Math 317 Week 01 Real Infinite Series

Transcription

1 Mth 37 Week 0 Rel Ifiite Series Lst Updte: Februry 7, 04 Tble of cotets Mi refereces Defiitios d properties Ifiite series Ifiite sum Covergece through prtil sum Criteri for covergece No-egtive series Covergece/divergece through compriso Typicl o-egtive series d their implictios Tests Rtio d root tests Rbe s test d Guss s test Advced Topics, Notes, d Commets Abel s formul Coditiolly coverget series There is o ultimte test Ifiite product Some fu exmples More exercises d problems Bsic exercises Covergece of ifiite series: Defiitio d properties Compriso Tests Abel s re-summtio More exercises Problems

2 Mth 37 Week 0 Rel Ifiite Series Mi refereces Bimore Bimore, K. G., Mthemticl Alysis: A Strightforwrd Approch, Cmbridge Uiversity Press, 977, Chpter 6. Folld Folld, Gerld B., Advced Clculus, Pretice-Hll, Ic., 00. Chpter 6. Klie Klie, Morris, Mthemticl Thoughts from Aciet to Moder Times, Oxford Uiversity Press, 97. Chpter 0. RIS Bor, Diel D., Khoury, Michel J., Rel Ifiite Series, The Mthemticl Associtio of Americ, 006. USTC3 He, Che, Shi, Jihui, Xu, Seli, ShuXueFeXi Mthemticl Alysis III, Higher Eductio Press, 985, Sectio 8.. PKU She, Xiechg, ShuXueFeXi Mthemticl Alysis II, Higher Eductio Press, 986, Chpter. Bby Rudi Rudi, Wlter, Priciples of Mthemticl Alysis, McGrw-Hill Compies, Ic., 3rd editio, 976. Chpter 3. Euler Sdifer, C. Edwrd, The Erly Mthemtics of Leohrd Euler, MAA, 007.

3 Lst Updte: Februry 7, Ifiite series Ifiite sum. Defiitios d properties Defiitio. Ifiite series Give sequece { } of rel umbers, the forml sum is clled ifiite series. = Remrk. Note tht is just other wy of writig the forml sum They me exctly the sme thig which, up to ow, is othig. Remrk 3. The summtio i of these ifiitely my rel umbers is forml becuse it is ot cler wht it mes to sy = s R. Essetilly, to defie the sum of,...,,... is to defie fuctio f: R R R R. 3 This turs out to be tricky tsk, s c be see from the followig exmple. Exmple 4. Some exmples of ifiite series: = si = si + si + 5 = = It is ituitively cler tht the vlue of should be while should be. It s ot cler whether the first two sums correspod to y vlue. I prticulr, my differet vlues c be resobly ssiged to the first sum. Exmple 5. RIS, 6.3, Fllcy 7 Let We hve be y ifiite series. Let S R be rbitrry. =S S ; =S S ;... 8 Everythig ccel i the sum except S. Therefore S =. Remrk 6. For more such fllcious results, check out Chpter 6 of RIS.. Meig: followig ccepted rules of ddig fiitely my umbers, such s groupig d re-rrgemet.

4 4 Mth 37 Week 0 Rel Ifiite Series.. Covergece through prtil sum Oe good wy of defiig ifiite sum is through the limit of prtil sums. Defiitio 7. Prtil sum d covergece The th prtil sum of ifiite series is defied s s = m= m. If the sequece {s } coverges to some rel umber s, the we sy the ifiite series coverges to s, or equivletly sy its sum is s, d simply write = s. 9 If s or, we sy the ifiite series diverges to or respectively d write Remrk 8. Note tht = or. 0 s =,s = +,s 3 = + + 3,... Recllig theorems for the covergece of sequeces, we hve Theorem 9. = s if d oly if for y ε > 0, there is N N such tht for ll > N, s m < ε; m= = if d oly if for y M R, there is N N such tht for ll >N, m >M; 3 m= = if d oly if for y M R, there is N N such tht for ll >N, Exmple 0. Prove tht Proof. By iductio we c prove tht m < M. 4 m= =. 5 s = 6 d the coclusio esily follows. Exmple. Prove tht + coverges d fid the vlue. Proof. By iductio we c prove tht s = + 7

5 Lst Updte: Februry 7, 04 5 Therefore =. + Exercise. Study the covergece/divergece of =0 Exercise. Prove the followig: i. A coverget series hs uique sum; +. Hit:. + ii. If =A R {±}, d c R, c=/ 0, the c = c A; iii. If =A, b = B, the + b = A +B;.3. Criteri for covergece Theorem. Cuchy A ifiite series coverges to some s R if d oly if for y ε > 0 there is N N such tht for ll m > >N, m k < ε. 8 k=+ Exercise 3. Prove the bove theorem. Hit: 3 Corollry 3. If coverges to s R the lim = 0. Equivletly, if lim does ot exist, or exists but is ot 0, the does ot coverge to y rel umber. Proof. For y ε >0, sice ll m >>N, coverges, it is Cuchy d there exists N N such tht for m k=+ k < ε. 9 Now tke N = N +. The for y >N, we hve > >N which gives 0 = k <ε. 0 k= Thus by defiitio of covergece of sequece lim = 0. Remrk 4. The bove corollry is very useful, however we should keep i mid tht:. The coverse is ot true. Tht is lim =0 does ot imply the covergece of.. It cot be pplied to coclude =/ or. Exercise 4. Give couterexmple to the followig clims:. = lim = 0= = s for some s R. =s R ext = lim 3. Recll Defiitio 7. The check the Cuchy criterio for ifiite sequece. = 0.

