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# Analysis I MTH 320

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This 40 page Class Notes was uploaded by Donny Graham on Saturday September 19, 2015. The Class Notes belongs to MTH 320 at Michigan State University taught by Richard Siefring in Fall. Since its upload, it has received 18 views. For similar materials see /class/207294/mth-320-michigan-state-university in Mathematics (M) at Michigan State University.

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Math 3207 Homework 5 Solutions 1 Exerl 324 Complete the proof of Theorem 325 by proving the following A 0 et I 6 R and let A C R Let an be a sequence with an 6 A and an I for all n 6 N and assume that I lim ant Prove that I is a limit point of A Proof Let 5 gt 0 Then by the assumption that an converges to I there exists an N 6 N so that an 7 1 lt 5 if n 2 N and in particular aN 7 1 lt 5 However aN 7 1 lt 5 is equivalent to 75 lt aN 7 z lt 5 which is equivalent to I 7 5 lt aN lt I 5 Moreover aN I by the assumption that an a I for all n 6 Ni Hence for any 5 gt 0 we can nd an a E A with a 6 VE and a f It Thus I is a limit point of Al D Exerl 3 26 Prove Theorem 3281 Consider a set F C R Prove that the following statements are equivalent ie a ltgt b a F is closed b If is a Cauchy sequence with In E F for all n 6 N then lim In 6 Fl Proof We rst assume that F is closed ie that F contains all its limit pointsl Let be a Cauchy sequence satisfying In 6 F for all n 6 N Then is convergent since every Cauchy sequence is convergent Theorem 264 Let L lim Int We claim that L 6 Fl If there is some n 6 N so that In L then L 6 F since we assume that In 6 F for all n 6 N If not ie if In L for all n 6 N then Theorem 3 25 implies that L is a limit point of Fl Therefore L E F since we are assuming that F is closed Now assume that b holdsl Let 1 be a limit point of Fl Then according to Theorem 325 there is a sequence converging to l and satisfying In E F and In l for all n 6 Ni Since the sequence In is convergent it is a Cauchy sequence Theorem 2 64 and hence the assumption that b is true implies that l 6 Fl We conclude that F contains all of its limits and hence is a closed set D 3 Exeri 3 28 Consider a set A C R and let L denote the set of limit points of Al a Prove that the set L is closed A Proof Let I be a limit point of L We will show that z 6 L ie that z is a limit point of A In order to show that z is a limit point of A we need to show that for every 5 gt 0 the set I 7 51 U 11 5 contains an element of Al Let 5 gt 0 Since I is assumed to be a limit point of L there is a l E I 7 51 L with 1 1 De ne 5 ll 7 ml and note that 5 gt 0 by the assumption that I ll Since 1 6 L l is a limit point of A and hence there is an a 6 l 7 5 l 5 N A with a y ll We claim that a y I If a I then ll 7 al ll 7 ml 5quot This contradicts that fact that a6 175l5 and hence ll7al lt 5quot We next claim that a 6 I 7 5 z 5 or equivalently that lz 7 al lt 5 Indeed using the triangle inequality and ll 7 al lt 5 7 zl we have t at a lliallIllal mmvw lt2ll7zli Next using that l 6 I 7 51 is equivalent to ll 7 ml lt 5 we conclude that lz 7 al lt a Therefore a 6 I 7 5 z 5 N A and a y I Since we can repeat this argument for any 5 gt 0 it follows that z is a limit point of Al B Let I be a limit point of the set A A U L Prove that z is a limit point of A Proof We will argue by contradictioni Let I be a limit point of A A U L and assume that z is not a limit point of Al Using the de nition of limit point the assumption that z is not a limit point of A implies that there exists an 5 gt 0 so that VE Q A Qi Moreover for any 5 satisfying 0 lt 5 8 the set V41 N A is a subset of VE N A so we conclude Q A Q for all 8 6 05 1 Now since we assume that z is a limit point of A A U L it follows that for every 5 gt 0 that the set N A U L is nonempty Using that V495 1 m A U L VEW 1 Al U VEI 1 Ll with l we conclude that N L y Q for all 8 E 0 5 and hence L y Q for all 5 gt 0 Therefore I is a limit point of Li By part a this implies that z E L and therefore I is a limit point of A in contradiction to our assumption that z is not a limit point of A D Explain why you can conclude that A A U L is closed Prop If every limit point of A A U L is a limit point of A this means that every limit point of A is an element of L and hence an element of A A U L Therefore A contains all of its limit points and is hence a closed set Math 3207 Homework 7 Solutions 1 Let f g z A 7 R be functions and let 5 6 R be a limit point of Al Assume that limgc C L and limincga Ml Use de nition 421 ie don t use Theorem 423 to prove gig fltzgtgltzgt 7 LM Proof Let 5 gt 0 We would like to show that there exists a 6 so that 7 LMl lt 5 provided lz 7 cl lt 6 and z 6 Al We consider four cases Casel LM0i In this case we choose 61 gt 0 so that if z 6 A and lz 7 cl lt 61 then lf1LllfIllt E which we can do by the assumption that