2.1LimitsofSequences.
CHAPTER2
2.1.0. a)True.If xn converges,thenthereisan M> 0suchthat |xn|≤ M .ChoosebyArchimedesan N ∈ N suchthat N>M/ε.Then n ≥ N implies |xn/n|≤ M/n ≤ M/N<ε
b)False. xn = √n doesnotconverge,but xn/n =1/√n → 0as n →∞
c)False. xn =1convergesand yn =( 1)n isbounded,but xnyn =( 1)n doesnotconverge.
d)False. xn =1/n convergesto0and yn = n2 > 0,but xnyn = n doesnotconverge.
2.1.1. a)BytheArchimedeanPrinciple,given ε> 0thereisan N ∈ N suchthat N> 1/ε.Thus n ≥ N implies |(2 1/n) 2|≡|1/n|≤ 1/N<ε.
b)BytheArchimedeanPrinciple,given ε> 0thereisan N ∈ N suchthat N>π2/ε2.Thus n ≥ N implies |1+
c)BytheArchimedeanPrinciple,given ε> 0thereisan N ∈ N suchthat N> 3/ε.Thus n ≥ N implies |3(1+1/n) 3|≡|3/n|≤ 3/N<ε.
d)BytheArchimedeanPrinciple,given ε> 0thereisan N ∈ N suchthat N> 1/√3ε.Thus n ≥ N implies |(2n 2 +1)/(3n 2) 2/3|≡|1/(3n 2)|≤ 1/(3N 2) <ε.
2.1.2. a)Byhypothesis,given ε> 0thereisan N ∈ N suchthat n ≥ N implies |xn 1| <ε/2.Thus n ≥ N implies |1+2xn 3|≡ 2 |xn 1| <ε.
b)Byhypothesis,given ε> 0thereisan N ∈ N suchthat n ≥ N implies xn > 1/2and |xn 1| <ε/4.In particular,1/xn < 2.Thus n ≥ N implies |(πxn 2)/xn (π 2)|≡ 2 |(xn 1)/xn| < 4 |xn 1| <ε.
c)Byhypothesis,given ε> 0thereisan N ∈ N suchthat n ≥ N implies xn > 1/2and |xn 1| <ε/(1+2e). Thus n ≥ N andthetriangleinequalityimply
|(x 2 n e)/xn (1 e)|≡|xn 1| 1+ e xn ≤|xn 1| 1+ e |xn| < |xn 1|(1+2e) <ε.
2.1.3. a)If nk =2k,then3 ( 1)nk ≡ 2convergesto2;if nk =2k +1,then3 ( 1)nk ≡ 4convergesto4.
b)If nk =2k,then( 1)3nk +2 ≡ ( 1)6k +2=1+2=3convergesto3;if nk =2k +1,then( 1)3nk +2 ≡ ( 1)6k+3 +2= 1+2=1convergesto1.
c)If nk =2k,then(nk ( 1)nk nk 1)/nk ≡−1/(2k)convergesto0;if nk =2k+1,then(nk ( 1)nk nk 1)/nk ≡ (2nk 1)/nk =(4k +1)/(2k +1)convergesto2.
2.1.4. Suppose xn isbounded.ByDefinition2.7,therearenumbers M and m suchthat m ≤ xn ≤ M forall n ∈ N.Set C :=max{1, |M |, |m|}.Then C> 0, M ≤ C,and m ≥−C.Therefore, C ≤ xn ≤ C,i.e., |xn| <C forall n ∈ N
Conversely,if |xn| <C forall n ∈ N,then xn isboundedaboveby C andbelowby C
2.1.5. If C =0,thereisnothingtoprove.Otherwise,given ε> 0useDefinition2.1tochoosean N ∈ N such that n ≥ N implies |bn|≡ bn <ε/|C|.Hencebyhypothesis, n ≥ N implies |xn a|≤|C|bn <ε.
Bydefinition, xn → a as n →∞.
2.1.6. If xn = a forall n,then |xn a| =0islessthananypositive ε forall n ∈ N.Thus,bydefinition, xn → a as n →∞
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2.1.7. a)Let a bethecommonlimitpoint.Given ε> 0,choose N ∈ N suchthat n ≥ N implies |xn a| and |yn a| areboth <ε/2.BytheTriangleInequality, n ≥ N implies |xn yn|≤|xn a| + |yn a| <ε.
