Suppose n1 < n2 < n3 < ··· is a strictly increasing sequence of indices, then (snk ) is a subsequence of (sn) We will Now we state some limit theorems Let (sn) be a sequence that converges to s ∈ R Applying the definition to ε = 1, we see
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Suppose n1 < n2 < n3 < ··· is a strictly increasing sequence of indices, then (snk ) is a subsequence of (sn) We will Now we state some limit theorems Let (sn) be a sequence that converges to s ∈ R Applying the definition to ε = 1, we see
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Convergent Sequences
Denition 1.A sequence of real numbers (sn) is said to converge to a real numbersif8" >0;9N2N;such thatn > Nimpliesjsnsj< ":(1)
When this holds, we say that (sn) is a convergence sequence withsbeing its limit, and write s n!sors= limn!1sn. If (sn) does not converge, then we say that (sn) is a divergent sequence. We rst show that one sequence (sn) can not have two dierent limits. Supposesn!s andsn!t. Let" >0. Then"2 >0. Sincesn!s, by denition there isN12Nsuch that forn > N1,jsnsj<"2 . Sincesn!t, by denition there isN22Nsuch that forn > N2, jsntj<"2 . Here we useN1andN2in the two statements because theNcoming from the two limits may not be the same. LetN= maxfN1;N2g. Ifn > N, thenn > N1andn > N2 both hold. Sojsnsj<"2 andjsntj<"2 , which by triangle inequality imply that jstj jsnsj+jsntj<"2 +"2 Nowjstj< "holds for every" >0. We then conclude thatjstj= 0 (for otherwisejstj>0, we then get a contradiction by choosing"=jstj). Sos=t, and the uniqueness holds. We will use the following tools to check whether a sequence converges or diverges. 1. the denition 2. basic examples 3. limit theorems 4. b oundednessand subsequences. We have stated the denition. Now we consider some examples. Example 1.Lets2R. Ifsn=sfor alln, i.e., (sn) is a constant sequence, then limsn=s. Proof.For any given" >0 we simply chooseN= 1. Ifn > N, thenjsnsj= 0< ".Example 2.We have1n !0. Proof.Let" >0. By Archimedean property, there isN2Nsuch that1N < ". Ifn > N, then j 1n 0j=1n <1N < ":Example 3.The following two sequences are divergent 1 (i)( sn) = ((1)n) = (1;1;1;1;1;1;:::); (ii) ( sn) = (n) = (1;2;3;4;5;6;:::). Proof.(i) We use the notation of subsequence and statement that will be proved later. Suppose n1< n2< n3 (sn). We will prove a theorem, which asserts that, if (sn) converges tos, then any subsequence of (sn) also converges tos. The sequence (sn) = ((1)n) contains two constant sequences (1;1;1;:::) (withnk= 2k) and (1;1;1;:::) (withnk= 2k1), which converge to dierent limits. So the original (sn) can not converge. (ii) We use the following theorem. If (sn) is convergent, then it is a bounded sequence. In other words, the setfsn:n2Ngis bounded. So an unbounded sequence must diverge. Since forsn=n,n2N, the setfsn:n2Ng=Nis unbounded, the sequence (n) is divergent.Remark 1.This example shows that we have two ways to prove that a sequence is divergent:
(i) nd two subsequences that convergent to dierent limits; (ii) show that the sequence is unbounded. Note that the (sn) in (i) is bounded and divergent. The (sn) in (ii) is divergent, but limsnactually exists, which is +1, and its every subsequence also tends to +1. We will dene that limit later. Now we state some limit theorems.