limgc C L 0 Similarly the assumption that limgc C 91 M 0 allows us to choose a 62 gt 0 so that for z 6 A satisfying lz 7 cl lt 62 we will have MI Ml l9rl lt r The above inequalities tell us that for z 6 A satisfying lz 7 cl lt 6 I min6162 that lf191 LMllfI9Illf1ll9Illt 8 Thus EmmiC 0 LMl Case2 L7 0M0i In this case we choose a 61 gt 0 so that 7Ll lt lLl for all z 6 A satisfying lz7 cl lt 61 and we choose 62 gt 0 so that lgz 7 Ml lt ll provided I 6 A and lz 7cl lt 62 Then for z E A satisfying lz 7 cl lt 6 2 min 61 62 we have that lf191 LMl lfI9Il lf1 L9I L9Il S WI Ll lLll9Il lt 2lLllWEH 5 lt ZlLl a Thus EmmiC 0 LMl Case 3 L 0 M 0 This is nearly identical to case 2 We omit the details Case4 L7 0M7 0i In this case we start by observing that for all z 6 A we have that lf191 LMl lf1 L L9I M M LMl WW LQI M 1 LM L9I MN Slf1Lll9IMllMllf1LllLll9IMl Choose 61 gt 0 so that 7 Ll lt min l slim for all z E A satisfying lz 7 cl lt 61 and we choose 62 gt 0 so that lgz 7 Ml lt min7 slim for all z 6 A satisfying lz 7cl lt 62 Then for z 6 A satisfying lz 7 cl lt 6 2 min 61 62 we have that lf191 LMl S WI Lll91 Ml lMl WI Ll W MI Ml lt mmWW 8 Thus lim c fzgz LM 2 Exeri 4i3i4b Let f z A 7 R be a function and let 5 6 A be an isolated point of Al Prove that f is continuous at of Proof Let c E A be an isolated point of A ie assume that c E A is not a limit point of Al Taking the logical negation of De nition 324 we see that if c E A is an isolated point then there exists an 50 gt 0 so t at V50 0 0 m A 3 or equivalently since 5 6 A that V50c A Let 5 gt 0 and choose 6 so Then if z 6 A and lz 7 cl lt 6 50 we can conclude from V50 0 N A c that z cl Consequently for all z 6 A satisfying lz 7 cl lt 6 we have that m7 molta We can therefore conclude that f is continuous at or D 3 Exeri 4 32 Consider functions f z A 7gt R and g z B 7gt R Assume that the range of f is contained in B that f is continuous at c 6 A and that g is continuous at 6 Bi Prove that the composition 9 o f z A 7gt R is continuous at or Proof 1 Use de nition of continuity Let 5 gt 0 Since 9 is continuous at fc there is a 61 so that y 6 B and ly7 lt 61 implies that 7 lt 5 1 Moreover since f is continuous at 5 there is a 6 gt 0 so that I 6 A and lI 7 cl lt 6 implies that 7 lt 61 Since E B we can apply 1 with y to conclude that for I E A satisfying lI 7 cl lt 6 that 7gfcl lt 5 Therefore 9 o f is continuous at or Proof 2 Use Theorem 432 v First assume that c 6 A is an isolated pointi Then 9 o f is continuous at c by problem We next assume that c 6 A is a limit point of Al We must consider two cases the case when is a limit point of B and that when is not a limit point of Bi Assume rst that is a limit point of Bi Since f is continuous at c E A Theorem 4i3i2iv tells us that if In is a sequence in A satisfying lim In c then limfIn Applying Theorem 432iv to the sequence yn the fact that g is continuous at implies that limgfIn limgyn in this case e next consider the case that c 6 A is a limit point of A but is not a limit point of Bi lf In 6 A is a sequence satisfying limIn c we can again use Theorem 4i3i2iv to conclude that limfIn Since is not a limit point of B we can nd an 50 gt 0 so that VEO N B Moreover since lim we can nd an N so that for n 2 N we will have that 6 VEO H B and consequently that for all n 2 Ni Therefore for all n 2 N and we can conclude that limgfIn In summary we have shown that if c E A is a limit point of A and In is a sequence in A which converges to c then converges to According to Theorem 4 3i2iv this means that g o f is continuous at or 4 Exeri 4 39 Consider a function f R 7gt R and assume that there is a c 6 01 so that lf1fyl 0l19l 2 for all z y 6 Rf a Show that f is continuous on R Proof Let y 6 R and let 5 gt 0 Choose 6 Eci Then for z 6 R satisfying 1 7 yl lt 6 50 we use 2 to conclude W 7M air 7 yl lt w 7 c2 7 a Thus f is continuous at y Since this proof applies to all y E R we can conclude that f is continuous on 1 A b Given a point yl 6 R de ne a sequence by yn1 Prove that yn is a Cauchy sequence and therefore convergent i Proof We rst claim that for any m k E N that lymk 7 ml S cm llymc 7 ml 3 We prove this by induction on mi lfm l the statement reduces to ly1k 7 yll Co y1k 7 yll ly1k 7 yll which is true Now assume that the inequality 3 holds for some given m 6 Ni hen we use yn1 with 2 to conclude that lym1k 7 ym1l lfltymk fyml S Clymk 7 ml and using our induction hypothesis lymk 7 yml g cm llyllrk 7 yll we continue S C Emillyl l k 7 yil Cm171lylk 7 yllv so indeed if 3 holds for some m 6 N it also holds for m 1 We conclude that 3 holds for all m 6 Ni We next claim that k71 lyk1 7 yil S 62gt ly2 7 yil 4 for all k 6 Ni We again argue by induction The statement reduces to lyg 7 yll lyg 7 yll if k 1 We next assume that 4 holds for some given k E N and attempt to prove that it also holds for k 1i We have that lyk11 7 yll lyk2 7 yk1 yk1 7 yil S lyk2 yk1l lyk1 yilA We can use 3 to conclude