Bydefinition, xn yn → 0as n →∞.
b)If n convergestosome a,thengiven ε =1/2,1= |(n +1) n| < |(n +1) a| + |n a| < 1for n sufficiently large,acontradiction.
c)Let xn = n and yn = n +1/n.Then |xn yn| =1/n → 0as n →∞,butneither xn nor yn converges.
2.1.8. ByTheorem2.6,if xn → a then xnk → a.Conversely,if xnk → a foreverysubsequence,thenit convergesforthe“subsequence” xn.
2.2LimitTheorems.
2.2.0. a)False.Let xn = n2 and yn = n andnotebyExercise2.2.2athat xn + yn →∞ as n →∞
b)True.Let ε> 0.If xn →−∞ as n →∞,thenchoose N ∈ N suchthat n ≥ N implies xn < 1/ε.Then xn < 0so |xn| = xn > 0.Multiply xn < 1/ε by ε/( xn)whichispositive.Weobtain ε< 1/xn,i.e., |1/xn| = 1/xn <ε
c)False.Let xn =( 1)n/n.Then1/xn =( 1)nn hasnolimitas n →∞.
d)True.Since(2x x) =2x log2 1 > 1forall x ≥ 2,i.e.,2x x isincreasingon[2, ∞).Inparticular, 2x x ≥ 22 2 > 0,i.e.,2x >x for x ≥ 2.Thus,since xn →∞ as n →∞,wehave2xn >xn for n large,hence
2 xn < 1 xn → 0 as n →∞
2.2.1. a) |xn|≤ 1/n → 0as n →∞ andwecanapplytheSqueezeTheorem.
b)2n/(n2 + π)=(2/n)/(1+ π/n2) → 0/(1+0)=0byTheorem2.12.
c)(√2n +1)/(n + √2)=((√2/√n)+(1/n))/(1+(√2/n)) → 0/(1+0)=0byExercise2.2.5andTheorem 2.12.
d)Aneasyinductionargumentshowsthat2n +1 < 2n for n =3, 4, ....Wewillusethistoprovethat n2 ≤ 2n for n =4, 5, ....It’ssurelytruefor n =4.Ifit’strueforsome n ≥ 4,thentheinductivehypothesisandthefact that2n +1 < 2n imply (n +1)2 = n 2 +2n +1 ≤ 2n +2n +1 < 2n +2n =2n+1 sothesecondinequalityhasbeenproved. Nowthesecondinequalityimplies n/2n < 1/n for n ≥ 4.HencebytheSqueezeTheorem, n/2n → 0as n →∞
2.2.2. a)Let M ∈ R andchoosebyArchimedesan N ∈ N suchthat N> max{M, 2}.Then n ≥ N implies n2 n = n(n 1) ≥ N (N 1) >M (2 1)= M .
b)Let M ∈ R andchoosebyArchimedesan N ∈ N suchthat N> M/2.Noticethat n ≥ 1implies 3n ≤−3 so1 3n ≤−2.Thus n ≥ N implies n 3n2 = n(1 3n) ≤−2n ≤−2N<M .
c)Let M ∈ R andchoosebyArchimedesan N ∈ N suchthat N>M .Then n ≥ N implies(n2 +1)/n = n +1/n>N +0 >M
d)Let M ∈ R satisfy M ≤ 0.Then2+sin θ ≥ 2 1=1implies n2(2+sin(n3 + n +1)) ≥ n2 1 > 0 ≥ M for all n ∈ N.Ontheotherhand,if M> 0,thenchoosebyArchimedesan N ∈ N suchthat N> √M .Then n ≥ N implies n2(2+sin(n3 + n +1)) ≥ n2 1 ≥ N 2 >M
2.2.3. a)FollowingExample2.13,
as n →∞.