Theorem 1(Theorem 9.1).Every convergent sequence is bounded. Proof.Let (sn) be a sequence that converges tos2R. Applying the denition to"= 1, we see that there isN2Nsuch that for anyn > N,jsnsj<1, which then implies thatjsnj jsj+1. Let M= maxfjs1j;js2j;:::;jsNj;jsj+ 1g:
The maximum exists since the set is nite. Then for anyn2N,jsnj M(consider the case nNandn > Nseparately), i.e.,MsnM. Sofsn:n2Ngis bounded.Theorem 2(Theorem 9.3).If(sn)converges tosand(tn)converges tot, then(sn+tn)
converges tos+t. Proof.Let" >0. Then"2
>0. Sincesn!s, there isN12Nsuch that forn > N1,jsnsj<"2 Sincetn!t, there isN22Nsuch that forn > N2,jtntj<"2 . LetN= maxfN1;N2g. If n > N, thenn > N1andn > N2both hold, and sojsnsj<"2 andjtntj<"2 , which together imply (by triangle inequality) that j(sn+tn)(s+t)j jsnsj+jtntj<"2 +"2 So we have the desired convergence.Theorem 3(Theorem 9.4).If(sn)converges tosand(tn)converges tot, then(sntn)converges
tost. 2 Discussion. We need to boundjsntnstjfrom above for bign. We write s ntnst=sntnsnt+sntst=sn(tnt) +t(sns): By triangle inequality, we get
jsntnstj jsn(tnt)j+jt(sns)j=jsnjjtntj+jtjjsnsj: Sincetn!tandsn!s, we know thatjtntjandjsnsjcan be arbitrarily small if we choose nbig enough. Thus, ifjsnjandjtjare not too big, then we can control the sum on the RHS (righthand side). In fact, the size ofjsnjcan be controlled because of Theorem 9.1. Proof.Since (sn) is convergent, by Theorem 9.1, there isM >0 such thatjsnj Mfor every n. We may chooseMbig such thatM jtj. Let" >0. Then"2M>0. Sincesn!s, there isN12Nsuch that forn > N1,jsnsj<"2M. Sincetn!t, there isN22Nsuch that for n > N 2,jtntj<"2M. LetN= maxfN1;N2g. Ifn > N, thenn > N1andn > N2both hold,
and sojsnsj<"2Mandjtntj<"2M, which together withjsnj Mandjtj Mimply that jsntnstj jsn(tnt)j+jt(sns)j=jsnjjtntj+jtjjsnsj Mjtntj+Mjsnsj< M"2M+M"2M=":Corollary 1.If(sn)converges tos,k2R, andm2N, then(ksn)converges toksandsmnconverges tosm.
Proof.For the sequence (ksn), we apply Theorem 9.4 to the sequence (tn) withtn=kfor all n. For the sequence (smn) we use induction. In the induction step, note thatsm+1n=snsmnand apply Theorem 9.4 totn=smnCorollary 2.If(sn)converges tosand(tn)converges tot, then(sntn)converges tost.
Proof.We writesn+tn=sn+ (1)tnand apply Theorem 9.3 and the previous corollary.From this corollary we see thatsn!sisns!0. By the Theorem below, the latter
statement is equivalent to thatjsnsj !0. Theorem 4.(a) Suppose two sequences(sn)and(tn)satisfy thattn!0andjsnj jtnjfor all but nitely manyn. Thensn!0. (b) For any sequence(sn),sn!0if and only ifjsnj !0. Proof.(a) LetN02Nbe such thatjsnj jtnjforn > N0. Let" >0. Sincetn!0, there is N 12Nsuch that forn > N1,jtn0j< ". LetN= maxfN0;N1g. Forn > N,jsnj jtnjand
jtn0j< ", which imply thatjsn0j=jsnj jtnj=jtn0j< ". (b) From (a) we know that ifjsnj=jtnjfor alln, thensn!0 itn!0. We then apply this result totn=jsnjand use thatjjsnjj=jsnj.3 Lemma 1(Lemma 9.5).If(sn)converges tossuch thats6= 0andsn6= 0for alln, then (1=sn)converges to1=s. Discussion. We need to boundj1=sn1=sjfrom above for bign. We write 1s n1s =ssns ns =jsnsjjsnjjsj: Sincesn!s,jsnsjcan be arbitrarily small if we choosenbig enough. Thus, ifjsnjandjsj are not too close to 0, then we can control the size of the RHS. This means that we need a positive lower bound of the setfjs1j;js2j;:::g. Proof.Sinces6= 0, we havejsj2
>0. Sincesn!s, applying the denition to"=jsj2 , we get N2Nsuch that forn > N,jsnsj , which then implies by triangle inequality that jsnj jsj jsnsj>jsj jsj2 =jsj2 . Letm= minfjs1j;js2j;:::;jsNj;jsj2 g. Thenmexists and is positive since the set is a nite set of positive numbers. Let" >0. Thenmjsj" >0. Sincesn!s, there isN02Nsuch thatn > N0implies that jsnsj< mjsj", which together withjsnj mfor allnimplies that 1s n1s =jsnsjjsnjjsjjsnsjmjsj s n6= 0for alln, then(tn=sn)converges tot=s. Proof.By Lemma 9.5, (1=sn) converges to 1=s. Applying Theorem 9.4 to the sequences (1=sn) and (tn), we get the conclusion.Example 4.Derive lim3n+17n4and lim4n3+3nn 36
Solution.We write
3n+ 17n4=3 + 1=n7 + (4)1=n;4n3+ 3nn
36=4 + 3(1=n)21 + (6)1=n:
We have shown that lim1=n= 0. So (i) lim(3 + 1=n) = 3 + 0 = 3 and lim(7 + (4)1=n) = 7 + (4)0 = 7, which imply that lim3n+17n4= lim(3 + 1=n)=lim(7 + (4)1=n) = 3=7; (ii)
lim(4 + 3(1=n)2) = 4 + 302= 4 and lim(1 + (6)1=n) = 1 + (6)0 = 1, which imply that lim 4n3+3nn
36= lim(4 + 3(1=n)2)=lim(1 + (6)1=n) = 4.We now state some theorems about the relation between limits and orders.