that lyk2 7 yk1l g cklyg 7 yll and we use our inductive hypothesis to conclude that lyk1 yil S 2113 CZ lyg 7 ml We thus continue k71 S CW2 yll 02 92 7 yll 0 k ly27y1t 0 We thus see that if 4 holds for some given 16 it also holds for k 1 We conclude that 4 holds for all k 6 Ni 1 Note that since the choice of 6 depends only on s and not on y this proof actually shows that f is uniformly continuous on R We now observe that 4 implies that l9 yll lt lyk1 yll 7 1 C 5 for all k 6 N Indeed the term 21101 CZ 1 c c2 ck l is the k 7 1 partial sum of a geometric seriesi Since 5 6 01 we know Example 275 that 2110ch g 220 CZ liar This observation used with 4 immediately yields Moreover using 3 and 5 gives us W2 7 ml lymk yml S Cm 1 6 for all k m E N We now show that yn is a Cauchy sequence If yg yl 6 implies that yn yl for all n E N and hence yn is a Cauchy sequence in this case Assume then that yg y yli Let 5 gt 0 Since 5 6 01 we know that limmx00 cm l 0 Example 253 Therefore there exists an N 6 N so that cm l lt 2135 for m 2 Ni Again using 6 we can conclude that for n m 2 N that W 7 ml S 8 Hence is a Cauchy sequence D c Let y lim yni Prove that yr Proof Recall that if yn converges to y then yn1 converges to y Exeri 2i4i2bi Also Theorem 4i3i2iv and the fact that f is continuous on R and hence at y tell us that the sequence converges to Thus the equation yn1 de ning the sequence tells us that y gigolo yn1 lim fyn f y n7gtoltgt so y as claimed D d Prove that the limit y is independent of the choice of y1 ie any sequence de ned by choosing an 11 and de ning zn1 converges to the y from part Proof Part b tells us that the sequence In convergesi Let I lim zni Then part c tells us that 1 satis es 1 Using z and y with y limyn in 2 then implies that lr 7 yl 7 lf17 fyl S Clr 7 yl or that lz 7 yl clz 7 This in turn implies that 17clz7yl 0 If I f y then lz 7 yl gt 0 so 17clz 7 yl gt 0 since we assume that c lt 1 This contradicts l 7 clz7 yl 0 so we can conclude that z y D Math 320 Homework 1 Solutions 1 Prove that there is no rational number T satisfying 7 2 5i PToof In order to complete the proof we will need the following fact Claim Let p be an integer if p is not divisible by 5 then p2 is not divisible by 5 Proof of Claim Assume that p is not divisible by 5 Using basic properties of the integers p can be written p 5n T where n is an integer and T 6 1 234 Squaring this we get 55n22nTl ifTl 55n22nT4 ifT2 2 2 2 25 10 p n mT 55n22nTl4 ifT3 55n22nT3l ifT4i In any case we see that p2 is not divisible by 5 Now we prove there is no rational number whose square is 5 Arguing by contradiction we assume there is a rational number T satisfying 7 2 5 and we write I7 T 7 q where p and q are integers with no common factors and where q f 0 The equation 7 2 5 then becomes 2 p7 5 gt p2 5qu We conclude that p2 is divisible by 5 and thus by the claim in the rst paragraph that p is also divisible by 5 Writing p 51 for k E Z the equation above becomes 5k2 5q2 25k2 542 5 5162 42 This now implies that L12 is divisible by 5 and arguing as above that q is divisible by 5 as well Thus we have shown that p and q are both divisible by 5 in contradiction to the assumption that p and 4 have no common factors This contradiction completes the proof D 2 a Let A By and C be setsi Prove that A B U C AB N A C Proof By de nition of set difference 16ABUC ltgt zeAandzj BUCi By de nition of union 1 6 B U C precisely when I 6 B or z 6 C so z B U C is equivalent to z B and I if C so 16ABUC ltgt zeAandz Bandz C ltgt zeAandz BandzeAandz C ltgt z ABandzeAC and using the de nition of intersection we conclude ltgt zeAB ACA Thus zeABUC ltgt zeAB AC which is equivalent to A B U C AB Q A G Let A and B1 for i E N be setsi Use induction to prove that AB1UB2UUBn AB1 AB2 ABn for all n 6 N Proof When n 1 the statement is trivial The case n 2 is done in part a For our inductive step we assume that for some k 2 2 that AB1UB2UUBkAB1 ABk and try to prove that AB1UB2UUBkUBk1AB1 ABk ABk1i Using the associativity of unions and part a we have AB1UB2U UBkUBk1AB1UB2U UBkUBk1gt ABlUB2UUBk ABk1 Aquot 1 and 39 39 39 ABl m ABkl ABk1 ABl quot39 ABk ABk1 and using our inductive of 39 t t39 we get so 1 let s us conclude AB1UB2UUBkUBk1 AB1 ABk ABk1 as desired 3 De ne a sequence of numbers recursively by a A 0quot yi 1 1 3 4V4 Use induction to prove that yn lt 4 for all n 6 Ni Proof When n 1 we have yl 1 lt 4 Next we assume that for some n 2 1 that yn lt 4 and try to prove that yn1 de ned by yn1 3yn 44 Eyn 1 satis es yn1 lt 4 We have that ynlt4 Eyn 3 yn1yn1lt314 so assuming yn lt 4 let s us conclude that yn1 lt 4 as well Thus by induction yn lt 4 for all n 6 Ni D Prove that yn1 gt yn for all n 6 Ni Proof 1 Use part a We have that yn1 7 on 2 1 yn 1iiyn By part a we know that yn lt 4 for all n 6 Ni We thus have ynlt4 7iyngti1 1iiyngtQ which with the above equation gives us yn1 yn gt 0 or yn1 gt yn as required D Proof 2 Use induction