b)FollowingExample2.13, n3 + n 2 2n3 + n 2 = 1+(1/n2) (2/n3) 2+(1/n2) (2/n3) → 1 2 as n →∞
c)Rationalizingtheexpression,weobtain
as n →∞ bythemethodofExample2.13.(Multiplytopandbottomby1/√
.) d)Multiplytopandbottomby1/√n toobtain
2.2.4. a)Clearly,
Thus
Since y =0, |yn|≥|y|/2forlarge n.Thus
as n →∞ byTheorem2.12iandii.HencebytheSqueezeTheorem, xn/yn → x/y as n →∞ b)Bysymmetry,wemaysupposethat x = y = ∞.Since yn →∞ implies yn > 0for n large,wecanapply Theorem2.15directlytoobtaintheconclusionswhen α> 0.Forthecase α< 0, xn >M implies αxn <αM Sinceany M0 ∈ R canbewrittenas αM forsome M ∈ R,weseebydefinitionthat xn →−∞ as n →∞.
2.2.5. Case1. x =0.Let > 0andchoose N solargethat n ≥ N implies |xn| < 2.By(8)in1.1, √xn < for n ≥ N ,i.e., √xn → 0as n →∞. Case2. x> 0.Then
Since √xn ≥ 0,itfollowsthat
Thislastquotientconvergesto0byTheorem2.12.HenceitfollowsfromtheSqueezeTheoremthat √xn → √x as n →∞
2.2.6. BytheDensityofRationals,thereisan rn between x +1/n and x foreach n ∈ N.Since |x rn| < 1/n, itfollowsfromtheSqueezeTheoremthat rn → x as n →∞
2.2.7. a)ByTheorem2.9wemaysupposethat |x| = ∞.Bysymmetry,wemaysupposethat x = ∞.By definition,given M ∈ R,thereisan N ∈ N suchthat n ≥ N implies xn >M .Since wn ≥ xn,itfollowsthat wn >M forall n ≥ N .Bydefinition,then, wn →∞ as n →∞. b)If x and y arefinite,thentheresultfollowsfromTheorem2.17.If x = y = ±∞ or x = y = ∞,thereis nothingtoprove.Itremainstoconsiderthecase x = ∞ and y = −∞.ButbyDefinition2.14(with M =0), xn > 0 >yn for n sufficientlylarge,whichcontradictsthehypothesis xn ≤ yn
2.2.8. a)Takethelimitof xn+1 =1 √1 xn,as n →∞.Weobtain x =1 √1 x,i.e., x2 x =0. Therefore, x =0, 1.
b)Takethelimitof xn+1 =2+ √xn 2as n →∞.Weobtain x =2+ √x 2,i.e., x2 5x +6=0.Therefore, x =2, 3.But x1 > 3andinductionshowsthat xn+1 =2+ √xn 2 > 2+ √3 2=3,sothelimitmustbe x =3.
c)Takethelimitof xn+1 = √2+ xn as n →∞.Weobtain x = √2+ x,i.e., x2 x 2=0.Therefore, x =2, 1.But xn+1 = √2+ xn ≥ 0bydefinition(allsquarerootsarenonnegative),sothelimitmustbe x =2. Thisproofdoesn’tchangeif x1 > 2,sothelimitisagain x =2.
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2.2.9. a)Let E = {k ∈ Z : k ≥ 0and k ≤ 10n+1y}.Since10n+1y< 10, E ⊆{0, 1, , 9}.Hence w :=sup E ∈ E.Itfollowsthat w ≤ 10n+1y,i.e., w/10n+1 ≤ y.Ontheotherhand,since w +1isnotthesupremumof E, w +1 > 10n+1y.Therefore, y<w/10n+1 +1/10n+1
b)Applya)for n =0tochoose x1 = w suchthat x1/10 ≤ x<x1/10+1/10.Suppose
Then0 <x sn < 1/10n,sobya)choose xn
,i.e.,
c)Combineb)withtheSqueezeTheorem.
d)Sinceaneasyinductionprovesthat9n >n forall n ∈ N,wehave9 n < 1/n.HencetheSqueezeTheorem impliesthat9 n → 0as n →∞.Hence,itfollowsfromExercise1.4.4canddefinitionthat
Similarly,
2.3TheBolzano–WeierstrassTheorem.
