f(bx) dx = f dx = dx l dx f(0) log b x a + l log b a 2ɛ log b a.

Transcription

1 Eercise 5 For y < A < B, we hve B A f fb B d = = A B A f d f d For y ɛ >, there re N > δ >, such tht d The for y < A < δ d B > N, we hve ba f d f A bb f d l By ba A A B A bb ba fb d f d = ba < m{, b}δ = f f < ɛ, A > m{, b}n = f l < ɛ B ba A bb B d ba ɛ A d, d bb ɛ d B f bb d B bb d = B d = log b, we coclude tht B f fb d f log b + l log b ɛ log b This implies tht lim A +,B + Eercise 5 B A f fb d = f l log b By the chge of vrible = b, we get y f + b b b d = f + y + y d = b y Thus by further chge of vrible t = b, we hve f + b d = = f + b + b d f t + 4bdt = f f d y + b dy y y f t + 4bdt

2 Eercise 54 d Assume b is the oly plce where improperess hppes The the covergece of b hd implies tht for y ɛ >, there is δ, such tht < b c < δ, < b c < δ = c c = ɛ < fd < ɛ, c c fd c c c c hd < ɛ gd c The coclusio is gd < ɛ By the Cuchy criterio, the itegrl c t b The covergece t the other improperess is similr c c b b hd < ɛ fd gd coverges Eercise 55 We try to prove tht there is icresig sequece stisfyig lim = + d lim f = The property is equivlet to tht, for y ɛ > d N, there is > N, such tht f < ɛ Sice f, the opposite of the property is tht, there re ɛ > d N, such tht f ɛ for ll > N However, by the divergece of ɛd d the compriso test, this implies tht shows tht the opposite is wrog Now we further ssume f is mootoe y > If f >, the the divergece of fd lso diverges The cotrdictio to the ssumptio If f is icresig, the f f for f d implies Therefore f cot be icresig uless it is costtly zero So the mootoe fuctio must be decresig, d lim + f = l eists f l for ll Therefore we must hve l = Eercise 56 We hve f mi{, f } + f coverges By the compriso test, If l >, the the compriso test implies b If b fd lso coverges f d coverges, the fd diverges We hve fd diverges b + f d Eercise 57 By lim + f =, we kow f for sufficietly big Sice f, this imples tht f f for sufficietly big By the covergece of test, f d coverges fd d the compriso

3 Eercise 58 Suppose f is positive d icresig o [, + Suppose F = d d coverges if d oly if f F coverges By f positive d icresig, we hve F = ftdt d By the compriso test, if f = 4 By f positive d icresig, we lso hve By the compriso test, if F = d F ftdt ftdt d f coverges, the = f f coverges, the f = f d f coverges ftdt Prove tht d F coverges Eercise 59 For y ɛ >, there is δ >, such tht δ implies f f < ɛ Moreover, sice f is itegrble o [, ], there is M, such tht f < M for ll [, b] Now we hve Sice t gf t d t tδ gfd g f f d t + ɛ ɛ t tδ tδ g f f d t g d + M g d + M t tδ t tδ g d g d g d coverges, by the Cuchy criterio, for the give, ɛ d the correspodig δ, there is T, such tht t > T implies t This proves tht gf t d t t t lim gf d t + t tδ g d < ɛ The t > T implies gfd ɛ t g d + M gfd =

4 Combied with we coclude tht t lim t + b gfd = lim f gd = f b + t lim t + gf t d = f gd gd, Eercise 5 By the secod Itegrl Me Vlue Theorem Theorem 44, we hve b si d p c si d p + b si d b p + p b p Lettig b +, we get + si d p By lim + q p =, we coclude lim + q By chgig to, we hve lim + si d = lim + c p si si d p = d = lim b b + b si d The fil limit is becuse > Similr rgumet c be mde for with chgig to Eercise 5 Give the ssumptio for the Abel test, we hve lim + f = l The lim + f + c l = Moreover the covergece of gd implies there is boud for gd for ll c [, + Thus we c pply Dirichlet test to coclude tht coverges The fgd = f lgd + l f lgd gd coverges Eercise 5 The tests pply to the improper itegrl of ubouded fuctio o bouded itegrl For itegrl b fgd tht is improper t b, the Dirichlet test verifies the followig f is mootoe er b d stisfies lim b f = c There is M, such tht gd < M for ll c [, b

