复旦大学数学分析习题课讲义.pdf

数学分析习题课讲义 分析十段天元手筋 张毅 by Dr Yi Zhang Institute of Mathematical Sciences Fudan University Shanghai 200433 P R China zhangyi math Lecture given at Fudan University 2006 The author holds the copyright of this lecture notes Any person s intending to copy a part or whole of the materials in the notes in a proposed publication must seek copyright release from the author 1 Analysisi Dedicate to Katie 创造的神秘 有如夜间的黑暗 是伟大的 而知识的幻影 不过如晨间之雾 iiLecture Notes at Fudan Contents Referencesii 1 Problems on Sets Sequences and Limits1 1 1 Elementary Technique1 1 2 Applications of the Stolz theorem11 2 Inequality14 3 Orders Estimate of Infi nitesimal20 3 1 Notations and Examples20 3 2 Exercises and homework23 4 Application of Infi nitesimal Wallis Formula and Stirling Formula27 5 Topic on the Gamma functions30 5 1 functions30 5 2 Exercises34 6 Working Technique in function theory36 6 1 Iteration technique36 6 2 Exercises and Homework39 7 Applications of Diff erential41 8 Treasures in Calculus42 8 1 Euler s identifi cation 2 1 1 22 1 n2 2 6 42 8 2 Irrationality of log2 2 3 43 8 3 The properties of Tchebychev s functions46 8 4 Mertens Theorem and Selberg s Inequality53 8 5 An Elementary Proof of The Prime Number Theorem by Selberg and Erd os58 8 6 Gauss s Proof of The Fundamental Theorem of Algebra59 9 Advanced Techniques in Analysis60 9 1 Preliminary Results in Analysis60 9 2 Scaling Technique and Schauder Estimate72 9 3 Campanato s characterization of L2functions to be H older continuous79 10 Advanced Topics in Analysis82 10 1 Topic on Riemann Zeta Function To be Continuous 82 10 2 Elementary on Nevanlinna Theory To be Continuous 83 10 3 Elementary on p adic Series To be Continuous 84 References F M 菲赫金哥尔茨 微积分学教程 VOL I II III Higher Education Press H G H Hardy A Course of Pure Mathematics Tenth Edition Combridge University Press 2002 HUA 华罗庚 数论导引 科学出版社1979 J J Jost Partial Diff erential Equations GTM 214 Spinger 2002 P Problems Selection in The William Lowell Putnam Mathematical Competitions P Y 潘承洞 于秀源 阶的估计 Analysisiii 曾子曰 大学之道 在明明德 在亲民 在止于至善 知止而后定 定而后能静 静而后能安 安而后能虑 虑而后能得 物有本末 事有始终 知所先后 则近道矣 Analysis1 1 Problems on Sets Sequences and Limits 1 1 Elementary Technique Exercise 1 1 Show that the set of all irrational number of R is uncountable Proof Suffi cient to show that the set of all irrational number in 0 1 is uncountable Otherwise the set of real numbers in 0 1 is countable ie 0 1 x1 x2 Cover each xi i 1 with the corresponding interval Ii xi 1 2 1 3 i xi 1 2 1 3 i Then there holds 0 1 x1 x2 i 1 Ii and so 1 l 0 1 X i 1 l Ii X i 1 l xi 1 2 1 3 i xi 1 2 1 3 i X i 1 1 3 i 1 2 It is a contradiction Example 1 2 Let R Q and 1 1 1 Let A n n N B n n N be two strictly increasing sequences of positive integers Show that A B N A B Proof We have 1 1 W L O G we assume 2 Then 1 A B 1 If a B 6 then there exist two integers m n such that m n q N Hence there holds that q m n q 1 A contradiction 2 If there is a positive integer p A B then there exist two integers m n