Base case yg Z gt 1 yli For the inductive step we assume that for some k 2 1 that yk1 gt yki Then we have yk1 gt yk j 3yk1 gt 3 gt3 k14 gt 3yk4 3 4 3 4 j 919 gt 911 j yk2 gt yk17 so yk1 gt yk gt yk2 gt yk1i Thus by induction we conclude that yn1 gt yn for all n 6 N D 4 Assume that A and B are nonempty sets that A is bounded above and that B C Al Prove that B is bounded above and that supB sup Al Proof We rst claim that any upper bound for A is also an upper bound for Bi To see this let u be an upper bound for Al Then u 2 a for all a 6 Al But since B is a subset of A every element of B is an element of A so it follows that u 2 b for all I 6 B as well Therefore u is an upper bound for B and indeed every upper bound for A is an upper bound for B as well Thus B is bounded above To see that supB sup A we note that since supA is an upper bound for A the previous paragraph tells us that supA is also an upper bound for Bi By the de nition of supremum this immediately implies that sup E g sup A D 5 Use the Completeness Axiom to prove that every nonempty set of real numbers which is bounded below has a greatest lower bound See 133 in the textbook for some guidance Proof Let A be a nonempty set of real numbers which is bounded belowi We seek to show that A has an in mumi De ne a set L to be the set of lower bounds of A ie L16Rll aforalla6Ai Note that L is nonempty by the assumption that A is bounded belowi We claim that L is bounded above Let a 6 A Then by the de nition of L it follows that l g a for all 1 6 L so we see that any element ofA is an upper bound for Li Using the Axiom of Completeness we conclude that L has a supremum since L is nonempty and bounded above We next claim that sup L is an element of L ie that sup L is a lower bound for Al Suppose noti Then there exists an element a E A satisfying a lt sup Li Choosing an 5 gt 0 satisfying 5 lt supL 7 a we know from Lemma 137 that there is an 1 6 L satisfying 1 gt sup L 7 5 Since 8 lt sup L 7 a we can conclude sup L 7 5 gt a and thus 1 gt a This contradicts the fact that l is a lower bound for A so we can conclude that supL is indeed a lower bound for A and thus supL 6 Li Finally we claim that supL is an in mum for Al Indeed we have shown that supL is a lower bound for A in the previous paragraph Moreover ifl is a lower bound for A then 1 6 L by de nition of L and thus 1 g supL by de nition of supi Using the de nition of in mum we conclude that sup L inf A and thus A has an in mum as claimed D Math 3207 Homework 10 Solutions 1 Exerl 627 Assume that converges uniformly to f on A and that each fn is uniformly contin ous on Al Prove that f is uniformly continuous on i Proof Let 5 gt 0 Since fn converges uniformly to f on Ay we can nd an N 6 N so that for n 2 N 7 g g for all z 6 A and in particular g g for all z 6 Al Since fN is assumed to be uniformly continuous on A we can then nd a 6 gt 0 so that for z y E A satisfying 1 7 y lt 6 mm 7 my lt g Then for z y E A satisfying 1 7 y lt 6 we nd that lf1 fyllf1 fNI fNI fNy fNy fyl S lf1 fNIl WNW fNyl lfNy fyl We can thus conclude that f is uniformly continuous on Al D 2 Exerl 61413 a Show that 91 2 C 1 A 0quot n 60 in continuous on all of R1 Proof Using that lcoszl g 1 for all z 6 R we have that cos 27 lcos2 zl 2n T gglinforallzeRl converges Example 275 the Weierstrass Mtest Corollary 61415 cos2quotm 2 is continuous on R we can conclude by Theorem 61412 that 91 oo 1 Slnce the ser1es En1 27 implies that the series 221 converges uniformly on R Moreover since each func tion Jena cos2quotm 2 Eco cos2quotm 711 2 is also continuous on R1 Show that hz 221 i is continuous on 711 Proof For all z 6 711 we have that 1 and thus that g 1 for any n 6 N1 Conse quently we can conclude that z z 1quot n2 11 n2 g for alle 7111 1 W W Since the series 221 51 converges Example 244 the Weierstrass M test implies that the series 221 g converges uniformly on 71 1 Since each function fn 95 is continuous on 71 1 Theorem 6142 allows us to conclude that hz 221 7 is continuous on 711 D Alternate proof At 1 1 the power series 221 becomes 221 517 absolutelyl Therefore Theorem 61512 implies that 221 since each function 7 is continuous on 711 we conclude by Theorem 61412 that D which converges g converges uniformly on 711 and hx 221 g is continuous on 711 3 Exeri 64quot Let a 1 hz E 7 711 12 n2 Show that h is a continuous function de ned on all of Rf Proof For any I E R we have that 12 Z 0 so 12 n2 2 n2 2 0 and consequently 1 7 lt 12n2 12n2 7 converges Example 244 we can use the Weierstrass Mtest to conclude that iQforallzeRi n Since the series 7712 21 52 converges uniformly on Rf Moreover each function 52 is continuous on R since 12 n2 2 n2 gt 0 for all z E R and n 6 Ni Therefore Theorem 642 allows us to D conclude that h R a R de ned by hz Zoo 1 is continuous on R nl 124 712 ls h differentiable If so is the derivative function h continuous Proof We consider the derived series 21 1 