6 6 Mth 37 Week 0 Rel Ifiite Series Hit: 4 Exmple 5. Let = c r for r,c R. The If r <, the = c r. b If r, the + c > 0 = 0 c= 0. c < 0 c If r, the does ot exist s exteded rel umber. Proof. We prove for c =. The geerliztio is trivil. We hve m= m = + +r = r r. 3 For y ε > 0, tke N N such tht N log r [ε r], the for y > N, r m= m = r r < r N < ε. 4 r b For y M R. Tke N N such tht N > M. The for every >N we hve m =>N > M M. 5 m= m= c Sice r,. Therefore by Corollry 3 does ot coverge to y rel umber. We still eed to show tht =/,. To do this, we show tht s = m= stisfies s 0 whe is odd d s 0 whe is eve. Clerly s = > 0, s =+r 0. For 3, clculte s = m= r m +r = r r As r, r which gives r + r r +r. 6 r r. 7 Therefore, s d r cot tke opposite sigs. Exmple 6. We hve = = ; 5 = ;.

7 Lst Updte: Februry 7, No-egtive series.. Covergece/divergece through compriso I most situtios, it is impossible t lest very hrd to explicitly clculte the prtil sum S := m= m d it is therefore ot possible to estblish covergece/fid the sum bsed o defiitio oly. Oe wy to overcome this is through the followig comprisio theorem. Theorem 7. Compriso Let d b be two ifiite series. Assume tht there re c >0 d N 0 N such tht cb for ll >N 0. The b coverges = coverges. b does ot coverge 5 = b =. Proof. Note tht d b re equivlet logicl sttemets 6, so we oly eed to prove. We show tht is Cuchy. For y ε > 0, sice b coverges, it is Cuchy d there is N N such tht for ll m > >N, m b k < ε c. 9 k=+ Tke N = mx {N,N 0 }. The for y m > >N, m k=+ k m k=+ k c m k=+ b k < ε. 30 So is Cuchy d therefore coverges. Exercise 5. Show tht the = i the bove cot be replced by. Also show tht the bsolute vlue is ecessry i c b. Hit: 7 Exercise 6. Let be o-egtive series. The it coverges it is bouded bove. Hit: 8 Exmple 8. It is cler by the bove theorem tht if coverges, so does. The coverse is ot true, s c be see from the followig exmple: Tke = +. The we clerly see tht S = = Note tht here it is ot ecessry for to be. 6. By the hypotheses i the Theorem, b 0. See Exercise Cosider = /, b = /. 8. Thus the prtil sum m= m is icresig d there re oly two cses: bouded bove or ot.

8 8 Mth 37 Week 0 Rel Ifiite Series coverges. O the other hd, we hve S + S 0 so S + coverges to the sme limit. From here it is esy to prove by defiitio tht S to the sme limit, which turs out to be l. Exmple 9. Prove tht Proof. We otice tht > = The coclusio follows from + + =. Exercise 7. Prove the followig limit compriso test. Let, b be two o-egtive series d further ssume b > 0. Assume tht The lim = L R {+}. 34 b i. If 0 < L <, the, b both coverge or both diverge; ii. If L = d coverges, the b coverges; iii. If L = 0 d diverges, the b diverges. The improve the result usig limsup d limif. Hit: 9 Exercise 8. Apply the limit compriso test to study Exercise 9. C we drop the o-egtive ssumptio i the bove limit compriso test? Hit: 0 Remrk 0. A sequece tht coverges but with = is clled coditiolly coverget. O the other hd, sequece such tht coverges is sid to be bsolutely coverget. Absolutely coverget sequeces c udergo y re-rrgemet d still coverge to the sme sum but coditiolly coverget sequeces do ot ejoy such property. Exercise 0. Prove tht if does ot coverge to some s R, the =. Remrk. The pl I light of the Theorem 7, it is importt to study o-egtive series, tht is ifiite series stisfyig 0 for ll N. Oce o-egtive sequece b is show to be coverget, we kow tht y stisfyig c b for some costt c is lso coverget. It is further possible to mke this compriso itrisic, tht is desig some criterio ivolvig oly while still gurtees the reltio c b. Such criteri re usully clled tests. We will study the simpliest tests i the followig sectio. 9. For exmple, whe 0 < L <, there is N N such tht for ll > N, L < < 3 L b. 0. No. Cosider = /, b = /.