5 The Abel test verifies the followig f is mootoe er b d bouded b gd coverges

6 Eercise 53 The prtil sum of + is s = + Thus the series coverges if d oly if lim + coverges, which is equivlet to lim coverges Eercise 54 Let s d t be the prtil sums of d y The the prtil sum of + by is s + bt The the clim o the sum of series + by follows from the correspodig properties of the sequeces Eercise 55 Let s be the prtil sum of the series The the prtil sum of the series k + k k+ is the subsequece s k The first prt bsiclly sys tht the covergece of s implies the covergece of s k For the secod prt, the sme sig coditio shows tht, if k < k+, the s lies betwee s k d s k+ The property implies tht the subsequece coverges if d oly if s coverges s k Eercise 56 Let the prtil sums of be s The the prtil sums of + re s The covergece of s implies the covergece of s Covergece, if s coverges to l, the s + = s + + lso coverges to l by lim = Therefore s lso coverges to l We coclude tht s coverges if d oly if s coverges Eercise 57 Sice is decresig, we hve I prticulr, we hve k k k + + k k+ k k+ d k k k= + i= i = k + + k k+ k= k k+ = + k k k= k= Sice >, this implies tht the prtil sum of is bouded if d oly if the prtil sum of is bouded Agi sice the terms re positive, the boudedess is equivlet to the covergece, so tht the covergece of the two series re equivlet For the specil cse =, p > p whe p, p = diverges, which hppes if d oly if p p For the specil cse = coverges if d oly if <, ie, p > p log, p log coverges if d oly if p log = p diverges, which hppes if d oly if p > log p p Eercise 58

7 By cos si y = si + y si y, si si y = cos + y + cos y, we hve si y si y cos + ky = k= = si + ky = k= = k= si k= si y si + k y si cos + k + cos + + y + cos + k y, y + cos y + k y y For =, we get cos ky = k= si + y si y + = si + y si y Itegrtig from π to gives k= si k k = y π cos kydy = k= π si + y si y dy π Tke f = si d g = si y i Eercise 43 Note tht g is bouded d cotiuous o [π, ] for fied, π The we hve + y si lim π si y dy = π si ydy π π si y dy = Therefore k= si k k = lim k= si k k = π Eercise 5

8 We hve + y = m{, y } + mi{, y } d y = m{, y } mi{, y } Therefore m{, y } d mi{, y } coverge if d oly if +y d y coverge This further implies tht +y d y coverge, which is equivlet to d y coverge Thus we coclude tht m{, y } d mi{, y } coverge implies d y coverge The coverse is ot true For emple, d + coverge However, { } m, + = d { } mi, + = diverge By + y = m{, y } + mi{, y } d the compriso test, d y bsolutely coverge if d oly if m{, y } d mi{, y } bsolutely coverge Eercise 5 [This is the discrete versio of Eercise 58] By icresig d >, we hve = By the compriso test, the covergece of implies the covergece of O the other hd, we hve =, By compriso test, coverges if d oly if Eercise 5 By >, we hve coverges < < By the compriso test, the covergece of implies the covergece of O the other hd, suppose The = + We hve m m + m m m + m By = +, for y fied m, we hve lim m+ m = > This implies tht the Cuchy criterio for the covergece of + + +, fils for ɛ = Eercise 53 The prtil sum s = is icresig The i= i i = i= s i s i s i i= s i s i s i s i = = < s i s i s s s i=