such that m p m 1 and n p n 1 Therefore m p p 1 m 1 m 1 and n p 1 n 1 n 1 Since 1 1 1 we have m n p p 1 1 then f 2k 2f k 2 1 f k 2 1 f k 2 f k 1 and so we have an increasing sequence of integer f k f 2k f 4k f 2lk But it contradict that f is a bounded function with all values in Z Therefore f Z 1 0 1 c Let f 1 cos with 2 0 We can show that f n cos n n Z by induction and the formula 2cos cos cos cos Example 1 4 Defi ne the function G N 0 Z by G 0 0 G n n G G n 1 n N To show that G n 5 1 2 n 1 Proof 1 Actually we have 1 G n n and G n 1 G n n 1 At fi rst G 1 1 G 2 1 By induction we assume 1 G k k and G k 1 G k 1 k n 1 Then we have 1 G n 1 n 1 and so G G n 1 G n 1 n 1 1 n n 1 G n n G G n n 1 n G n G n 1 1 G G n 1 G G n 2 0 since 1 G n 2 G n 1 n 1 2 By induction we can show G n 1 G n 1 or 0 n 3 Defi ne F n n 1 where 5 1 2 Defi ne S n F n F F n 1 We are going to show that S n n and so F n G n Let K n and so n K with 0 1 Then F n n 1 K K F F n 1 F n K 1 Since 2 1 we have K 1 n 1 n 2 1 n K S n F n F F n 1 n 1 Analysis3 Let T 1 Then 1 T 0 1 T an and an N Defi ne bn a1 a2 an the least common multiple of a1 an To show that X n 1 1 bn Proof Let d n p p N p n Then d n 2 n and so we have n d bn 2 p bn By the previous exercise we obtain the result Example 1 6 To show lim n n X k 1 3 r 1 k n2 1 1 6 Proof Let k 3 r 1 k n2 1 Then 0 k 1 and k 1 3 k ON the other hand 1 k n2 1 k 3 1 3 2 k 3 k 3 k 1 4 2 k 3 k 1 4 k 3n2 2 3 k and so n X k 1 k 3n2 n X k 1 k n X k 1 4k2 27n4 k 3n2 4Lecture Notes at Fudan Since n X k 1 k n n 1 6 and n X k 1 k2 n n 1 2n 1 6 we have lim n n n 1 6n2 2n n 1 2n 1 3 27n4 lim n n X k 1 k lim n n n 1 6n2 Remark Let g x n X i 1 i 1 x i Then g x 1 x g x n X i 1 1 x i n 1 x n 1 and so g x 1 x 1 x n 1 x2 n 1 x n 1 3x Comparing the coeffi cient of the term x of the polynomial g x we have n X i 1 i2 C3 n 1 nC 2 n 1 n n 1 2n 1 6 Example 1 7 Let an bn be two sequence satisfying 1 an bn bn6 0 n 1 2 2 X n 1 an bn A X n 1 an bn 2 B Then the sum X n 1 an an bn is convergent Proof By the condition 2 for any R with 0 0 for any n l N we have an bn X i l ai bi and X i l ai bi 2 and so 1 7 1 an an bn an bn 1 an bn an bn 1 an bn 1 an bn 2 an bn an bn 2 1 X k 1 an bn 2k n N 1 7 2 n X k 1 an bn 2k an bn 2 1 an bn 2 2 an bn 2 2 2 n N Analysis5 1 7 3 X i l ai bi 3 X i l ai bi 3 v u u t X i l ai bi 2 v u u t X i l ai bi 4 X i l ai bi 2 l N On the other hand for l Nwe have X n l an an bn X n l an bn an bn 2 1 X k 1 an bn 2k 2 X n l an bn an bn 2 X k 1 an bn 2k But by 1 7 2 and 1 7 3 we have the following X n N an bn an bn 2 X k 1 an bn 2k X n l an bn n X n l an bn 2 n X n l an bn 3 1 an bn 2 2 3 X n l an bn 3 1 an bn 2 2 X n l an bn 3 2 At all we obtain X n l an an bn 0 and bn6 0 n N we have 0 0 and so bn an bn has a positive upper bounded uniformly for n N Therefore the sum X n 1 an an bn an bn is convergent Example 1 8 Let an be a sequence satisfying an 1 2 an 1 n N To shown that lim n an exists and lim n an 1 6Lecture Notes at Fudan Hint Let bn 1 an n 1 2 We have a new