2.3.0. a)False. xn =1/4+1/(n +4)isstrictlydecreasingand |xn|≤ 1/
4as n →∞
b)True.Since(n 1)/(2n 1) → 1/2as n →∞,thisfactorisbounded.Since | cos(n2 + n +1)|≤ 1,itfollows that {xn} isbounded.HenceithasaconvergentsubsequencebytheBolzano–WeierstrassTheorem.
c)False. xn =1/2 1/n isstrictlyincreasingand |xn|≤ 1/2 < 1+1/n,but xn → 1/2as n →∞
d)False. xn =(1+( 1)n)n satisfies xn =0for n oddand xn =2n for n even.Thus x2k+1 → 0as k →∞,but xn isNOTbounded.
2.3.1. Supposethat 1 <xn 1 < 0forsome n ≥ 0.Then0 <xn 1 +1 < 1so0 <xn 1 +1 < √xn 1 +1and itfollowsthat xn 1 < √xn 1 +1 1= xn.Moreover, √xn 1 +1 1 ≤ 1 1=0.Hencebyinduction, xn is increasingandboundedaboveby0.ItfollowsfromtheMonotoneConvergenceTheoremthat xn → a as n →∞ Takingthelimitof √xn 1 +1 1= xn weseethat a2 + a =0,i.e., a = 1, 0.Since xn increasesfrom x0 > 1, thelimitis0.If x0 = 1,then xn = 1forall n.If x0 =0,then xn =0forall n. Finally,itiseasytoverifythatif x0 = for = 1or0,then xn = forall n,hence xn → as n →∞
2.3.2. If x1 =0then xn =0forall n,henceconvergesto0.If0 <x1 < 1,thenby1.4.1c, xn isdecreasing andboundedbelow.Thusthelimit, a,existsbytheMonotoneConvergenceTheorem.Takingthelimitof xn+1 =1 √1 xn,as n →∞,wehave a =1 √1 a,i.e., a =0, 1.Since x1 < 1,thelimitmustbezero.
Finally, xn+1 xn = 1 √1 xn xn = 1 (1 xn) xn(1+ √1 xn) → 1 1+1 = 1 2
2.3.3. Case1. x0 =2.Then xn =2forall n,sothelimitis2.
Case2. 2 <x0 < 3.Supposethat2 <xn 1 ≤ 3forsome n ≥ 1.Then0 <xn 1 2 ≤ 1so √xn 1 2 ≥ xn 1 2, i.e., xn =2+ √xn 1 2 ≥ xn 1.Moreover, xn =2+ √xn 1 2 ≤ 2+1=3.Hencebyinduction, xn isincreasing andboundedaboveby3.ItfollowsfromtheMonotoneConvergenceTheoremthat xn → a as n →∞.Taking thelimitof2+ √xn 1 2= xn weseethat a2 5a +6=0,i.e., a =2, 3.Since xn increasesfrom x0 > 2,the limitis3.
Case3. x0 ≥ 3.Supposethat xn 1 ≥ 3forsome n ≥ 1.Then xn 1 2 ≥ 1so √xn 1 2 ≤ xn 1 2,i.e., xn =2+ √xn 1 2 ≤ xn 1.Moreover, xn =2+ √xn 1 2 ≥ 2+1=3.Hencebyinduction, xn isdecreasing
andboundedaboveby3.ByrepeatingthestepsinCase2,weconcludethat xn decreasesfrom x0 ≥ 3tothe limit3.
2.3.4. Case1. x0 < 1.Suppose xn 1 < 1.Then xn 1 = 2xn 1 2 < 1+ xn 1 2 = xn < 2 2 =1.
Thus {xn} isincreasingandboundedabove,so xn → x.Takingthelimitof xn =(1+ xn 1)/2as n →∞,wesee that x =(1+ x)/2,i.e., x =1. Case2. x0 ≥ 1.If xn 1 ≥ 1then
Thus {xn} isdecreasingandboundedbelow.RepeatingtheargumentinCase1,weconcludethat xn → 1as n →∞.