221 Let R gt 0 If I E 7RR we nd that 721 2 lt lt 12n22 I2n22 7 I2n22 7 n4 where we ve used that 12 n22 14 2127 n4 2 n4 for all z 6 R and n 6 Ni Since the series 221 7714 converges Corollary 247 so does 221 37142 Theorem 2i7i1i and so the Weierstrass M test tells us that 221 21252 then implies that h is differentiable on 7R R and h z 221 Moreover since the series 221 21252 7RRi Since R is arbitrary we can conclude that h is differentiable at any I E R that h is D continuous at any I 6 R and that h z 221 0521252 for any I 6 i converges uniformly on 7R Rli Theorem 6 43 converges uniformly on 7R R we can conclude that h is continuous on 4 Exeri 6511 Consider the function 9 de ned by the power series 13 I4 7Iim 12 700 n1i i 911 n z 2 a 1s 9 de ned on 71 1 If so is it continuous on this set A 0quot V A 0 V Proof At 1 1 the series becomes 22171 n1ii Since is a decreasing sequence which converges to 0 the series 220171 1 converges by the alternating series test Theorem 2717 Therefore Theorem 61511 implies that the series 221 71 Jr1757 converges absolutely for any I with lt 1 Moreover Theorem 61517 implies that the series is 91 22171 1 D is continuous on 71 1 1s 9 de ned on 71 1 If so is it continuous on this set Proof Since we ve show in part a that the series converges at z 1 and for all z 6 711 it follows that the series converges and hence g is de ned on 7111 Theorem 6517 again implies that g is continuous on 7111 1s 9 de ned on 711 If so is it continuous on this set Proof No At 1 71 the series become 271n1 113 2 1 n 712n1 TL 00 711 Since the series 221 is divergent Example 245 so is 221 1 by Theorem 21711 n 1s 9 de ned for any I with gt 1 Prove your answer is correct Proof No Suppose that there were an 10 with 110 gt 1 at which the series converged Then Theorem 6511 would imply that the series converges for all z with lt 110 In particular we could conclude the series converges at z 71 This is a contradiction since we have observed in part c that this series diverges at z 71 Math 3207 Homework 2 Solutions 1 Prove that every nonnegative real number has a square root adapt the proof of Theorem 145 in the textbook Proof Let a 6 R and assume that a 2 0 We would like to show that there is a real number a E R which satis es a2 a If a 0 then 02 0 a so a has a square root and there is nothing more to prove Assume then that a gt 0 Following the example of Theorem 145 we de ne a set TaIER 12ltai Then 02 0 lt a so 0 6 Ta and thus Ta is not empty We claim that Ta is bounded above Indeed if a 6 01 then we have that a lt l gt a2 lt a lt 1 So since any number I Z 1 satis es 22 2 b 2 l gt a we can conclude that b 2 1 implies that b if Tai Thus 1 is an upper bound for Ta if a lt 1 On the other hand if a Z 1 then a2 2 a 2 ll Moreover if b gt a 2 1 we have that b2 gt 12a 2 a2 so no such I can be in Ta and we see that a is an upper bound for Ta if a 2 1 Thus Ta is bounded above Since Ta is nonempty and bounded above the axiom of completeness tells us that Ta has a supremumi De ne a sup Tai Note that a Z 0 since 0 6 Ta and a is assumed to be an upper bound for Tai We would like to show that a2 a We argue by contradictioni If we assume a2 lt a then gt 0 Choose an no 6 N so that 70 lt a 7042 h h Mod w ic we can do by Theorem 142 We then nd that an7o 27a2i7j 1 7 no and using no 2 l gt i lt l 5 71 g i we continue no nO no 12i7jn71O 2 a nio2al i 1042 39 and usmg our assumption that no lt 2a1 we continue 2 7 2 lt a 301 2a 1 ai Thus a 77102 lt a which means that a 6 Tai However a n70 gt a sup Ta which is a contradicts the fact that sup Ta is an upper bound for Tai Thus we must have a2 Z a 2 If we assume a2 gt a then 03 gt 0 We choose an no 6 N satisfying 7710 lt min0 0 7a 2 57 20 that 77710 gt 70 a and n70 lt We then nd that gt or equivalently so 12 272i 1 a a no5g 2 l gta7n702a 2 i 7 a 7a and usmg our assumption 7 no gt 20 we continue gt a 7 tuna a So with our choice of no we have that a72a7gt0mda72gtai Thus if z gt a 7 7710 then 12 gt a so I if Ta and thus a 7 70 is an upper bound for Tai This is a contradiction since a 7 7710 lt a sup Tai With this contradiction we can conclude that a2 a D 1 2 2 Let A C R be a set of real numbers For C 6 R de ne the set CA by CAIERlzCaforsomea6Ai a Prove that if A is bounded above and C 2 0 then CA is bounded above and that supCA Csup Al Proof We rst address the case C 0 In this case if A is nonempty which it must be for supA to be de ned then CA 0 and it follows easily that supCA 0 Csup Al lndeed 0 Z 0 so 0 Z a for all a 6 Al Thus 0 is an upper bound for Moreover if z lt 0 I can t be an upper bound for CA by de nition of upper bound There 0 is indeed the supremum of 0 CA We now assume C gt 0 In this case we rst claim that CsupA is an upper bound for CA Assume to the contrary that CsupA is not an upper bound for CAi Then by de nition of upper bound there exists an I 6 CA so that z gt Csup Al But by de nition of CA if z 6 CA there exists