sequence bn with the relation 1 bn 1 bn 1 1 i e bn 1 bn 1 bn We fi nd bn 0 bn 1 0 If one bm 0 then bn 0 n m and limn an 1 No we assume bn6 0 n 1 2 Then we have 1 bn 1 1 bn 1 1 and so 1 bn 1 n 1 b1 i e bn 1 b1 nb1 1 0 n Example 1 9 Let an be a sequence satisfying 1 X n 1 an 1 2 for all n N 0 an X k n 1 ak Then for any x 0 1 there exists a subsequence ank of an such that X k 1 ank x Proof Defi ne the n1 n2 by induction 1 Let n1 min n an x It is well defi ned since by the condition 1 limn an 0 Then by the condition 2 we have 0 x then we defi ne n2 min n n1 an1 an n1 1 since by step 1 and then by the condition 2 we have an1 an2 1 an1 X k n2 ak x then we defi ne n3 min n n1 an1 an2 an nk k X i 1 ani an nk 1 since 1 9 1 By the condition 2 there must hold an1 an2 ank ank 1 1 k X j 1 anj X k nk 1 ak x It is a contradiction to the defi nition of nk 1 Example 1 10 1 Let f x x2 c Compute the iteration f n x f f f z n Solute Let x x2 g x x c Then we have f 1 g and f n x 1 g n where 1is the inverse function of g n x x nc and so f n x x2 nc 8Lecture Notes at Fudan 2 Let f x x 3 x3 c Compute the iteration f n x f f f z n Solute Let x x3 g x x x c Then we have f 1 g and f n x 1 g n where 1is the inverse function of By induction we have g n x x Pn 1 i o ci x cn Then f n x x 3 q Pn 1 i o ci x3 cn Exercise 1 11 a Let f x x3 6x2 12x 6 Compute the iteration f n x f f f z n b Let f x x 4 2x 1 Compute the iteration f n x f f f z n c Let f x 2x 1 x2 Compute the iteration f n x f f f z n Analysis9 Exercise 1 12 The sequence a0 a1 an is defi ned by a0 1 2 ak ak 1 1 na 2 k 1 To show that 1 1 n an 1 Hint By induction to show for any 1 k n we have n 1 2n k 2 ak n 2n k and let k n we get n 1 n 2 an 1 Exercise 1 13 Let a 1 a2 an be a sequence of diff erent positive integers To show that n X k 1 ak k2 n X k 1 1 k Hint Let Sk a1 ak Then Sk k X i 1 i k k 1 2 and we have n X k 1 ak k2 n X k 1 Sk 1 k2 1 k 1 2 Sn n2 n X k 1 k k 1 2 1 k2 1 k 1 2 n 1 2n Exercise 1 14 Let a1 a2 anand b1 b2 bn To show n X k 1 ak n X k 1 bk n n X k 1 akbk Proof Let D X k 1 akbk X k 1 ak X k 1 bk We have D1 0 and D 1 D X k 1 ak ak 1 bk bk 1 0 Exercise 1 15 1 To show that log n 1 n X k 1 1 k logn 1 10Lecture Notes at Fudan 2 Consider N N X k 1 1 k R To show that lim N N 2nfor all n N Exercise 1 18 If the following is true for all n N Bn A An 1 show that B A2 Analysis11 Hint lim n n n 1 2 Applications of the Stolz theorem Theorem Stolz theorem Let yn is strictly increasing sequence i e yn 1 yn and yn If lim n xn xn 1 yn yn 1 a a Then lim n xn yn a Example 1 19 Let an bn be two sequences satisfying that bn 1 an 918an 1 n N Then we have lim n bnexists lim n anexists Proof We only prove the part Let b limn bn We defi ne two new sequences n b 919 an n b bn 918 n 1 2 Let 1 918 then the sequences n n satisfy n 1 n n 1 By induction we have n 1 n 1 X i 1 i n 1 i n 1 0 Pn 1 i 1 i 1 i 0 1 n 1 and so n 1 Pn 1 i 1 i 1 i 1 1 n 1 By the Stolz theorem we have lim n Pn 1 i 1 i 1 i 1 1 n lim n n 1 1 n 1 1 n 1 1 0 1 20 Exercises 