9 Therefore coverges Eercise 54 For m >, by i decresig, we hve B m i m i= i m = i= i m i= For fied, we my tke lim m d get B i= i lim m m = i = i= i i= Therefore the prtil sum is bouded, d the series coverges Eercise 55 If for some costt < d sufficietly big, the for sufficietly big By <, coverges By the compriso test, coverges If for ifiitely my, the for ifiitely my This implies tht does ot coverge to, d therefore diverges If lim <, the we pick stisfyig lim < <, d we hve for sufficietly big By pplyig the first prt of the root test, we fid tht coverges If lim >, the by the chrcteristio of upper limit, there re ifiitely my terms stisfyig > By pplyig the secod prt of the root test, we fid tht diverges Eercise 56 Suppose + = N y + for N The N+ N y N+ N+ N y N+ y N+ y = Cy, C = N y N y N+ y y N The covergece of y implies the covergece of Cy The by the compriso test, we get the covergece of Eercise 57 If + for some costt < d sufficietly big, the + y + for y = y d sufficietly big By the covergece of ] d Eercise 56, we coclude tht coverges If + for sufficietly big, the + for sufficietly big This implies tht does ot coverge to, d therefore diverges

10 + If lim <, the we pick stisfyig lim + < <, d we hve + for sufficietly big By pplyig the first prt of the test bove, we fid tht coverges + If lim >, the by the chrcteristio of lower limit, we hve + > for sufficietly big By pplyig the secod prt of the test bove, we fid tht diverges Eercise 58 If lim < l, the there is turl umber N, such tht N implies < l By Eercise 3, this further implies tht N l N l, or N l Sice ln N lim l =, we coclude tht lim N l The fct tht lim < l implies lim l is equivlet to lim lim The iequlity lim lim c be proved similrly From the iequlity, we c deduce the rtio test from the root test i the coverget cse We lso see tht the two tests re the sme if the rtio hs the limit, i which cse the four limits re the sme Eercise 59 The coditio for the covergece uder the Rbe test is + p for some p > d sufficietly big This is the sme s p + If lim + >, the this will be stisfied Thus we see tht lim + > implies the covergece of For the divergece uder the Rbe test, the coditio is + for sufficietly big This is the sme s + for sufficietly big The coditio will be stisfied if lim + < Eercise 53 The coditio + p + o for some p > d sufficietly big is the sme s + p = + p + o + o

11 So if + + p for some p > d sufficietly big, the coverges The limit versio is tht if lim + >, the coverges The coditio for divergece is similr Eercise 53 If -digit umber Z Z Z does ot coti 9, the ech digit Z i must be chose from {,,, 3, 4, 5, 6, 7, 8}, d Z cot be Therefore ech Z i hs 9 choices, d Z hs 8 choices This gives the totl umber 8 9 Let A be the collectio of turl umbers tht do ot cotis digit 9 The the prtil sum i/ A, i< i = i/ A, i< i + i/ A, i< i + + i i/ A, i< = 8 Sice the prtil sum is bouded, the series coverges Similr rgumet pplies if the other digits is prohibited Actully ectly the sme estimtio pply if the digit is ot Moreover, if the bse is chged to N, the similr rgumet gives the boud NN, d the series still coverges For the series, the similr estimtio gives ip i p 8 + p 8 p p p p i/ A, i< O the other hd, we lso hve i p 8 + p 8 p p p p i/ A, i< Therefore i/ A 9 coverges if d oly if coverges This mes 9 < p ip p Eercise 53 By f f o [, ], we hve Therefore d is decresig d d = f fd

12 By fk + k= k+ f d = fk Eercise 533 k fd fk, we hve k+ k fd + f fk fk + + f = f k= The fuctio log is strictly icresig, so tht log k log! = log + log + + log < log + = log d = < log + log log = log! log d + By log + = log, the iequlity is the sme s e Eercise 534 We hve log! = log = d wish to show tht the right side coverges to k+ k 3 log d logk + d log d <! < e e log + log + + log, log d log d The right side ppers to be the Riem sum SP, log for the prtitio P cosistig of i = i d smple poits i = i However, we cot use the defiitio of Riem itegrl becuse the itegrl here is improper Usig the fct tht log is icresig d the ide of the proof of Propositio 5, we hve d log + log + + log = log + log + + log log + log + + log = log + log + = log + log d, log d = log + log log log d, log d