Theorem 6(Exercise 8.9).(a) If(sn)converges tos, and there isN02Nsuch thatsn0 for alln > N0, thens0. (b) Suppose(sn)converges tosand(tn)converges tot. If thereN02Nsuch thatsntn for alln > N0, thenst. 4 Proof.(a) We prove by contradiction. Supposes <0. Let"=jsj=s >0. Sincesn!s, there isN2Nsuch that forn > N,jsnsj< ", which implies thatsn< s+"= 0. Let n= maxfN;N0g+ 1. Thenn > N0andn > N. Fromn > N0we getsn0; fromn > Nwe getsn<0. This is the contradiction. (b) Applying (i) to the sequence (tnsn) we conclude that its limittsis nonnegative.Forx2[0;1) andn2N, the power rootx1=nis dened as the uniquey2[0;1) such that
y n=x. The uniqueness of suchyfollows from the fact that if 0y1< y2, thenyn1< yn2. The existence follows from the \Intermediate Value Theorem" for continuous functionf(x) =xn, which will be stated and proved later. We now just accept the existence ofx1=nfor any x2[0;1). It is clear that 0x1< x2implies that 0x1=n 1< x1=n
2. We restrict our attention
to [0;1) although in the case thatnis an odd number, we can also denex1=nforx <0. Whenn= 2,x1=2is often written aspx. We have the following theorem. Theorem 7(Example 5).Suppose(sn)converges tosandsn0for alln. Then(ps n) converges tops. DiscussionWe want to boundjps
npsjfrom above for bign. It is useful to note the equality ps nps)(ps n+ps) = (ps n)2(ps)2=sns: Taking absolute value, we get
j ps npsj jps n+psj=jsnsj: If ps >0, then j ps npsj=jsnsjps n+ps jsnsjps Proof.By Theorem 6,s0. First supposes >0. Thenps >0. Let" >0. Thenps" >0. Sincesn!s, there isN2Nsuch that forn > N,jsnsj
forsn=n,n2N, the setfsn:n2Ng=Nis unbounded, the sequence (n) is divergent.Remark 1.This example shows that we have two ways to prove that a sequence is divergent:
(i) nd two subsequences that convergent to dierent limits; (ii) show that the sequence is unbounded. Note that the (sn) in (i) is bounded and divergent. The (sn) in (ii) is divergent, but limsnactually exists, which is +1, and its every subsequence also tends to +1. We will dene that limit later.Now we state some limit theorems.
Theorem 1(Theorem 9.1).Every convergent sequence is bounded. Proof.Let (sn) be a sequence that converges tos2R. Applying the denition to"= 1, we see that there isN2Nsuch that for anyn > N,jsnsj<1, which then implies thatjsnj jsj+1. LetM= maxfjs1j;js2j;:::;jsNj;jsj+ 1g:
The maximum exists since the set is nite. Then for anyn2N,jsnj M(consider the casenNandn > Nseparately), i.e.,MsnM. Sofsn:n2Ngis bounded.Theorem 2(Theorem 9.3).If(sn)converges tosand(tn)converges tot, then(sn+tn)
converges tos+t.Proof.Let" >0. Then"2
>0. Sincesn!s, there isN12Nsuch that forn > N1,jsnsj<"2 Sincetn!t, there isN22Nsuch that forn > N2,jtntj<"2 . LetN= maxfN1;N2g. If n > N, thenn > N1andn > N2both hold, and sojsnsj<"2 andjtntj<"2 , which together imply (by triangle inequality) that j(sn+tn)(s+t)j jsnsj+jtntj<"2 +"2So we have the desired convergence.Theorem 3(Theorem 9.4).If(sn)converges tosand(tn)converges tot, then(sntn)converges
tost. 2 Discussion. We need to boundjsntnstjfrom above for bign. We write s ntnst=sntnsnt+sntst=sn(tnt) +t(sns):By triangle inequality, we get
jsntnstj jsn(tnt)j+jt(sns)j=jsnjjtntj+jtjjsnsj: Sincetn!tandsn!s, we know thatjtntjandjsnsjcan be arbitrarily small if we choose nbig enough. Thus, ifjsnjandjtjare not too big, then we can control the sum on the RHS (righthand side). In fact, the size ofjsnjcan be controlled because of Theorem 9.1. Proof.Since (sn) is convergent, by Theorem 9.1, there isM >0 such thatjsnj Mfor every n. We may chooseMbig such thatM jtj. Let" >0. Then"2M>0. Sincesn!s, there isN12Nsuch that forn > N1,jsnsj<"2M. Sincetn!t, there isN22Nsuch that for n > N2,jtntj<"2M. LetN= maxfN1;N2g. Ifn > N, thenn > N1andn > N2both hold,
and sojsnsj<"2Mandjtntj<"2M, which together withjsnj Mandjtj Mimply that jsntnstj jsn(tnt)j+jt(sns)j=jsnjjtntj+jtjjsnsjMjtntj+Mjsnsj< M"2M+M"2M=":Corollary 1.If(sn)converges tos,k2R, andm2N, then(ksn)converges toksandsmnconverges tosm.