9 Lst Updte: Februry 7, Typicl o-egtive series d their implictios Exmple. Geometric series We hve see tht cosequece, if other series stisfies for some c > 0 d for ll > some N 0 N, the r coverges whe 0 r <. As cr 35 coverges. Exmple 3. Geerlized hrmoic series/p-series The series coverges whe > d diverges whe. Proof. Whe, we hve therefore it suffices to show the divergece of the followig trick: Therefore we hve recll s is the prtil sum. We use >, > =, > =, 38 s k > k + 39 where k = k = k+. Therefore s is ot bouded d the series diverges to. Whe >, we use the followig trick: ; < = ; < 4 4 = ; 4 We see tht N < = < This proof is ttributed to Nicole Oresme c , whose motivtio my be If God were Ifiity, the diverget series would be His gels flyig higher d higher to rech Him. The result ws re-proved my times, icludig oe i the book Trcttus de Seriebus Ifiitis Tretise o Ifiite Series by Jcob Beroulli, who commeted So the soul of immesity dwells i miuti. Ad i rrowest limits o limits ihere. Wht joy to discer the miute i ifiity! The vst to perceive i the smll, wht diviity! Quotes from Clifford A. Pickover, The MαTH βook, p.04. Accordig to Klie the proof ppered i Oresme: Questioes Super Geometrim Euclidis c. 360.

10 0 Mth 37 Week 0 Rel Ifiite Series Thus this o-egtive series is bouded from bove d therefore coverges. Remrk 4. Usully the covergece/divergece of geerlized hrmoic series is studied through the followig reltio betwee o-egtive series d improper itegrls. Theorem. Itegrl test Cosider series. Assume there is fuctio f:[, R + such tht f=. Further ssume f to be itegrble d decresig. The coverges if d oly if fx dx 44 coverges. Furthermore i the cse of covergece, we hve = < fx dx <. 45 We see here tht it c be doe directly. It is lso worth metioig tht the techique i the bove proof is specil cse of the so-clled Cuchy s Codestio Test. Exercise. Prove the Itegrl test theorem. Hit: Remrk 5. The sme method c show tht log +, log + log[log + +], coverges whe > d diverges whe. Exercise. Prove tht log + 48 is coverget for > d diverget for usig two methods: Directly d through comprig gist certi improper itegrls. Hit: 3. Notice 3. For the direct proof, whe >, show tht Thus this positive series is bouded bove. + fxdx k logk + =. 49 k= +

11 Lst Updte: Februry 7, Rtio d root tests 3. Tests The followig two itrisic covergece tests bsed o compriso with geometric series re the simplest d most populr tests for covergece/divergece. Theorem 6. Rtio test Let N. The If limsup + <, the series coverges. be ifiite series. Further ssume tht =/ 0 for ll If limif + >, the series diverges. Proof. Assume limsup + = L <. Set r = L + limsup + = lim d ε 0 = L. By defiitio { sup } k+ k k 50 therefore there is N N such tht for ll >N, sup k+ k k L < ε 0 5 which gives sup + >N <L+ε 0=r <= 0 < + <r <. 5 This gives, for ll >N +, < N+ r N = N+ r N r. 53 Note tht sice N is fixed, we hve for ll > N d cosequetly <cr 54 coverges. Assume limif + =L>. Set ε0 =L. The by defiitio, similr to the limsup cse bove, there is N N such tht for ll >N, + > 55 which mes for ll N + N+ 56 As cosequece 0. By Corollry 3 we kow tht diverges. Exmple 7. Prove tht Proof. We hve! coverges. + = 57 +

12 Mth 37 Week 0 Rel Ifiite Series therefore So the series coverges. limsup + =0<. 58 Exercise 3. Let be ifiite series. Let b be the series obtied by droppig ll 0 s from { }. Prove tht = s b = s 59 for s R ext. Hit: 4 Now use this result to geerlize the rtio test to ll ifiite series without ssumig =/ 0. Note tht both re limsup! Root test is very specil i this sese. Theorem 8. Root test Let be ifiite series. The If limsup / <, the the series coverges. If limsup / >, the the series diverges. Proof. Assume limsup / = L <. Set r = L + d ε 0 = L. The by defiitio, s i the proof of the bove rtio test, there is N N such tht for ll > N, Therefore coverges. / <r <= <r. 60 Assume limsup / >. The proof is left s exercise. Remrk 9. It turs out tht for y sequece {x }, limif x + x limif x / limsup x / limsup x + x. 6 Therefore the root test is shrper th the rtio test, i the sese tht y series tht psses the rtio test for covergece or divergece will lso pss the root test. Exmple 30. Cosider the ifiite series = { 3 where = k = k 3 k. It c be esily verified tht = k limif + = 0; limsup so the rtio test does ot pply. O the other hd so the root test tells us tht the series coverges. 3.. Rbe s test d Guss s test + = 6 limsup / = 63 Note tht either the rtio test or the root test works for the geerlized hrmoic series. Exercise 4. Verify this clim. Hit: 5 4. The prtil sums of b form subsequece of the sequece of prtil sums of.

13 Lst Updte: Februry 7, 04 3 Theorem 3. Rbe s Test 6 Let coverges if diverges if Exercise 5. Cosider be positive ifiite series. The limif > 64 + limsup < d Prove tht either the rtio test or the root test pplies. Apply Rbe s test to prove tht the first oe diverges but the secod oe coverges. Exercise 6. Let p >0. Prove tht either the rtio test or the root test pplies. Apply Rbe s test to prove tht pp + p+ 67! diverges. Exercise 7. Let x > 0. Study the series Hit: 7! x + x Proof. We prove the covergece prt d leve the other hlf s exercise. Sice limif >, there is β >0 such tht + > + β 70 + for ll > some N 0 N. Thus we hve for such, We prove tht there is γ >0 such tht for lrge. By MVT we hve, for some ξ >+ + β β >, +, + +γ 7 + +γ = + + γ ξ γ + γ <+γ. 73 Now y γ such tht γ + γ <+ β suffices lim /. = ; lim 6. Joseph Ludwig Rbe, We hve = x x Therefore the series coverges if x > d diverges if x <. Whe x = we hve = so the series lso diverges i + this cse.