2.3.5. Theresultisobviouswhen x =0.If x> 0thenbyExample2.2andTheorem2.6, lim n→∞ x 1/(2n 1) =lim m→∞ x 1/m =1.
If x< 0thensince2n 1isodd,wehavebythepreviouscasethat x1/(2n 1) = ( x)1/(2n 1) →−1as n →∞
2.3.6. a)Supposethat {xn} isincreasing.If {xn} isboundedabove,thenthereisan x ∈ R suchthat xn → x (bytheMonotoneConvergenceTheorem).Otherwise,givenany M> 0thereisan N ∈ N suchthat xN >M Since {xn} isincreasing, n ≥ N implies xn ≥ xN >M .Hence xn →∞ as n →∞. b)If {xn} isdecreasing,then xn isincreasing,soparta)applies.
2.3.7. ChoosebytheApproximationPropertyan x1 ∈ E suchthatsup E 1 <x1 ≤ sup E.Sincesup E/ ∈ E, wealsohave x1 < sup E.Suppose x1 <x2 < <xn in E havebeenchosensothatsup E 1/n<xn < sup E ChoosebytheApproximationPropertyan xn+1 ∈ E suchthatmax{xn, sup E 1/(n +1)} <xn+1 ≤ sup E Thensup E 1/(n +1) <xn+1 < sup E and xn <xn+1.Thusbyinduction, x1 <x2 < ...andbytheSqueeze Theorem, xn → sup E as n →∞
2.3.8. a)ThisfollowsimmediatelyfromExercise1.2.6. b)Bya), xn+1 =(xn + yn)/2 < 2
< ··· <x1.Similarly, yn = y2 n < √xn
= yn+1 implies xn+1 >yn+1 >yn >y1.Thus {xn} isdecreasingandboundedbelowby y1 and {yn} isincreasing andboundedaboveby x1. c)Byb), xn
Hencebyinductionanda),0 <xn+1 yn+1 < (x1 y1)/2n d)Byb),thereexist x,y ∈ R suchthat xn ↓ x and yn ↑ y as n →∞.Byc), |x y|≤ (x1 y1) · 0=0.Hence x = y
2.3.9. Since x0 =1and y0 =0,
Noticethat x1 =1= y1.If
x
as n →∞,itfollowsthat xn/yn →±√2as n →∞.Since xn,yn > 0,thelimitmustbe √2.
2.3.10. a)Notice x0 >y0 > 1.If xn 1 >yn 1 > 1then y2 n 1 xn 1yn 1 = yn 1(yn 1 xn 1) > 0so yn 1(yn 1 + xn 1) < 2xn 1yn 1.Inparticular, xn = 2xn 1yn 1 xn 1 + yn 1 >yn 1
Itfollowsthat √xn > √yn 1 > 1,so xn > √xnyn 1 = yn > 1 1=1.Hencebyinduction, xn >yn > 1forall n ∈ N
Now yn <xn implies2yn <xn + yn.Thus xn+1 = 2xnyn xn + yn <xn.
Hence, {xn} isdecreasingandboundedbelow(by1).ThusbytheMonotoneConvergenceTheorem, xn → x for some x ∈ R
Ontheotherhand, yn+1 isthegeometricmeanof xn+1 and yn,sobyExercise1.2.6, yn+1 ≥ yn.Since yn is boundedabove(by x0),weconcludethat yn → y as n →∞ forsome y ∈ R
b)Let n →∞ intheidentity yn+1 = √xn+1yn.Weobtain,fromparta), y = √xy,i.e., x = y.Adirect calculationyields y6 > 3.141557494and x7 < 3.14161012.
2.4Cauchysequences.
2.4.0. a)False. an =1isCauchyand bn =( 1)n isbounded,but anbn =( 1)n doesnotconverge,hence cannotbeCauchybyTheorem2.29.
b)False. an =1and bn =1/n areCauchy,but an/bn = n doesnotconverge,hencecannotbeCauchyby Theorem2.29.
c)True.If(an + bn) 1 convergedto0,thengivenany M ∈ R, M =0,thereisan N ∈ N suchthat n ≥ N implies |an + bn| 1 < 1/|M |.Itfollowsthat n ≥ N implies |an + bn| > |M | > 0 >M .Inparticular, |an + bn| divergesto ∞.Butif an and bn areCauchy,thenbyTheorem2.29, an +bn → x where x ∈ R.Thus |an +bn|→|x|, NOT ∞.
d)False.If x2k =log k and xn =0for n =2k,then x2k x2k 1 =log(k/(k 1)) → 0as k →∞,but xk does notconverge,hencecannotbeCauchybyTheorem2.29.
