[PDF] Homework  { corrections

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Math 320: Analysis I Spring 2015 By the Completeness Axiom, supfs n: n > Ngand supft n: n > Ngexist and are real numbers Since supfs n: n > Ngand supft n: n > Ngare the least upper bounds for fs n: n > Ngand ft n: n > Ng, respectively, the last inequality in (1) implies s n+ t n supfs n: n > Ng+ supft n: n > Ng: (3)

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Homework #6 { corrections

Gabriel Roisenberg Rodrigues

March 4, 2015

Problems from textbook:

(11.8) Use Denition 10.6 and Exercise 5.4 to prove liminfsn=limsup(sn) for every sequence (sn). Proof.First, recall that ifSis a nonempty subset ofRandS:=fs:s2Sg, infS=sup(S). Now, letSN=fsn:n > Ng, for someN2N; thenSN= fsn:n > Ng. Notice that bothSNandSNare nonempty subsets ofR. By the result mentioned above, we have infSN=sup(SN), that is, inffsn:n > Ng=supfsn:n > Ng:(1) By Denition 10.6 in the textbook, we have liminfsn= limN!1inffsn:n > Ng; but since equation (1) holds, we get liminfsn= limN!1inffsn:n > Ng = lim

N!1(supfsn:n > Ng)

= lim

N!1(1supfsn:n > Ng)

= lim

N!1(1)limN!1supfsn:n > Ng

=limN!1supfsn:n > Ng =limsup(sn); which is the equality we were looking for. (12.4) Show limsup(sn+tn)limsupsn+ limsuptnfor bounded sequences (sn) and (tn).

Hint: First show

supfsn+tn:n > Ng supfsn:n > Ng+ supftn:n > Ng:

Then apply Exercise 9.9(c).

Proof.Since (sn) and (tn) are bounded, there are upper boundsU1;U22Rsuch that s nU1andtnU2for alln2N; in particular, since these inequalities are true for alln2N, they are also true for all natural numbersn > N, withN2N. Moreover, s nU1andtnU2are equivalent toU1sn0 andU2tn0, respectively.

Hence, for alln > N,

(U1sn) + (U2tn)0,U1+U2(sn+tn)0 ,sn+tnU1+U2:(2) 1

Math 320: Analysis I Spring 2015

By the Completeness Axiom, supfsn:n > Ngand supftn:n > Ngexist and are real numbers. Since supfsn:n > Ngand supftn:n > Ngare the least upper bounds for fsn:n > Ngandftn:n > Ng, respectively, the last inequality in (1) implies s n+tnsupfsn:n > Ng+ supftn:n > Ng:(3) On the other hand, since inequality (2) holds for alln > N, it follows that the set R N=fsn+tn:n > Ngis bounded above as well; thus, by the Completeness Axiom, R Nhas a supremum supRN2R. IfTis the set of all upper bounds forRN, then (supfsn:n > Ng+supftn:n > Ng)2T, and supRN= supfsn+tn:n > Ngis the least upper bound forRN. Therefore, we get supfsn+tn:n > Ng supfsn:n > Ng+ supftn:n > Ng:(4) SinceSN= supfsn:n > Ng=s2RandTN= supftn:n > Ng=t2Rfor all n2N, the Limit Theorem applies, and we get lim

N!1(SN+TN) = limsupSN+ limsupTN:

Consequently, by the Comparison Theorem, inequality (3) implies limsupfsn+tn:n > Ng limsupfsn:n > Ng+ limsupftn:n > Ng; as desired.

Additional Problem:

(1) For each sequence (sn) below, nd limsupsnand liminfsnusing denitions. This means that you need to compute supfsn:nNgand inffsn:nNgfor everyN. (a)sn=n,n2N

Solution.

inffsn:nNg= inffn2N:nNg=N) liminfsn= limN!1inffn2N:nNg= limN!1N=1: supfsn:n > Ng= supfn2N:nNg=1 ) limsupsn= limN!1supfn2N:nNg= limN!11=1: (b)sn=1n ,n2N inffsn:nNg= inf1n :nN= 0) liminfsn= limN!1inf1n :nN= limN!10 = 0: supfsn:nNg= sup1n :nN=1N limsupsn= limN!1sup1n :nN= limN!11N = 0:11 Notice that this is precisely what we would expect: sincesn!0, we should have liminfsn= limsupsn= limsn= 0. 2

Math 320: Analysis I Spring 2015

(c)sn= sinn3 ,n2N

Observe that

(sn) = (p3 2 ;p3 2 ;0;p3 2 ;p3 2 ;0;p3 2 ;p3 2 ;0;p3 2 ;p3 2 ;0;:::) is a sequence with period 6, i.e., thesnrepeats itself after every 6 steps. So for everyN, the setfsn:nNgcontains exactly 3 elements:p3 2 ;0;p3 2 , i.e., fsn:nNg=fp3 2 ;0;p3 2 g.

Thus, inffsn:nNg= minfp3

2 ;0;p3 2 g=p3 2 liminfsn= limN!1infsinn3 :n2N = lim N!1 p3 2 =p3 2 supfsn:nNg= minfp3 2 ;0;p3 2 g=p3 2 limsupsn= limN!1supsinn3 :n2N = lim N!1 p3 2 p3 2 3quotesdbs_dbs15.pdfusesText_21