1 Let p1 p2 pn be positive numbers If lim n pn p 0 then lim n n p 1p2 pn p 12Lecture Notes at Fudan 2 Let a0 a1 a2 an be positive numbers If lim n an 1 an p then lim n n a n p 3 To show for any k N lim n Pn i 1i k nk 1 1 k 1 Proof By Stolz theorem we compute lim n Pn 1 i 1 ik Pn i 1i k n 1 k 1 nk 1 lim n n 1 k n 1 k 1 nk 1 lim n n 1 k nk n 1 k 1 nk 1 nk On the other hand n 1 k nk 1 O 1 n n 1 k 1 nk 1 nk k 1 O 1 n 4 To show for any k N lim n n Pn i 1i k nk 1 1 k 1 1 2 Proof n Pn i 1i k nk 1 1 k 1 1 2 k 1 Pn i 1i k nk 1 k 1 nk By Stolz theorem it is suffi cient to compute that lim n k 1 n 1 k n 1 k 1 nk 1 k 1 n 1 k nk On the other hand k 1 n 1 k n 1 k 1 nk 1 k 1 k X i 0 Ci kn i k X i 0 Ci k 1n i k 1 X i 0 k 1 Ci k C i k 1 n i So k 1 n 1 k n 1 k 1 nk 1 k 1 n 1 k nk 1 2k k 1 O 1 n k k 1 O 1 n 5 Let x1 0 1 and xn be a sequence with xn 1 xn 1 xn n N Then we have lim n nxn 1 Analysis13 Proof Since x2 x1 1 x1 1 2 2 1 by induction we have xn 0 1 n N On the other hand xn 1 xn 1 xn 1 so xn is a strictly decreasing sequence with a fi nite bound and then lim n xn A 0 1 Moveover by A A 1 A we have A 0 By Stolz theorem lim n nxn lim n n 1 xn lim n n n 1 1 xn 1 xn 1 lim n 1 xn 1 6 let ak be a sequence satisfying that lim n An lim n n X k 1 ak A 0 with 1 p 1 q 1 Then we have n X k 1 akbk n X k 1 ap k 1 p n X k 1 bq k 1 q Moreover the inequality becomes to be an equality if and only if there exists a t R such that ap k tbq k k 1 n The proof of the H older inequality depends heavily on the following lemma Lemma 2 2 Let A B 0 then for any 0 1 we have A B1 A 1 B and the inequality becomes to be an equality if and only if A B Proof W L O G we assume A B 0 6 0 Then we have A B ZA B 1 x 1dx 1 0 with 1 p 1 q 1 Then we have AB 0 Exercise 2 4 Minkowski Inequality Let ak bk 0 k 1 n p 1 Then we have n X k 1 ak bk p 1 p n X k 1 ap k 1 p n X k 1 bp k 1 p Moreover the inequality becomes to be an equality if and only if there exists a t R such that ak tbk k 1 n Theorem 2 5 Arithmetic Mean Geometry Mean Inequality Let a1 anbe positive real numbers and a1 anbe positive real numbers with p1 pn 1 Then we have Gn n Y i 1 api i n X i 1 piai An moreover the inequality becomes to be an equality if and only if a1 a2 an Proof W L O G we assume a1 a2 an Then there exists an integer k 1 n 1 such that ak Gn ak 1 Analysis15 Consider the following formula An Gn 1 n X i 1 pi ai Gn Gn n X i 1 pi logai logGn Gn k X i 1 pi Gn ai Gn logGn logai n X i k 1 pi ai Gn Gn logai logGn k X i 1 pi Z Gn ai 1 Gn 1 t dt n X i k 1 pi Z ai Gn 1 Gn 1 t dt k X i 1 pi Z Gn ai 1 t 1 Gn n X i k 1 pi Z ai Gn 1 Gn 1 t 0 Moreover the inequality becomes to be an equality if and only if a1 a2 an Gn 2 6 Exercises 1 Let ak bk 0 k 1 n with 1 p 1 q 1 with 0 p n X k 1 ap k 1 p n X k 1 bq k 1 q Hint We have 0 1 1 p 1 0 1 q p 1 k 0 We defi ne Mp a Pn k 1a p k n 1 p To show that for 0 r s we always have Mr a 0 with 1 p 1 q 1 Assume P n 1a p n and P m 1b q n Then we have X m 1 X n 1 anbm m n sin p X n 1 ap n 1 p X m 1 bq m 1 q Hint