2.4.1. Since(2n2 +3)/(n3 +5n2 +3n +1) → 0as n →∞,itfollowsfromtheSqueezeTheoremthat xn → 0 as n →∞.HencebyTheorem2.29, xn isCauchy.
2.4.2. If xn isCauchy,thenthereisan N ∈ N suchthat n ≥ N implies |xn xN | < 1.Since xn xN ∈ Z,it followsthat xn = xN forall n ≥ N .Thusset a := xN
2.4.3. Suppose xn and yn areCauchyandlet ε> 0.
a)If α =0,then αxn =0forall n ∈ N,henceisCauchy.If α =0,thenthereisan N ∈ N suchthat n,m ≥ N implies |xn xm| <ε/|α|.Hence
| <ε for n,m ≥ N
b)Thereisan N ∈ N suchthat n,m ≥ N implies |xn xm| and |yn ym| are <ε/2.Hence |xn + yn (xm + ym)|≤|xn xm| + |
for n,m ≥ N
c)ByrepeatingtheproofofTheorem2.8,wecanshowthateveryCauchysequenceisbounded.Thuschoose M> 0suchthat |xn| and |yn| areboth ≤ M forall n ∈ N.Thereisan N ∈ N suchthat n,m ≥ N implies |xn xm| and |yn ym| areboth <ε/(2M ).Hence |xnyn (xmym)|≤|xn xm||
for n,m ≥ N
2.4.4. Let sn = n 1 k=1 xk for n =2, 3, ....If m>n then sm+1 sn = m k=n xk.Therefore, sn isCauchyby hypothesis.Hence sn convergesbyTheorem2.29.
2.4.5. Let xn = n k=1( 1)k/k for n ∈ N.Suppose n and m areevenand m>n.Then S := m k=n ( 1)k k ≡ 1 n 1 n +1 1 n +2 1 m 1 1 m
Eachterminparenthesesispositive,sotheabsolutevalueof S isdominatedby1/n.Similarargumentsprevail forallintegers n and m.Since1/n → 0as n →∞,itfollowsthat xn satisfiesthehypothesesofExercise2.4.4. Hence xn mustconvergetoafiniterealnumber.
2.4.6. ByExercise1.4.4c,if m ≥ n then |xm+1 xn| = | m k=n (xk+1 xk)|≤ m k=n 1 ak = 1 1 am (1 1 an ) 1 a 1
Thus |xm+1 xn|≤ (1/an 1/am)/(a 1) → 0as n,m →∞ since a> 1.Hence {xn} isCauchyandmust convergebyTheorem2.29.
2.4.7. a)Suppose a isaclusterpointforsomeset E andlet r> 0.Since E ∩ (a r,a + r)containsinfinitely manypoints,sodoes E ∩ (a r,a + r) \{a}.Hencethissetisnonempty.Conversely,if E ∩ (a s,a + s) \{a} isalwaysnonemptyforall s> 0and r> 0isgiven,choose x1 ∈ E ∩ (a r,a + r).Ifdistinctpoints x1, ,xk havebeenchosensothat xk ∈ E ∩ (a r,a + r)and s :=min{|x1 a|, , |xk a|},thenbyhypothesisthereis an xk+1 ∈ E ∩ (a s,a + s).Byconstruction, xk+1 doesnotequalany xj for1 ≤ j ≤ k.Hence x1, ... ,xk+1 are distinctpointsin E ∩ (a r,a + r).Byinduction,thereareinfinitelymanypointsin E ∩ (a r,a + r).
b)If E isaboundedinfiniteset,thenitcontainsdistinctpoints x1,x2, ....Since {xn}⊆ E,itisbounded.It followsfromtheBolzano–WeierstrassTheoremthat xn containsaconvergentsubsequence,i.e.,thereisan a ∈ R suchthatgiven r> 0thereisan N ∈ N suchthat k ≥ N implies |xnk a| <r.Sincethereareinfinitelymany xnk ’sandtheyallbelongto E, a isbydefinitionaclusterpointof E
2.4.8. a)Toshow E :=[a,b]issequentiallycompact,let xn ∈ E.BytheBolzano–WeierstrassTheorem, xn hasaconvergentsubsequence,i.e.,thereisan x0 ∈ R andintegers nk suchthat xnk → x0 as k →∞.Moreover, bytheComparisonTheorem, xn ∈ E implies x0 ∈ E.Thus E issequentiallycompactbydefinition.
b)(0, 1)isboundedand1/n ∈ (0, 1)hasnoconvergentsubsequencewithlimitin(0, 1).
c)[0, ∞)isclosedand n ∈ [0, ∞)isasequencewhichhasnoconvergentsubsequence.