13 Therefore by the covergece of log d, we hve log log d log + log + + log The covergece of lim log = Therefore log d tells us lim lim log + log + + log = log d = log d log d log d = Moreover, we lso kow Eercise 535 Usig the ide of the proof of Propositio 5, we my estimte s follows log used i cse p = The + d = p p [ p + p p + p ], + d = p p [ + p p ] + p [ + p + 3 p p + p + p + p ] = [ p + p p pp + 3 p 9 pp + p pp + p + p p ] + + o = p 4p + o + p [ + p + p p + p + p ] The we coclude = p p [ + 4 p + p p + p + p p + p pp + p + + o ] = p 4p + o

14 If p >, the is decresig for sufficietly lrge, d lim = Thus coverges, d [ p ] coverges If p <, the is icresig for sufficietly lrge Thus does ot coverge to, so tht diverges, d [ p ] diverges 3 If p =, the [ + ] = Therefore diverges, d ] [ p diverges Eercise 536 Give the coditio of the Abel test, the sequece coverges to ξ The ξy stisfies the coditio of the Dirichlet test Therefore the series ξy coverges Combied with the covergece of y, we fid the series y coverges Eercise 537 The Leibiz test c be derived from the Dirichlet test by keepig d tkig y = The prtil sum of is either,, or Eercise 538 If p >, the by the covergece of p bsolutely si d the compriso test, p coverges If < p, the decreses d coverges to, d the the prtil sum of si is p si bouded by Emple 54 The by the Dirichlet test, the series coverges p si O the other hd, for d < < π, we hve the estimtio similr to the p oe i Emple 54 m k = si p k i= si m i m i p k si si iπ = p π p The divergece of si the implies the diverges of We coclude tht the series p p coditiolly coverges for < p Eercise 539 For ech k, cosider those stisfyig 4k k + π = k π 6 4 < < k + 4 π = 4k + k + π 6 i= k i= i p

15 These stisfy si >, d the umber of such is kπ Therefore si kπ p k 4 π< <k+ 4 π The right side is > b for costt b d sufficietly big k kp Eercise 539 By Eercise 4, for f = si, we hve Sice stisfies f + + p p+ p si d si p p f = For p >, the series p+ covergece of the followig re equivlet + si d p si d p 3 si + si + p + p 4 si p k + p π 4 p + si + ω [,+]f + p 8 cos p si p+ o [, +], we hve ω [,+] f + p This proves the estimtio p+ coverges Therefore the estimtio bove shows tht the Moreover, the ture of the covergece bsolute, coditiol, divergece re lso equivlet Eercise If p, the estimtio i the first prt implies tht the Cuchy criterio fils, d the series diverges

16 If p >, the the secod prt shows tht the covergece for the series is the sme s the si coverges for the improper itegrl d By Eercise 4876, the series coditiolly p coverges for < p d bsolutely coverges for p > Eercise 54 By Eercise 53, we hve k = k = log + γ + r where γ is the Euler-Mscheroi costt d lim r = The Therefore m = m = k k = log m + γ + r m k = log + γ + r log + γ + r = log + log + γ + r r σ m, = m 4 = log + log m + λ m,, where λ m, = r m r + r d lim m, λ m, = For the rerrgemet of p positive terms followed by q egtive terms, we hve the specil prtil sums s kp+kq = σ kp,kq = log + log p q + λ kp,kq, s kp+k q = σ kp,k q = log + log kp k q + λ kp,k q Both specil prtil sums coverges to log + log p Sice geerl prtil sum is sdwiched q betwee the specil prtil sum, the whole series coverges to log + log p q e s Now for y rel umber s, = is the solutio of log + log = s The we my fid strictly icresig sequeces m k d k, such tht m =, =

