100%(1)1 out of 1 people found this document helpful
This preview shows page 1 - 3 out of 3 pages.
Homework 4H´ector Guillermo Cu´ellar R´ıosFebruary 21, 20066.2(a) Ifsn→0, then for everyε >0 there existN∈Rsuch thatn > Nimpliessn< ε. True. According to Definition 16.2, ifsn→0, thenfore eachε >0 there exists a real numberNsuch that for alln∈N,n > Nimplies that|sn-0|< ε, then|sn|< ε, clearlysn< ε.(b) If for everyε >0 there existsN∈Rsuch thatn > Nimpiessn< ε,thensn→0. False, the sequencesn=-nis always less thanεand itdoes not converge to any number.(c) Given sequences (sn) and (an), if for somes∈R,k >0 andm∈Nwehave|sn-s| ≤k|an|for alln > m, then limsn=s. False. Considersn= (-1)n,an=nkands= 0, none of the sequences converge,furthermore|sn-0| ≤1≤knk=nfor alln.(d) Ifsn→tandsn→t, thens=t. True by Theorem 16.14.16.6 . Using only Definition 16.2, prove the following.(a) For an real numberk,limn→∞(kn)= 0.ChooseN=|k|ε. Then by Definition 16.2kn-0 =kn=|k|n<|k|N=ε(b) For any real numberk >0, limn→∞(1nk)= 0LetN=1ε1/k, then1nk<1Nk=1(1ε1/k)k=ε(c) lim3n+ 1n+ 2= 3LetN=5εthen3n+1n+2-3 =-5n+2<5n<5N=ε.16.7 Using any of the results in this section, prove the following.(b) lim4n2-72n3-5= 0To apply the Theorem 16.8 we need to find an upper bound for4n2-72n3-5-01
whennis sufficiently large. the numerator is easy, since|4n2-7|<5n2for alln. Similarly the denominator|2n3-5|> n3for alln >2. Thenwe have:Ifn≥2, then|4n2-7|<5n2and|2n3-5|> n3, so that4n2-35n2-2n-0 =4n2-35n2-2n<5n2n3= 51nSince lim1n= 0, Theorem 16.8 implies thatlim4n2-35n2-2n=45.(d) lim√nn+ 1= 0√nn+ 1-0 =√nn+ 1<√nn=1√n.Since lim1√n= 0 by Practice 16.4, Theorem 16.8 implies that lim√nn+1=016.8 Show that each of the following sequences is divergent.(a)an= 2nTo prove that the sequencean= 2nis divergent, let us suppose thatanconverges to some real numbers. Lettingε= 1 in the definitionof convergence, we find that there exist a numberNsuch thatn > Nimplies that|2n-s|<1, we obtain the inequality 1-2n < s <1+2n. Since 2n > Nthen|2(2n)-s|<1 which give us the inequality1-4n < s <1 + 4nwhich is clearly a contradiction.(c)cn=cosnπ3(cn) = (12,-12,-1,-12,12,1, . . .) As in the previous excercise let us sup-pose thatcnconverges to some real numbers. Lettingε=14and bythe definition of convergence, there exist some numberNsuch thatn > Nimplies that|cosnπ3-s|<14. Let us choose ann > Nsuchthatnmod 6 = 3 then| -1-s|<14and this is equivalent to theinequality-54< s <-34. Similarly choose ann > Nsuch thatnmod6 = 0 and|1-s|<14or34< s <54which is a contradiction. Thus thesequencecnis divergent.16.12(a) Suppose that limsn= 0. If (tn) is a bounded sequence, prove that lim(sntn) = 0.If (tn) is a bounded sequence then|tn| ≤Mfor alln. Then|sntn-0|=|sntn|=|sn||tn| ≤M|sn|. Then by Theorem 16.8 lim(sntn) = 0.2