2.5Limitssupremumandinfimum.
2.5.1. a)Since3 ( 1)n =2when n isevenand4when n isodd,limsupn→∞ xn =4andliminf n→∞ xn =2.
b)Sincecos(nπ/2)=0if n isodd,1if n =4m and 1if n =4m +2,limsupn→∞ xn =1andliminf n→∞ xn = 1.
c)Since( 1)n+1 +( 1)n/n = 1+1/n when n isevenand1 1/n when n isodd,limsupn→∞ xn =1and liminf n→∞ xn = 1.
d)Since xn → 1/2as n →∞,limsupn→∞ xn =liminf n→∞ xn =1/2byTheorem2.36.
e)Since |yn|≤ M , |yn/n|≤ M/n → 0as n →∞.Therefore,limsupn→∞ xn =liminf n→∞ xn =0byTheorem 2.36.
f)Since n(1+( 1)n)+ n 1(( 1)n 1)=2n when n isevenand 2/n when n isodd,limsupn→∞ xn = ∞ and liminf n→∞ xn =0.
g)Clearly xn →∞ as n →∞.Therefore,limsupn→∞ xn =liminf n→∞ xn = ∞ byTheorem2.36.
2.5.2. ByTheorem1.20, liminf n→∞ ( xn):=lim n→∞ (inf k≥n ( xk))= lim n→∞ (sup k≥n xk)= limsup n→∞ xn. Asimilarargumentestablishesthesecondidentity.
2.5.3. a)Sincelimn→∞(supk≥n xk) <r,thereisan N ∈ N suchthatsupk≥N xk <r,i.e., xk <r forall k ≥ N b)Sincelimn→∞(supk≥n xk) >r,thereisan N ∈ N suchthatsupk≥N xk >r,i.e.,thereisa k1 ∈ N suchthat xk1 >r.Suppose kν ∈ N havebeenchosensothat k1 <k2 < <kj and xkν >r for ν =1, 2, ,j.Choose N>kj suchthatsupk≥N xk >r.Thenthereisa kj+1 >N>kj suchthat xkj+1 >r.Hencebyinduction,there aredistinctnaturalnumbers k1,k2, ...suchthat xkj >r forall j ∈ N
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2.5.4. a)Sinceinf k≥n xk +inf k
n yk isalowerboundof xj + yj forany j ≥ n,wehaveinf k≥n xk +inf k≥n yk ≤ infj≥n(xj + yj ).Takingthelimitofthisinequalityas n →∞,weobtain
Note,weusedCorollary1.16andthefactthatthesumontheleftisnotoftheform ∞−∞.Similarly,foreach j ≥ n,
Takingtheinfimumofthisinequalityoverall
.Therefore,
TheremainingtwoinequalitiesfollowfromExercise2.5.2.Forexample,
b)Itsufficestoprovethefirstidentity.ByTheorem2.36anda),
Toobtainthereverseinequality,noticebytheApproximationPropertythatforeach n ∈ N thereisa jn >n suchthatinf k≥n(xk + yk) >xjn 1/n + yjn .Hence
forall n ∈ N.Takingthelimitofthisinequalityas n →∞,weobtain
c)Let xn =( 1)n and yn =( 1)n+1.Thenthelimitsinfimumareboth 1,thelimitssupremumareboth1, but xn + yn =0 → 0as n →∞.If xn =( 1)n and yn =0then
(xn + yn)= 1 < 1=limsup n
xn +liminf n→∞ yn
2.5.5. a)Forany j ≥ n, xj ≤ supk≥n xk and yj ≤ supk≥n yk.Multiplyingtheseinequalities,wehave xj yj ≤ (supk≥n xk)(supk≥n yk),i.e., sup j≥n xj yj ≤ (sup k≥n xk)(sup k≥n yk).