an a E A so that z Cal We thus have Ca gt C sup A which since C gt 0 is equivalent to a gt sup A This is a contradiction since supA is an upper bound for Al Thus CsupA is an upper bound for CA so CA is bounded above further by de nition of supCA this implies that C sup A Z supCAi We next claim that C 1 supCA is an upper bound for Al If not there exists an a 6 A so that a gt C 1 supCAi Since C gt 0 this is equivalent to Ca gt sup CA This is a contradiction since a E A implies that Ca 6 CA and thus we must have Ca supCA since supCA is an upper bound for CAi Thus C 1 supCA is an upper bound for A as claimed By the de nition of sup A this implies that supA 671 sup CA or equivalently since C gt 0 that C sup A g supCAi Inequalities l and 2 then tell us that supCA CsupA as claimed 3 4 b Prove that if A is bounded above and that C lt 0 then CA is bounded below and that infCA Csup A Proof This proof proceeds similarly to part a We rst claim that CsupA is a lower bound for CAi This will imply that CA is bounded below and by de nition of infCA that infCA 2 C sup Al To see that CsupA is a lower bound for CA we argue by contradiction and assume that this is not the case Then there exists an I 6 CA satisfying 1 lt Csup A By de nition of CA there exists an a 6 A so that z Ca so this becomes Ca lt C sup Al Using that C lt 0 this is equivalent to a gt sup Al This is a contradiction since sup A Z a for all a 6 Al We thus conclude that Csup A is a lower bound for CA and that 3 holds Next we claim that C 1 infCA is an upper bound for A If not then there is an a 6 A with a gt C 1 infCAi Since C lt 0 this is equivalent to Ca lt inf CAi By de nition of CA Ca 6 CA since we assume a 6 A so this inequality contradicts the fact that infCA a for all a E CAi Thus C linfCA is an upper bound for A as claimed and using the de nition of sup A we get sup A g 671 infCAi Again using C lt 0 this is equivalent to C sup A Z infCAi lnequalities 3 and 4 then tell us that CsupA infCA as desired D 3 Assume that A and B are countable setsi Prove that the Cartesian product A X B is countable Proof sketch The argument here is very similar to the one 1 gave in class to show that the positive rationals are countable Since A and B are both countable sets we can nd bijective functions f z N 7gt A and g z N 7gt Br De ne an and b1 901 By de nition of Cartesian product every element I 6 A X B is of the form I an 121 for some kl 6 Ni We can then plot every element of A X B in the rst quadrant of the I 7 y plane by putting element an 122 over the point Is 1 Considering the diagonals k Z constant gives us a method for tracing a path through these points that reaches every element of A X B and thus gives a way for putting the elements of A X B in bijective correspondence with the natural numbers D 4 a Use the de nition of convergence of a sequence to prove that limnn00 271 3 Proof We need to show that for any 5 gt 0 we can nd an N 6 N so that 7 31 lt 5 for all n 2 Ni Given 5 gt 0 choose N 6 N satisfying N gt Then for n 2 N we have that n2 2 N2 gt 1N52 15 or equivalently 51 lt 5 We nd then that 1 Z7 1 2 3nn1 7 3 i 2 S S lt b Use the de nition of convergence of a sequence to prove that the sequence de ned by 1 ifnp2 forsomepEN I 7 n 0 otherwise does not converge Proof To show that In doesn t converge to L we must show that there exists an 5 gt 0 so that for all N 6 N there is an n 2 N for which lIn 7 Ll Z 5 To show that In doesn t converge we must show that previous statement holds for all L E R1 lfL 0 choose 5 Given N E N choose n NW N Then In IN2 1 so lIn7Oll1l12a 1f L f 0 choose 5 Given N 6 N choose n N2 1Z Ni Since N2 1 is not a perfect square1 we conclude that In IN21 0 so lrn7Ll10LllLlZlLl8 1 Assume that N2 1 P2 for some P e N with P gt N Then P2 7 N2 1 which we can rewrite P 7 NP N 1 Since P and N are assumed to be distinct natural numbers we must have that P N Z 2 and P 7 N Z 1 This leads to the contradiction 1 Z 2 Math 3207 Homework 9 Solutions 1 Exerl 5l2l2c Use the de nition of derivative Def 521 to prove the quotient rule Theorem 5l2i4iv Let f 9 9712 4gt R be differentiable at c 6 97b and assume that 95 y 0 Show that hm 83 7 wltcgtgltcgt 7 feycl ma 1 7 c 9c2 Proof For all z y c satisfying 91 y 0 we have 55 7 83 1 fltzgtgltcgt 79mm mic gltzgtgltcgt If e 1 f190 090f09091f0 9I90 I i 0 96 f 0 7 250 9I90 Since f and 9 are assumed to be differentiable at c we have that liminc 72 fc and limgcnC 72750 9 cl Moreover the since 9 is assumed to be differentiable at 5 Theorem 5 23 implies thaf9 is continuous at c Therefore lim Ac9z 9cl Since 95 y 0 the Algebraic Limit Theorem Corollary 424 used with the above equations allows us to conclude that 8 7 g 9a hch 9 21 e 1 1mm 7quot 5 z 5 95 limzao y f c 7 fcy c 2 96 lim 1 2 Exeri 535 Let f z a b 4 R be differentiable on a b and assume that y 1 for all z 6 a7 12 Show that there is at most one point c 6 ab satisfying of Proof We argue by contradiction