Cauchy sequence converges tox, then the sequence itself converges tox. Proof.Let(an)be a Cauchy sequence, and let(ank)be a convergent subsequence. Set x= limk!1ank:

We wish to prove thatan!xasn! 1.

Given >0. There existsKsuch that for allk > K,

jankxj< =2:

Also, since(an)is Cauchy, there existsNsuch that

janamj< =2 for allm;n > N. Pick` > Klarge enough so thatn`> N. Then for alln > N, we have janxj=j(anan`) + (an`x)j janan`j+jan`xj< which concludes the proof. Thus to conclude the proof of the Cauchy Criterion, it remains only to prove the Monotone Subse- quence Theorem. We will do this at the start of next lecture.quotesdbs_dbs19.pdfusesText_25