Consider the following arguments a N X m 1 N X n 1 aman m n N X m 1 N X n 1 m n 1 pq am m n n m 1 pq an m n Analysis17 b m n 1 q 1 m n 1 m n m 1 q 1 n m Zn m n 1 m dx 1 x x 1 q Example 2 9 Hardy Landan inequality Let p and ak k 1 n be positive numbers To show for p 1 we have n X k 1 a1 ak k p p p 1 p n X k 1 ap k Hint Let Ak a1 ak k Then we have a Ap k p p 1A p 1 k ak Ap k p p 1A p 1 k Ak Ak 1 Ap k 1 kp p 1 k 1 p p 1 Ap 1 k Ak 1 Ap k p p 1A p 1 k ak Ap k 1 kp p 1 k 1 p 1 p 1 Ap k Ap k 1 1 p 1 k 1 A p k 1 kA p k b Thus we have n X k 1 Ap k p p 1A p 1 k ak p p 1 n X k 1 Ap 1 k ak again use H older inequality Example 2 10 Carleman inequality Let ak k N be positive numbers If X k 1 ak 0 k 1 n Denote Tl Pl k 1tk Show that n Y k 1 Atk k 1 Tn 1 Tn n X k 1 tkAk Hint Consider the following argument Let Lm m X k 1 Atk k Then we have Lm 1 Tm Atm m 1 Tm L 1 Tm 1 m 1 Tm 1 Tm A tm Tm m Tm 1 Tm L 1 Tm 1 m 1 tm Tm Am By induction we prove the statement 2 To show the Carleman inequality by using the Hardy Landan inequality Proof By the Hardy Landan inequality for all p 1 we have n X k 1 a1a2 ak 1 k n X k 1 a 1 p 1 a 1 p 2 a 1 p k k p p p 1 p n X k 1 ak Since lim p p p 1 p e we get n X k 1 a1a2 ak 1 k e n X k 1 ak 2 12 Appendix Computing the integral Z 0 dx 1 x x 1 q with residue theory Since 1 q 1 p 1 Then Z 0 dx 1 x x 1 q Z 0 x 1 p 1 1 xdx Analysis19 Let f z 1 1 z f z is a holomorphic at C 0 We have 2 1Resz 1 z 1 p 1f z Z R x 1 p 1f x dx Z R e 1 p 1 2 1 x 1 p 1f x dx Z z z 1 p 1f z dz Z z R z 1 p 1f z dz Resz 1 z 1 p 1f z e 1 p 1 logz z 1 e 1 p 1 log 1 1arg 1 e 1 p 1 1 e 1 p 1 On the other hand f z 1 z 1 Therefore we have Z z z 1 p 1f z dz 1 p 12 0 0 Z z R z 1 p 1f z dz R 1 p 1 R 2 R 0 R Z R x 1 p 1f x dx Z R e 1 p 1 2 1 x 1 p 1f x dx 1 e 1 p 1 2 1 Z R x 1 p 1f x dx At all we have 1 e 1 p 1 2 1 Z 0 x 1 p 1f x dx 2 1e 1 p 1 and so Z 0 dx 1 x x 1 q 2 1e 1 p 1 1 e 2 p 1 2 1 e 1 p 1 e 1 p 1 sin p 20Lecture Notes at Fudan 3 Orders Estimate of Infinitesimal 3 1 Notations and Examples 3 1 Notations 1 If lim x xo f x g x 0 we denote f x o g x It is obvious that o h o g o h g In particular let an be a sequence if lim n an 0 we denote an o 1 n 2 If lim x xo f x g x 1 we say f x and g x is equivalent for x x0 and we denote f x g x x x0 3 Let g x 0 if there is a constant A 0 such that f x Ag x x a b then we denote f x O g x x a b 4 Here we have some well know elementary functions For n even we denote n n n 2 n 4 4 2 For n odd we denote n n n 2 n 4 3 1 a sinx X k 0 1 k x2k 1 2k 1 x R and cosx X k 0 1 k x2k 2k x R b arctanx X k 0 1 k x2k 1 2k 1 x 1 c log 1 x X k 1 1 k 1 xk k x 1 d 1 x X k 0 k xk x 1 e ex X k 0 xk k x R Analysis21 f arctanx X k 1 n x2n 1 2n 1 x R g arcsinx x 1 3 1 2 x 3 1 5 3 4 x 5 1 2n 1 2n 1 2n x2n 1 x R Theorem In the neighborhood of x0 if f n x exists and f n x0 M then f x n 1 X k 0 f k x0 k x x0 k O x x0 n For examples we have a sinx x x3 3 O x 5 x R and cosx 1 x2 2 O x 4 x R b log 1 x x x2 2 O x 3 x 1 c 1 x 1 x O x2 R x 0 and A be any two constants Then for