14 4 Mth 37 Week 0 Rel Ifiite Series Now we hve < d coverges follows. Exercise 8. Prove the other hlf: +γ +γ < < < c +γ 74 limif diverges if limsup + <. 75 Remrk 3. Note tht the coverget prt of Rbe s test c be tured ito covergece test for geerl series without requirig >0 s + but the diverget prt cot, tht is limsup + >= coverges 76 <= diverges 77 is ot true, for the followig reso. The covergece of implies the covergece of ; But the divergece of does ot imply the divergece of, s the exmple + shows Note tht Rbe s Test is shrper th the rtio/root tests, due to the fct tht coverges/diverges slower th the geometric series. If we cosider eve slower coverget series, such s, we c obti eve shrper test such s the followig. log + Theorem 33. Guss s Test 8 Let write + where {θ } is bouded sequece d κ > 0. The i. If β >, the ii. If β, the coverges; diverges. be positive ifiite series. Further ssume we c = β + θ +κ 78 Proof. See Theorem. of RIS. Note tht the oly ew situtio here is β =. Remrk 34. It is possible to desig eve fier tests usig the covergece for > /divergece for of log + loglog + + log log The results form the so-clled Bertrd s Tests. See.4 of RIS. 8. Krl Friedrich Guss

15 Lst Updte: Februry 7, Advced Topics, Notes, d Commets 4.. Abel s formul Exmple 35. Cosider We show tht it ctully coverges. Cosider the prtil sum Deote A k = k l= m k=+ sik k si = si + si + 80 S m = m k=+ si l, d B k =. The we hve k = m k=+ A k A k B k sik k. 8 = [A + B + A B + ] +[A + B + A + B + ] + +[A m B m A m B m ] = [A m B m A B + ]+ +[A + B + B + +A + B + B A m B m B m ] m = [A m B m A B + ]+ [A k B k B k+ ] = [ Am m A ] + k=+ m [ + k=+ A k k k + Now otice tht {A } is i fct bouded sequece: ]. 8 A = si + si + +si si [si + si + +si] = si cos cos + +cos cos + + +cos cos + = si [cos0 + cos+ + cos ] [cos+cos3+ +cos +] = si cos0+cos cos cos + =. 83 si Now it is cler tht A for ll N. si Bck to 8: m sik k A m m m + A + + A k k k + k=+ [ k=+ si m + m + + k ] k + k=+ = [ ] 4 = si + + si. 84

16 6 Mth 37 Week 0 Rel Ifiite Series Now we re redy to show the series is Cuchy: For y ε >0, tke N N such tht N + 4, the for y m >>N, we hve ε si m si k k 4 + si < 4 ε. 85 N + si k=+ Therefore the series coverges. Exercise 9. Prove the followig Abel summtio formul: Let A k = k l= m k=+ k B k = [A m B m A B + ] m k=+ Drw logy to the formul of itegrtio by prts: b fx Gx dx=[fb Gb F G] where Fx = x ft dt, Gx= x gt dt. Exercise 0. Prove tht if coverges, the so does Theorem 36. Cosider with 0. If i. There is N 0 N such tht for ll >N 0, + ; ii. lim =0; the coverges to some s R. Exercise. Prove the theorem. Hit: 0 b l, B k = k l= b l, the A k b k+ 86 Fx gx dx 87. Hit:9 4.. Coditiolly coverget series Recll tht series is clled coditiolly coverget if it is coverget, but =. Otherwise it is clled bsolutely coverget. Exmple 37. Groupig Uless 0 or ll 0 d coverges, the order of summtio cot be chged. For exmple let = +. If we re llowed to group terms together d sum them first, we would hve both =+ ++ =+[ +] +[ +] + =+0+0+ =; 89 =+ + + =[ + ] +[ + ] + = =0. 90 Defiitio 38. Rerrgemet A rerrgemet of ifiite series ifiite series m= m where m: m is bijectio from N to N. is other 9. Deote S := m. The there is M >0 such tht for ll N, S <M. Now we c show the series is Cuchy through m k k k=+ = m Sk S k = k k=+ S + + S m m m + S k k k + k=+ S + + S m m m + S k k. 88 k + k=+ 0. Apply Abel s formul to the prtil sums.