We assume to the contrary that there exist distinct points cl CQ 6 ab satisfying fcl cl and ag 02 Without loss of generality we can assume that 02 gt all Since f is assumed differentiable on ah the mean value theorem implies that there is a point c 6 cl7 Cg satisfying f 02 if Ci W lt gt lt gt c 7 cl Substituting fcl cl and ag Cg on the right hand side we nd that 52 7 Cl 7 1 f c 62 7 Cl This contradicts the assumption that f 1 for all z 6 ab Thus there can be at most one point c 6 ab satisfying or D 3 Exeri 5 38 Consider a function 9 a7 12 4gt R Assume that 9 is differentiable at some point c 6 a7 12 and that 9c 0 Show that there is a 6 gt 0 so that 9I 9c for all I 6 V5 cc a7 12 Proof 1 Argue by contradiction If not then for every 6 gt 0 there is an I 6 V5c a7b satisfying 9I 9ci In particular for each n E N there is an In E V c 1 ab satisfying 9In 9ci Moreover In 6 V 0 implies that lInicl lt i so we can conclude that limngt00 In cl Thus using the de nition of derivative and the sequential characterization of limits we have that 9 hm 91 7 96 hm 9In 7 96 Igtc I 7 c ngtoltgt I 7 5 Since 9In 90 we can thus conclude that 9c 0 which contradicts the assumption that 9c 0 This contradiction allows us to conclude that there is a 6 gt 0 so that 9I y 9c for all I 6 V5c 1 ab D Proof 2 Direct argument Since 9 is a assumed to be differentiable at c and 9c 0 we can use the de nitions of derivative and limit with 5 l9 cl to nd a 6 gt 0 so that if I 6 ab and 0lt lIicl lt6then 91 7 95 l 3H m lt 2w Using this with the triangle inequality we nd that for any I 6 VHO a7 12 that Mal we 7 91 l I j regime 796 wife lt gigcl lg Iijlcl We conclude that 91 7 96 1 gt E l9 CM for all I 6 V5c ab and thus that was 79W gt g z 7 cl we gt o for all I 6 V5c a7bi This implies that 9I a 95 for all I 6 V5c 1 ab D 4 For each of the following sequences of functions nd the pointwise limit and determine whether ire prove or disprove the sequence converges uniformly a y on 0 00 Solution Using the Algebraic Limit Theorem we have for any I 6 0 00 that 1 1 lim 73320 W W n m 71320 so converges pointwise to 1 We claim that does not converge uniformly to 0712 on 0 If we assume to the contrary that converges uniformly to f on 000 then we could nd an N 6 N so that on 0 7 lt for all z 6 000 However given such an N 6 N we can choose IN TIN 6 000 and we would have that i 1 1 N1 N 1 lfNIN fINl 2 IN This is a contradiction so we conclude that is not uniformly convergent on 0 sinnx 1mc Solution We claim that converges uniformly and thus pointwise to 0 on 12 on 12 Let 5 gt 0 Choose N gt if 1 Then for n 2 N and z 2 1 we have that 11 and hence sinnz new WM 7 med 7 sinnz 7 1 nz lt 1 7 1 nz lt lt a 7 1 N Therefore converges uniformly to 0 on 1 2 D Math 3207 Homework 6 Solutions 1 Exerl 3213 In this problem you will prove that the only subsets of R that are both open and closed are R and the empty set Let E be a nonempty subset of R and assume that E is both open and closed Since E is nonempty there is an element a E El De ne the set NaEIgt0a7IaIQEl a Explain why Na is nonemptyi Proof Since E is assumed to be an open set there is an 5 gt 0 so that V5a a75a5 C E Therefore 5 6 NaE and hence NaE is nonemptyl D b Prove that if I 6 NaE then a 7 Ia I Q El A know that a 7 I a I C El De ne for all n 6 N we can conclude that Proof Let I 6 NaEl Then a sequence In a I 7 Q ag In lt aI for all n 6 N so In 6 a7IaI C E for all n 6 Ni Since limIn aI we can conclude from Theorem 325 that a I is a limit point of E and therefore that a I 6 E since E is assumed to be closed Applying a similar argument to the sequence In a7 Il 7 l we can conclude that a 7 I is also a limit point of E and hence that a 7 I 6 E as well again since E is assumed to be closed Thus a 7 Ia I C El D by de nition of NaE we 1 39 c 7 lt l Prove that if a 7 Ia I Q E then there is a y 6 Na satisfying y gt It Proof Assume that a 7I a I Q El Since E is assumed to be open there exists an 51 gt 0 so that V51a I a I 7 51 a I 51 C E and there exists an 52 gt 0 so that V 2 a 7 I a7I752a7I52 C El Since a7IaI C E we can conclude that a7I752 aI51 C El If we de ne 5 min 5152 then we can conclude that a 7 I 7 5 a I 5 C E and hence thatI5 NaEl D d Show that NaE is not bounded above argue by contradictioni A Proof Assume that NaE is bounded above Since Na is nonempty the axiom of com pleteness implies that NaE has a supremumi Let s sup We claim that s 6 NaE that is that a 7 sa s C El If not then a 7 sas gZ E which implies that there exists a point z 6 a 7 s a s with I if El De ne 5 s 7 lz 7 al and note that 5 gt 0 since I E a 7 sa 8 is equivalent to lz 7 al lt 8 By Lemma 137 there exists a y 6 NaE with y gt sup NaE 7 8 s 7 6 We can conclude that lz7al lt lz7al5 h7d7h7ab ilsllz7al 2 2 7s7 e7w7aw 378 lt9 so lz7alltyi This is equivalent to z 6 a7 y ayi Since y 6 Na E a7 y ay C E so we can conclude that z 6 E in contradiction to the fact that I if El This contradiction