17 m k re positive odd umbers k re positive eve umbers 3 lim k m k+ m k = lim k k+ k = 4 lim k m k k = For emple, i cse = p q is rtiol umber, we my tke m k = kp d k = kq The costruct the sequece = k= m k + + m k m k+ k + k + 4 k+ The prtil sums = m 4 + m + + m m s mk + k = m k 4 = log + k log m k + λ m k +, k k s mk+ + k = m k+ 4 = log + k log m k+ + λ m k+ +, k k Therefore lim s m k + k = log + k log lim m k = s, lim s mk+ + k = log + log lim k k mk+ m k m k = s k Moreover, we hve s mk + k s s mk+ + k for m k + k m k+ + k d s mk+ + k s s mk+ + k+ for m k+ + k m k+ + k+ The we coclude tht lim k s = s Eercise 54 The coditio k < B implies tht ki i B + B B i= i= I either cse tht or k coverge, we hve lim The for the fied B, we hve lim B + B B = Therefore lim i= k i coverges if d oly if lim i= i coverges, d the two limits re the sme Eercise 54 Let s be the prtil sum of p Let t be the prtil sum of k k p

18 For y ɛ >, there is N, such tht > N = k < ɛ p This mes tht, for > N, the differece betwee the prtil sums s d t is sum of some terms i with ɛ p < i < + ɛ p Therefore i p s t [+ɛ p ]+ [ ɛ p ] i +ɛ p p [ ɛ p ] + d p ɛ + p p [ ɛ p ] + p Here the estimtio by the itegrl is ispired by the proof of the itegrl compriso test We hve lim [ ɛ p ] = lim p [ ɛ p ] + = p For p =, we hve For < p <, we hve +ɛ p ɛ p +ɛ ɛ d = log + ɛ ɛ = log + ɛ ɛ d = p p [ + ɛp p ɛ p p ] = p p [ + ɛp p ɛ p p ] = p p p[ɛp + oɛ p ] = [ɛ + oɛ] I both cses, we c hve +ɛ p ɛ p d s smll s possible whe ɛ is smll eough Therefore +ɛ p d lim ɛ =, p p d we get lim s t = This implies tht s coverges if d oly if t coverges, d they hve the sme limit vlue Eercise 544 For the squre rrgemet, we hve k y i = i y i i= i= i= k + y + for k + + d k + y i = i y i + y + i + + i= i= i= i= i= i i= ++ k y i

19 for + + k + Sice d y coverge, we hve lim = lim y =, d k i= i, i= ++ k y i re bouded Therefore we get k lim y i = lim i y i = i y i k i= i= i= i= i= Eercise 545 The digol rrgemet hs the followig property: If i j, the i y i is rrged before i y j The squre rrgemet lso hs this property We will prove tht if bsolutely coverges d y coverges, the y rrgemet stisfyig the property coverges to y Let s be the prtil sum of y Let y = lim s = s For y ɛ >, there is N, such tht N = s s < ɛ; m > N = m + m+ + + < ɛ The secod implictio implies tht N ɛ The specil property of the rrgemet implies tht y prtil sum of the product series is of the form k y i = s + s + + l s l i= For N foud bove, there is K, such tht K i= y i cotis ll the terms i y j with i, j N The for k > K, we hve N, N,, N > N This implies s s < ɛ, s s < ɛ,, s N s < ɛ Let B be the boud for ll s The we hve k y i N s i= = s s + s s + + N s N s + N+ s N+ + + l s l N ɛ + N+ + N+ + + l B N + Bɛ Therefore k y i s i= k y i N s N s i= N + B + s ɛ Sice the coefficiet N +B+ s is bouded, we coclude tht lim k k i= y i = s

20 Cosider = y =, By k k we hve y + y + + y = k k = Therefore y + y + + y, so tht the digol rrgemet diverges by the Cuchy criterio I fct, the series y + y + + y lso diverges