17 Lst Updte: Februry 7, 04 7 Exmple 39. A exmple of rerrgemet of is Exercise. Prove tht if the sme sum. Hit: = s is bsolutely coverget, the y re-groupig or re-rrgemet give Exmple 40. Rerrgemet Cosider the sequece with = +. If we re llowed to freely rerrge tht is choose the order of summtio, the for y s R {, }, there is rerrgemet such tht it coverges to s. Proof. Cosider the cse s R. The cses s=, re left s exercises. Cosider the rerrgemet b defied s follows: Let k 0 be such tht k 0 s but <s. Set k 0 3 b =, b = 3,...,b k 0 = k 0 ; 9 The cse k 0 = is whe s. The we just set b = d tur to the ext step. Let k be such tht d set k 0 b k k= Let k be such tht k 0 +k b k + s k= d set Ad so o. Now set + + k b k0 +=, s, k k 0 +k b k0 +k += k 0 k= b k + + < s 93 k b k 0 +k = k k s, 0 +k b k + k= k k 0 + k 3 < 95 k 0 +,..., b k 0 +k +k = k 0 + k +, 96 k 0 +k + +k l S l = b k. 97 k= The we see tht if [k 0 + +k l, k 0 + +k l+ ], the s = m= b m 98. For re-groupig, ot tht the covergece of the re-grouped series is equivlet to the covergece of subsequece S k of the prtil sums S. The coclusio follows from, for y > k, S S k m=k+ m. 9 For re-rrgemet, first prove tht y positive coverget series c be re-rrged without chgig the sum. The cosider + d.

18 8 Mth 37 Week 0 Rel Ifiite Series is lwys betwee S l d S l+. Filly otice tht by costructio, S l s < l. Thus for y ε>0, tke L N such tht L>ε. Now set N = k k L. For y > N, there is l L such tht [k k l, k k l+ ]. Therefore we hve s s mx { S l s, S l+ s } l <ε. 99 L Tht is b s by defiitio. Remrk 4. Note tht the bove proof depeds o the fct tht k= k =, k= k =. 00 Exercise 3. Prove the bove two fcts. Hit: Theorem 4. Let be series whose terms go to 0. The exctly oe of the followig holds: coverges bsolutely; All re-rrgemets diverge to ; All re-rrgemets diverge to ; The series c be re-rrged to sum to y r R {, }. Proof. Problem There is o ultimte test We prove here tht ultimte test does ot exist. More specificlly, we cot fid the lrgest coverget series or the smllest diverget series. Theorem 43. RIS Gem 47, 48, CMJ, 8:4, pp Let be coverget series of positive terms. The there exists other coverget series A tht decreses more slowly i the sese tht A lim =. 0 b Let D be diverget series of positive terms. The there exists other diverget series d tht is smller i the sese tht Proof. Let r := +. The we hve r ց 0. Now defie. Compriso with. D lim =. 0 d A := r α. 03

19 Lst Updte: Februry 7, 04 9 for y α 0,. The we hve N A = N α r r r + Sice this boud is uiform i N, we see tht 0 r tα dt. 04 A coverges. b Let S :=D + +D, the S sice D diverges. Defie d = D S. The we hve m k=+ d k m k=+ D k = S m S = S. 05 S m S m S m Sice lim S =, for y N, there is m N such tht S < S m. Thus d is ot Cuchy d cot coverge Ifiite product Cosider the product of ifiitely my rel umbers: p := p p p Note the s=/ 0! Defiitio 44. We sy the ifiite product p coverges if d oly if P = p p coverges to s R, s=/ 0. We sy p diverges if d oly if it does ot coverge. Exercise 4. Prove tht diverges. Exercise 5. Prove tht, if p coverges, the lim p =. Hit: 3 Exercise 6. Prove tht, if p coverges, the there is N N such tht for ll > N, p > 0. Hit: 4 Exmple 45. =. Proof. We hve P = + + = The coclusio follows. Exercise 7. Let x <. Prove tht +x = x. 08 Now we cosider writig p = + with >. From the exercises followig Defiitio 44 we see tht this is resoble. Exercise 8. Prove tht if + coverges, the lim = Tke l or prove directly. 4. Otherwise P oscilltes.

20 0 Mth 37 Week 0 Rel Ifiite Series Theorem 46. The ifiite product + coverges if d oly if the ifiite series l + coverges. Furthermore i this cse we hve [ ] + = exp l Exercise 9. Prove the theorem. Theorem 47. If there is N N such tht > N, >0, the + coverges coverges. 0 Proof. Whe >0 we hve > l + >0 which gives =. For = we otice tht the covergece of + gives lim = 0 which implies l + lim =. From this it is esy to obti l + coverges = coverges. Exercise 30. Prove tht, if there is N N such tht > N, < 0, the + coverges coverges. Defiitio 48. Sy + bsolutely coverget, if d oly if + coverges. Exercise 3. If Exercise 3. If 4.5. Some fu exmples + is bsolutely coverget, it is coverget. + is bsolutely coverget, the y re-rrgemet coverges to the sme vlue. Exmple 49. We kow tht the hrmoic series obtied by droppig ll terms ivolvig 9: diverges to. Now cosider the sequece Does this series coverge? As it is o-egtive sequece, we oly eed to check whether it s bouded bove. We hve, + + <8, < I geerl, there re 8 9 k terms betwee 0 k d 9 k. 8 0 Overll the sum is bouded bove by 9 8 = 8 0 =0 9 0, so their sum is bouded bove by 0 k+ = 80. 6