any 0 we have 1 xA o 1 x x 2 logxA o x x 3 f x A o e f x for any increasing function f x with lim x f x Proof Let n x be the integer part of x and m A 1 If x then n We have 1 x 1 n Cm 1 n m 1 m 1 m 1 n 2 m 1 if n 2m 1 Thus if x we have 1 x xA m 1 2m 1 m 1 nm 1 1 n m Let x y we get 1 Let e 1 x logy we get 2 3 is obvious Example 3 3 Let an bn be two sequences with an o bn n and bn 0 If P n 1bn then we have N X n 1 an o N X n 1 bn N 22Lecture Notes at Fudan Proof For any 0 there exists M 0 such that for any n M we always have an M we have N X n 1 an M X n 1 an N X n M 1 bn M X n 1 an N X n M 1 bn Since P n 1bn there exists N0 M such that if N N0 N X n 1 bn M X n 1 an So N X n 1 an 2 N X n 1 bn Example 3 4 Let ak be a sequences with ak 0 and P k 1ak L if X k n an O an then we have n X k 1 1 ak O 1 an Proof Let An X k n an then an An An 1 1 such that An Aan We therefore have An An An 1 A An 1 A 1 A An A 1 A l 1An l Thus n X k 1 1 ak An an n X k 1 A Ak A an n X k 1 An Ak A an n 1 X l 0 A 1 A l yn and yn If lim n xn xn 1 yn yn 1 a a Then lim n xn yn a Analysis23 Proof In case of a there is a N N such that xn 1 xn yn 1 ynfor n N and xn and lim n yn yn 1 xn xn 1 0 In case of a let tn xn Thus these two cases are reduced to the case fi nite constant a Now we assume that a is a fi nite constant Then we have xk xk 1 a yk yk 1 o yk yk 1 sum with k we get n X k 2 xk xk 1 n X k 2 a yk yk 1 n X k 2 o yk yk 1 Since yn 1 yn n N and yn we have n X k 2 o yk yk 1 o n X k 2 yk yk 1 o yn y1 o yn Thus xn x1 a yn y1 o yn 3 2 Exercises and homework 3 6 Exercises 1 To show lim n n n n 1 logn 1 Proof n n exp logn n 1 logn n O log 2 n n2 n n n 1 logn O log2n n Another proof given by a student Let t n n 1 Then t 0 n and n 1 t n and so n n n 1 logn nt nlog 1 t t log 1 t 1 log 1 1 1 t 1 t loge 1 n 2 To prove for x 0 lim n n2 n x n 1 x logx 24Lecture Notes at Fudan Proof n2 n x n 1 x n2x1 n 1 x 1 n n 1 n2x1 n 1 exp logx n n 1 exp logx n n 1 1 logx n n 1 O 1 n4 so we have n2 n x n 1 x n2x1 n logx n n 1 O 1 n4 x1 nlogx O 1 n2 3 To prove lim n cosn x n e x2 2 Proof cos x n n 1 x2 2n O 1 n2 n exp nlog 1 x2 2n O 1 n2 Since log 1 x2 2n O 1 n2 x2 2n O 1 n2 we have cos x n n e x2 2 1 O 1 n Another proof given by a student cos x n n 1 sin2 x n n 2 1 x2 n n x2 x2 2 e x2 2 n 4 To prove for 0 2 we have lim x r x q x x x 1 2 Proof r x q x x x 1 r 1 x x 2 2 1 2 x 1 1 2 x p 1 x 2 1 O 1 x thus r x q x x x 1 2 1 1 2x 2 1 O x 2 O 1 x 5 To show lim x 0 x tanx 1 x2 e 1 3 Analysis25 Proof x tanx xcosx sinx x 1 x2 2 O x4 x x3 6 O x5 1 x2 2 O x4 1 x2 6 O x4 1 x2 3 O x4 x tanx 1 x2 exp log1 x2 3 O x4 x2 exp x2 3 O x4 x2 e 1 3 O x 2 3 7 Homework 1 Euler constant To show that lim n n X k 1 1 k logn 0 bn 1 bn 1 n 1 log n 1 logn Z n 1 n 1 n 1 1 t dt 0 Thus lim n an lim n bn k we have bn 1 1 1 2 1 1 n 1 3 1 1 n 1 2 n 1 k 1 1 n 1 2 n 1 k n 26Lecture Notes at Fudan Thus e lim n 1 1 1 2 1 1 n 1 3 1 1 n 1 2 n 1 k 1 1 n 1 2 n 1 k n ak Therefore bk ak e k N and solim k ak e 3 To show that for 1 X k n 1 k O 1 n 1 n X