Takingthelimitofthisinequalityas n →∞ establishesa).Theinequalitycanbestrictbecauseif xn =1 yn = 0 n even 1 n odd thenlimsupn→∞(xnyn)=0 < 1=(limsupn→∞ xn)(limsupn→∞ yn). b)Bya), liminf n→∞ (xnyn)= limsup n→∞ ( xnyn) ≥− limsup n→∞ ( xn)limsup n→∞ yn =liminf n→∞ xn limsup n→∞ yn
2.5.6. Case1. x = ∞.Byhypothesis, C :=limsupn→∞ yn > 0.Let M> 0andchoose N ∈ N suchthat n ≥ N implies xn ≥ 2M/C andsupn≥N yn >C/2.Thensupk≥N (xkyk) ≥ xnyn ≥ (2M/C)yn forany n ≥ N and supk≥N (xkyk) ≥ (2M/C)supn≥N yn >M .Therefore,limsupn→∞(xnyn)= ∞
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Case2. 0 ≤ x< ∞.ByExercise2.5.6aandTheorem2.36, limsup n→∞ (xnyn) ≤ (limsup n→∞ xn)(limsup n→∞ yn)= x limsup n→∞ yn
Ontheotherhand,given > 0choose n ∈ N sothat xk >x for k ≥ n.Then xkyk ≥ (x )yk foreach k ≥ n, i.e.,supk≥n(xkyk) ≥ (x )supk≥n yk.Takingthelimitofthisinequalityas n →∞ andas → 0,weobtain limsup n→∞ (xnyn) ≥ x limsup n→∞ yn.
2.5.7. Itsufficestoprovethefirstidentity.Let s =inf n∈N(supk≥n xk).
Case1. s = ∞.Thensupk≥n xk = ∞ forall n ∈ N sobydefinition, limsup n→∞ xn =lim n→∞ (sup k≥n xk)= ∞ = s.
Case2. s = −∞.Let M> 0andchoose N ∈ N suchthatsupk≥N xk ≤−M .Thensupk≥n xk ≤ supk≥N xk ≤ M forall n ≥ N ,i.e.,limsupn→∞ xn = −∞
Case3. −∞ <s< −∞.Let > 0andusetheApproximationPropertytochoose N ∈ N suchthat supk≥N xk <s + .Sincesupk≥n xk ≤ supk≥N xk <s + forall n ≥ N ,itfollowsthat s <s ≤ sup k≥n xk <s + for n ≥ N ,i.e.,limsupn→∞ xn = s
2.5.8. Itsufficestoestablishthefirstidentity.Let s =liminf n→∞ xn
Case1. s =0.ThenbyTheorem2.35thereisasubsequence kj suchthat xkj → 0,i.e.,1/xkj →∞ as j →∞
Inparticular,supk≥n(1/xk)= ∞ forall n ∈ N,i.e.,limsupn→∞(1/xn)= ∞ =1/s.
Case2. s = ∞.Then xk →∞,i.e.,1/xk → 0,as k →∞.ThusbyTheorem2.36,limsupn→∞(1/xn)=0=1/s
Case3. 0 <s< ∞.Fix j ≥ n.Since1/ infk≥n xk ≥ 1/xj implies1/ infk≥n xk ≥ supj≥n(1/xj ),itisclearthat 1/s ≥ limsupn→∞(1/xn).Ontheotherhand,given > 0and n ∈ N,choose j>N suchthatinf k≥n xk + >xj , i.e.,1/(inf k≥n xk + ) < 1/xj ≤ supk≥n(1/xk).Takingthelimitofthisinequalityas n →∞ andas → 0,we concludethat1/s ≤ limsupn→∞(1/xn).
2.5.9. If xn → 0,then |xn|→ 0.ThusbyTheorem2.36,limsupn→∞ |xn| =0.Conversely,iflimsupn→∞ |xn|≤ 0,then 0 ≤ liminf n→∞ |xn|≤ limsup n→∞ |xn|≤ 0, impliesthatthelimitssupremumandinfimumof |xn| areequal(tozero).HencebyTheorem2.36,thelimitexists andequalszero.