21 Lst Updte: Februry 7, 04 Therefore the ew series coverges. Remrk 50. Obviously we c try to study the sequece resulted from deletig ll terms ivolvig other digits, or sequece of umbers. For exmple we c delete ll terms ivolvig the combitio 43, tht is 435 is deleted while is ot. We c eve ply some silly gmes such s deletig ll 4537 terms ivolvig, or someoe s birthdy. The resultig sequeces re ll coverget. Exmple 5. RIS Gem 87 For ech positive iteger, let f be the umber of zeros i the deciml represettio of. Choose > 0 d cosider the series This series coverges for 0< < 9 d diverges for 9. f. 7 Proof. Fix k N. There re exctly m k 9 m+ k umbers with f = k. Now we re-rrge the series such tht ll umbers with the sme umber of digits d the sme f re summed first. This is OK sice the series is positive. If we sum ll umbers with m + digits, we hve f m m 9 k m+ k k 0 m+ = 9 m k=0 Thus we prove the divergece for 9. For the other hlf, we use d proceed similrly. f m m 9 k m+ k k m m = k=0 Exmple 5. Fermt s Lst Theorem This is dpted from the blog of Terece To of UCLA 5. We ll kow tht Fermt s Lst Theorem clims tht x + y = z, x, y, z N 0 does ot hve y solutio whe 3. O the other hd, it is well-kow tht whe =, there re ifiitely my solutios. But why? Wht s the differece betwee = d >? It turs out tht we c revel some differece through kowledge of covergece/divergece of ifiite series. Let s cosider the chce of three umbers,b,+b re ll the th power of turl umber. If we tret s typicl umber of size, the it s chce of beig th power is roughly / /. Igorig the reltio betwee,b,+b, we hve the followig probbility for,b,+b solvig the equtio 0: Now cosider ll umbers, b, we sum up the probbilities: I := = b +b. [ b= b +b ] 5. The probbilistic heuristic justifictio of the ABC cojecture, lik t

22 Mth 37 Week 0 Rel Ifiite Series d pply the followig ituitio bsed o the so-clled Borel-Ctelli Lemm i probbility: If I <, the the chce of 0 hvig solutio is very low, while if I =, the chce is very high. We otice tht I hs perfect symmetry betwee d b, which mes I = = [ b= b +b ] + =. 3 It is cler tht the secod series coverges for ll so c be igored for our purpose. Now cosider J = [ b +b ]. 4 = b= Whe b, we hve <+b< therefore which gives b <J < b 5 = b= = b= b < J < b. 6 = b= = b= So filly ll we eed to study is the covergece/divergece of We hve Therefore K = b = [ b ]. 7 = b= = b= b= b x dx. 8 K = 3 9 which is coverget whe 4 while diverget whe =. Remrk 53. The cse = 3 is bit tricky here. The series = 3 becomes the Hrmoic series = which is the borderlie betwee covergece d divergece. Our rgumet does ot provide y isight o why x 3 + y 3 = z 3 should ot hve solutios. Exercise 33. RIS, 6., Puzzle 3 Desert D is tryig to cross 500-mile desert i jeep. Ufortutely, his jeep could hold oly eough fuel to trvel 50 miles. However, Desert D hs ifiite collectio of cotiers tht he c use to leve cches of fuel for himself to use o lter trip. Further, o the eterig edge of the desert there is boudless supply of fuel, so tht Desert D c retur to the eterig edge of the desert s ofte s he wishes. C he cross the desert? Hit: Footote Desig strtegy so tht i trips Desert D c trvel miles.

23 Lst Updte: Februry 7, More exercises d problems Note. My of the followig problems re from RIS Chpters 4 d 5. Plese ote tht I chose those problems tht re ot prticulrly tricky d my help uderstdig cocepts d estblishig ituitios. Plese check out RIS if you would like to chllege your bri power. 5.. Bsic exercises 5... Covergece of ifiite series: Defiitio d properties Exercise 34. Fid diverget series such tht for every fixed k N, lim + + +k = 0. Hit: 7 Exercise 35. Let be ifiite series. Let b be the series obtied from by deletig ll the zeroes. For exmple would be come Prove tht coverges if d oly if b coverges, d whe coverge they hve the sme sum. Hit: See this footote. 8 Exercise 36. Let be ifiite series. Let be y groupig of the series. Prove tht if this ew series does ot coverge some fiite umber, the does ot coverge either. Hit: 9 Exercise 37. RIS Prove Hit: Compriso k + 3 k + =. 30 k +! k=0 Exercise 38. Assume d b coverge. Prove tht + b coverges. Exercise 39. Assume d b coverge. Prove tht + b coverges. Hit: 3 Exercise 40. Putm 940 Assume d b coverge. Let p. Prove tht b p coverges. Hit: 3 Exercise 4. RIS If, b re o-egtive d coverget, the coverget. Does the clim still hold if o-egtivity is dropped from either or b? Hit: 33 +b Exercise 4. RIS Suppose coverges but diverges. Prove tht diverges. is lso Exercise 43. Rtio compriso test Let, b be positive series. Assume there is N 0 N such tht > N 0, + b+. The b i. If b coverges, the coverges; ii. If diverges, the b diverges. Exercise 44. Prove or disprove: If > 0 d coverges, the t coverges. Hit: 34 Exercise 45. Prove tht 7. = /. si si + 8. The prtil sums of b is subsequece of tht of. 9. Cuchy criterio Or defiitio, otice tht the ew prtil sums form subsequece of the prtil sums of. 30. k + 3 k + =k + k +. 3., b 0 therefore, b b for lrge. 3. If b the b p b. 33. No. Cosider =, b = Compriso. 3