Proof.For the sequence (ksn), we apply Theorem 9.4 to the sequence (tn) withtn=kfor alln. For the sequence (smn) we use induction. In the induction step, note thatsm+1n=snsmnand apply Theorem 9.4 totn=smnCorollary 2.If(sn)converges tosand(tn)converges tot, then(sntn)converges tost.
Proof.We writesn+tn=sn+ (1)tnand apply Theorem 9.3 and the previous corollary.From this corollary we see thatsn!sisns!0. By the Theorem below, the latter
statement is equivalent to thatjsnsj !0. Theorem 4.(a) Suppose two sequences(sn)and(tn)satisfy thattn!0andjsnj jtnjfor all but nitely manyn. Thensn!0. (b) For any sequence(sn),sn!0if and only ifjsnj !0. Proof.(a) LetN02Nbe such thatjsnj jtnjforn > N0. Let" >0. Sincetn!0, there is N12Nsuch that forn > N1,jtn0j< ". LetN= maxfN0;N1g. Forn > N,jsnj jtnjand
jtn0j< ", which imply thatjsn0j=jsnj jtnj=jtn0j< ". (b) From (a) we know that ifjsnj=jtnjfor alln, thensn!0 itn!0. We then apply this result totn=jsnjand use thatjjsnjj=jsnj.3 Lemma 1(Lemma 9.5).If(sn)converges tossuch thats6= 0andsn6= 0for alln, then (1=sn)converges to1=s. Discussion. We need to boundj1=sn1=sjfrom above for bign. We write 1s n1s =ssns ns =jsnsjjsnjjsj: Sincesn!s,jsnsjcan be arbitrarily small if we choosenbig enough. Thus, ifjsnjandjsj are not too close to 0, then we can control the size of the RHS. This means that we need a positive lower bound of the setfjs1j;js2j;:::g.Proof.Sinces6= 0, we havejsj2
>0. Sincesn!s, applying the denition to"=jsj2 , we getN2Nsuch that forn > N,jsnsj , which then implies by triangle inequality that jsnj jsj jsnsj>jsj jsj2 =jsj2 . Letm= minfjs1j;js2j;:::;jsNj;jsj2 g. Thenmexists and is positive since the set is a nite set of positive numbers. Let" >0. Thenmjsj" >0. Sincesn!s, there isN02Nsuch thatn > N0implies that jsnsj< mjsj", which together withjsnj mfor allnimplies that 1s n1s =jsnsjjsnjjsjjsnsjmjsj s n6= 0for alln, then(tn=sn)converges tot=s. Proof.By Lemma 9.5, (1=sn) converges to 1=s. Applying Theorem 9.4 to the sequences (1=sn) and (tn), we get the conclusion.Example 4.Derive lim3n+17n4and lim4n3+3nn 36
Solution.We write
3n+ 17n4=3 + 1=n7 + (4)1=n;4n3+ 3nn
36=4 + 3(1=n)21 + (6)1=n:
We have shown that lim1=n= 0. So (i) lim(3 + 1=n) = 3 + 0 = 3 and lim(7 + (4)1=n) = 7 + (4)0 = 7, which imply that lim3n+17n4= lim(3 + 1=n)=lim(7 + (4)1=n) = 3=7; (ii)
lim(4 + 3(1=n)2) = 4 + 302= 4 and lim(1 + (6)1=n) = 1 + (6)0 = 1, which imply that lim 4n3+3nn
36= lim(4 + 3(1=n)2)=lim(1 + (6)1=n) = 4.We now state some theorems about the relation between limits and orders.