24 4 Mth 37 Week 0 Rel Ifiite Series coverges. Hit: Tests Exercise 46. Putm 94 Is the followig series coverget or diverget? Hit: ! ! ! Abel s re-summtio Exercise 47. Prove tht if coverges the coverges. Hit: 37. Exercise 48. Let λ be y positive umber. Prove tht if coverges the e λ lso coverges. Hit: 38 Exercise 49. Prove the followig Dirichlet s Test: Cosider b. Assume tht b decreses to 0 d the prtil sums of re uiformly bouded i. The b coverges. Exercise 50. Prove the followig Abel s Test: Let be coverget d {b } positive d decresig lim b my ot be 0. The b coverges. 5.. More exercises Exercise 5. Let be positive d decresig. The coverges = lim =0. Further prove tht positive d decresig re both ecessry. Also prove tht the coverse is flse. Hit: 39 Exercise 5. Let >0 d coverget. Set r := k= k. Prove tht r diverges. Hit: 40 Exercise 53. RIS Gem, CMJ, 6:, p, 79 The series p 33 coverges if d oly if p >. Hit: See this footote. 4 Exercise 54. RIS / +/ diverges. Hit: See footote 4. Exercise 55. Putm 950 Study the covergece/divergece of the followig series Hit: Footote. 43 log! + + log! + ; / 3 3 / /3 3 Exercise 56. RIS For y positive itegers d p, we hve Hit: 44 m= m m+ m + p = p + + p MVT. 36. Rtio test. 37. Abel resummtio. 38. e λ is decresig for lrge. 39. N m N m for every m > N. 40. Cuchy criterio Set b = 3 5 p. Prove tht b < < b. 4. Compre with diverget series. 43. First oe: compre with ; Secod oe, + +/ log. log

25 Lst Updte: Februry 7, 04 5 Exercise 57. RIS Proce tht but Hit: = = log Exercise 58. Let b be positive series. If b diverges, the so does b + b. Hit: Footote. 46 Exercise 59. Prove tht / =. Hit: Footote. 47 Exercise 60. Let { be coditiolly coverget. Let b ={ > d c 0 = <0. Prove tht b = ; c =. 38 Hit: 48 Exercise 6. Putm 956 Give T =, T + =T T +, >0. Prove =. Hit: Footote 49. T Exercise 6. Putm 988 Prove tht if is coverget series of positive umbers, the so is /+ Note tht this gives other proof of the fct tht there c be ot lrgest coverget series Hit: 50 Exercise 63. Folld Let be coverget series. Let b be such tht The b lso coverges d hs the sme sum s. Exercise 64. Folld Suppose > for ll. b =, b =, b 3 = 4, b 4 = 3, is bsolutely coverget if d oly if l + is bsolutely coverget. b Fid coditiolly coverget but l + diverges. Hit: 5 Exercise 65. Putm 994 Prove tht if stisfy 0< < + + for ll, the diverges. Hit: Footote 5. Note tht this lmost gives other proof of divergece of. Exercise 66. Cuchy s Codestio Test Let be positive, d ssume there is N 0 N such tht > N 0, +. The d 40 either both coverge or both diverge. Hit: 53 Apply Cuchy s Codestio Test to /. p 44. m m + m + p = m + p m m m + m + p. 45. Group 3 terms together. 46. Discuss two cses b bouded or ot. [ ] log 47. exp exp0. MVT. 48. Assume otherwise. The either oe of b, c coverges, or both coverge. I the ltter cse is bsolutely coverget, i the former cse cot be coverget. 49. T+ = T. T 50. Prove / For, first otice 0. The cotrol the rtio betwee d l+ ; For b, pick such tht diverges. 5. Prove tht if =/ 0 the the sequece cot be Cuchy. 53. We hve + k k. 4

26 6 Mth 37 Week 0 Rel Ifiite Series Exercise 67. Kummer 54 Let be positive ifiite series, d let {λ } be y sequece of positive umbers. If there is N 0 N such tht >N 0, λ + λ + k > 0 4 for some costt k, the coverges. Hit: 55 Exercise 68. Kummer s Test Let be positive ifiite series, let be positive d diverget, d defie κ := d + d The if there is N 0 N such tht > N 0, κ > k >0 for some costt k, the =0 if there is N 0 N such tht > N 0, κ 0, the =0 diverges. Exercise 69. Write Kummer s test i the limit form usig limsup d limif. coverges. Exercise 70. USTC3 Let p d q be coverget ifiite products. Discuss the covergece of the followig ifiite products: p + q, p q, Exercise 7. USTC3 Prove the followig = 3 ; [ ] = + 3 ; = = [ =0 p q, + ] = ; p q. 44 cos x = si x x ; 45 + = π Exercise 7. USTC3 Discuss the covergece of the followig ifiite products. ; + + ; p ; ; + /; + + = [ l +x ] / Problems Problem. Putm 964 Prove tht there is costt K such tht the followig iequlity holds for y sequece of positive umbers,,...: K Hit: 56. b I the bove exercise, cosider = k k=. We c esily show tht lim =. This would give k+ Does this cotrdict the coclusio of the bove exercise? Hit: 57 Problem. Putm 966 Show tht if the series p 54. Edurd Kummer k λ + λ Bsiclly, + + b where b is the medi of,...,. 57. No. Becuse 49 oly holds whe is lrge. k k + coverges, where p R +, the the series p + + p p 50

27 Lst Updte: Februry 7, 04 7 is lso coverget. Hit: 58. Problem 3. Cosider the re-rrgemet of with = + terms, the p positive terms, d so o. The the sum is ito p positive terms, the q egtive Problem 4. Prove tht for x > 0 l + l p l q. 5 Hit: 59 limif x + x limif x / limsup x / x + limsup. 5 x Problem 5. Schlomilch 60 Let >0 d ssume there is N 0 N such tht >N 0, +. Let < < be strictly icresig sequece of positive itegers such tht k+ k / k k is bouded s fuctio of k. The d k+ k k 53 k= either both coverge or both diverge. The pply this result to. Problem 6. Defie H :=. Write k= k γ := H l. 54 The lim γ exists. The use this to prove + = l. Problem 7. Let > 0. Is it possible to obti Covergece Test by studyig This is ispired by Exercise 65. limsup + + Problem 8. Discuss the covergece/divergece of where > 0. d limif Problem 9. Let H = Let p := mi { N, H > p}. The Solutio: Gem 55 of RIS. + +? p+ lim = e. 57 p p Problem 0. Let be coditiolly coverget. Let r R be rbitrry. The there is fuctio f: N {0, } such tht f =r. Problem. Prove the followig vrit of Guss s Test: Let be positive ifiite series. Assume tht it stisfies = β l + R 58 where lim l R =0. The the series coverges if β > d diverges if β <. Hit: 6 Problem. Prove Theorem 4: 58. H := p + + p. Ad use H >H H. Derive qudrtic iequlity for the prtil sum through Cuchy-Schwrz. 59. x = x x x x x x. x 60. Oscr Xvier Schl omilch Try to prove for some α > 0 d lrge eough. + α l + > l

28 8 Mth 37 Week 0 Rel Ifiite Series Let be series whose terms go to 0. The exctly oe of the followig holds: coverges bsolutely; All re-rrgemets diverge to ; All re-rrgemets diverge to ; The series c be re-rrged to sum to y r R {, }. Problem 3. Assume coverges. The + coverges coverges. 60 Note tht there is o ssumptio o the sig of. Hit: 6 Problem 4. Euler I his 734 pper De progressioibus hrmoicis observtioes Observtios o hrmoic progressios 63, Leohrd Euler did the followig mipultios o hrmoic-type series. Commet o his methods d results. If the results re right while the methods re ot, give correct proofs of the results. Let, b, c > 0. Let i=. Deote Now we hve Itegrte d tke i= we hve s = c + c +b + Now replcig i by i d usig the fct tht c +b + + c + i b. 6 δs δi = c + i b. 6 s = C + c l +i b. 63 b we hve l = + i b +ib i = /i +b /i+b = 0 + b 0+b = i Now re-rrge. we hve l = I prticulr: Now l l 4=0 gives b Cotiuig, Euler obtied: l = ; 4 67 l 4 = = = l 3 = l 4 3 = l Problem 5. There re more th oe wys to ssig umber s to ifiite sum + +. The bove is the most populr oe. The followig is other wy. 6. Prove tht coverges = [ l+ ] coverges. 63. Comm. Acd. Sci. Imp. Petropol /5 740, 50-6.

29 Lst Updte: Februry 7, 04 9 Let,...,,... R. Defie s= 73 if d oly if for every ε>0, there is fiite subset I N, such tht for every fiite subset N J I, we hve s j < ε. 74 j J Prove tht if s = by this defiitio, the s= by Defiitio 7; b Fid tht is coverget by Defiitio 7 but ot coverget by the bove defiitio; c Prove tht if s= by the bove defiitio, the s is lso the sum of y rerrgemet of { }. d Prove tht if s= of { }. by Defiitio 7, d N, 0, the s is lso the sum of y rerrgemet e Fid tht is coverget by Defiitio 7, but fter rerrgemet coverge to differet sum. f Prove tht if s = by the bove defiitio, the y re-groupig gives the sme sum. More specificlly, if i I S i = N d i =/ j = S i S j =, the = k 75 i I where ll three sums re defied s i this problem. g Prove tht is coverget by the bove defiitio if d oly if is coverget by Defiitio 7 This ew defiitio is closely relted to the cocept of et, which c ssig limits to objects without sequece structure, for exmple the covergece of Riem sums of fuctio to its Riem itegrl is covergece i the sese of et. Remrk. Whe we use the defiitio i the bove problem, we ofte write N isted of to emphsize tht order does ot mtter here. N is clled uordered series. It is cler tht the bove defiitio for covergece of uordered series c be esily exteded to the sum of ucoutbly my umbers. However it turs out tht for ucoutbly my umbers to hve fiite sum, t most coutbly my c be ozero. k Si