Theorem 6(Exercise 8.9).(a) If(sn)converges tos, and there isN02Nsuch thatsn0 for alln > N0, thens0. (b) Suppose(sn)converges tosand(tn)converges tot. If thereN02Nsuch thatsntn for alln > N0, thenst. 4 Proof.(a) We prove by contradiction. Supposes <0. Let"=jsj=s >0. Sincesn!s, there isN2Nsuch that forn > N,jsnsj< ", which implies thatsn< s+"= 0. Let n= maxfN;N0g+ 1. Thenn > N0andn > N. Fromn > N0we getsn0; fromn > Nwe getsn<0. This is the contradiction. (b) Applying (i) to the sequence (tnsn) we conclude that its limittsis nonnegative.Forx2[0;1) andn2N, the power rootx1=nis dened as the uniquey2[0;1) such that
y n=x. The uniqueness of suchyfollows from the fact that if 0y1< y2, thenyn1< yn2. The existence follows from the \Intermediate Value Theorem" for continuous functionf(x) =xn, which will be stated and proved later. We now just accept the existence ofx1=nfor any x2[0;1). It is clear that 0x1< x2implies that 0x1=n 1< x1=n
2. We restrict our attention
to [0;1) although in the case thatnis an odd number, we can also denex1=nforx <0. Whenn= 2,x1=2is often written aspx. We have the following theorem. Theorem 7(Example 5).Suppose(sn)converges tosandsn0for alln. Then(ps n) converges tops. DiscussionWe want to boundjps
npsjfrom above for bign. It is useful to note the equality ps nps)(ps n+ps) = (ps n)2(ps)2=sns: Taking absolute value, we get
j ps npsj jps n+psj=jsnsj: If ps >0, then j ps npsj=jsnsjps n+ps jsnsjps Proof.By Theorem 6,s0. First supposes >0. Thenps >0. Let" >0. Thenps" >0. Sincesn!s, there isN2Nsuch that forn > N,jsnsj
=jsnsjjsnjjsjjsnsjmjsj (b) Applying (i) to the sequence (tnsn) we conclude that its limittsis nonnegative.Forx2[0;1) andn2N, the power rootx1=nis dened as the uniquey2[0;1) such that
Solution.We write
3n+ 17n4=3 + 1=n7 + (4)1=n;4n3+ 3nn
36=4 + 3(1=n)21 + (6)1=n:
We have shown that lim1=n= 0. So (i) lim(3 + 1=n) = 3 + 0 = 3 and lim(7 + (4)1=n) = 7 + (4)0 = 7, which imply that lim3n+17n4= lim(3 + 1=n)=lim(7 + (4)1=n) = 3=7; (ii)
lim(4 + 3(1=n)2) = 4 + 302= 4 and lim(1 + (6)1=n) = 1 + (6)0 = 1, which imply that lim 4n3+3nn
36= lim(4 + 3(1=n)2)=lim(1 + (6)1=n) = 4.We now state some theorems about the relation between limits and orders.
Theorem 6(Exercise 8.9).(a) If(sn)converges tos, and there isN02Nsuch thatsn0 for alln > N0, thens0. (b) Suppose(sn)converges tosand(tn)converges tot. If thereN02Nsuch thatsntn for alln > N0, thenst. 4 Proof.(a) We prove by contradiction. Supposes <0. Let"=jsj=s >0. Sincesn!s, there isN2Nsuch that forn > N,jsnsj< ", which implies thatsn< s+"= 0. Let n= maxfN;N0g+ 1. Thenn > N0andn > N. Fromn > N0we getsn0; fromn > Nwe getsn<0. This is the contradiction. 1< x1=n
2. We restrict our attention
to [0;1) although in the case thatnis an odd number, we can also denex1=nforx <0. Whenn= 2,x1=2is often written aspx. We have the following theorem. Theorem 7(Example 5).Suppose(sn)converges tosandsn0for alln. Then(ps n) converges tops. DiscussionWe want to boundjps
npsjfrom above for bign. It is useful to note the equality ps nps)(ps n+ps) = (ps n)2(ps)2=sns: Taking absolute value, we get
j ps npsj jps n+psj=jsnsj: If ps >0, then j ps npsj=jsnsjps n+ps jsnsjps Proof.By Theorem 6,s0. First supposes >0. Thenps >0. Let" >0. Thenps" >0. Sincesn!s, there isN2Nsuch that forn > N,jsnsj