shows that s 6 NaEi Since we have shown that s 6 Na E we can conclude from part b that a 7 s a s C El Then part c let s us conclude that there exists a y 6 Na satisfying y gt s sup Na This contradicts the de nition of supremumi We therefore conclude that the set Na is not bounded above D Prove that E Ri Proof Assume that E Ri Then there exists a b E R with 12 El Then if z gt lb7 al it follows that b 6 a7 z az and consequently a 7 z az gZ E so z We can conclude that lb 7 al is an upper bound for Na so Na is bounded above This contradicts part We therefore conclude that E Ri D 2 Exeri 331 Show if K C R is compact and nonempty1 then supK and ian both exist and are elements of K i Proof Since K is compact the HeineBorel theorem implies that K is closed and bounded Since K is bounded and nonempty the axiom of completeness implies that supK exists and problem 5 from homework assignment 1 implies that ian existsi By Lemma 137 for every 5 gt 0 there is an I E K so that I gt supK 7 a In particular for every n E N there is an In E K satisfying 1 supK77ltIn supKi l n Using the squeeze theorem we can conclude from 1 that In converges to sup Ki Since K is a closed set it follows from Theorem 328 that supK is an element of Ki Similarly applying problem 6 from homework assignment 1 we can nd for any n 6 N an element In 6 K satisfying 1 iang In lt ian 7 n Again the squeeze theorem implies that limIn inf K and since K is closed we can conclude by Theorem 328 that ian 6 Ki D 1 The word nonempty was missing from the initial statement of the problem and is missing from Exer 1331 in the book but it is necessary for stated the conclusions to hold true 3 Exeri 334 Show that if K is compact and F is closed then K N F is compact Proof 1 Use the HeineBord Theorem Since K is compact K is closed and bounded by the Heine Borel theoremi Since K is bounded there is an M gt 0 so that lIl g Mfor allIEKi We claim that K N F is also boundedi Indeed if I 6 K N F then I 6 K so lt M and therefore K F is bounded Moreover since K is closed and F is closed K H F is closed by Theorem 3214 Therefore K H F is closed and bounded and the HeineBorel theorem therefore implies that K H F is compact D Proof 2 Use the de nition of compact Let be a sequence with In 6 K F for all n 6 Ni Then In 6 K for all n 6 Ni Since K is compact has a subsequence In which converges to some element I 6 Ki But since In 6 F for all n 6 N we know that In 6 F for all j 6 Ni Since F is closed we also have I limjn00 In E F by Theorem 328 so I 6 K H Fl We have shown that every sequence in K F has a subsequence which converges to something in K HF and hence K F is compact D Math 320 Homework 8 Solutions 1 Exerl 4l4i8a Assume that f 000 7 R is continuous at every point of its domain Show that if there exists a b gt 0 so that f is uniformly continuous on the set 12 0 then f is uniformly continuous on 0 Proof Let 5 gt 0 Since f is assumed to be uniformly continuous on 1200 there exists a 61 gt 0 so that if z y 6 1200 and z 7 y lt 61 then 7 lt 52 1 Since f is assumed to be continuous on 0 then f is continuous on 012 and since 012 is compact if follows that f is uniformly continuous on 0bi Therefore there exists a 62 gt 0 so that if z y 6 012 and z 7 y lt 62 then 7 lt 52 2 Choose 6 min 6162 Let I y 6 0 gt0 and assume that z 7 y lt 6 Without loss of generality we can assume that y 2 I We consider three possible cases Casel szandyZbi In this case we have that I7 y lt 6 implies that z 7y lt 61 so it follows from 1 that 1290 fy lt 52 lt 5 Case2 ngandyltbi In this case we have that z 7 y lt 6 implies that z7y lt 62 so it follows from 2 that 7 lt 52 lt 5 Case 3 I g b and y Z 12 Assume that z 7 y lt 6 Then we have that z7b b7zgy7zz7y lt6lt62l Since I 6 012 and b 6 0 b it follows from 2 that was 7 fbl 52 We further have that b7y y7b y7zz7y lt6lt61l Since I 6 1200 and y 6 1200 it follows from 1 that W 7 w s 52 Using the triangle inequality we therefore conclude that if I g b and y 2 b satisfy 1 7 y lt 6 then 161 fy 1600 fb 165 fy S 161 fb WI fy lt 52 52 a Putting the three cases together we ve found a 6 gt 0 so that if z y 6 0 gt0 satisfy 1 7 y lt 6 then 7 lt 5 Therefore f is uniformly continuous on 0 D 2 Exeri 41419 A function f z A 4gt R is said to be Lipschz39tz if there exists a real number M gt 0 so f17fy SM I 7 y for all z yEAwithzy yi a Show that if f z A a R is Lipschitz then f is uniformly continuous on Al Proof If we assume that f is Lipschitz there is an M gt 0 so that f0 7 y 9 I gM for all z y E A with z y y This implies that lfIfyl Seriyl for all z y E A since when I y both sides of the above inequality are 0 Given then 5 gt 0 choose 6 EMi Then for z y E A with z 7 y lt 6 we have that lfrfy Mhiy ltM6gi Therefore f is uniformly continuous on Al D b Is the converse true ie if f z A 4 R is uniformly continuous on A is f Lipschitz Prove or disprove Proof No uniformly continuous does not imply Lipschitzi As a counterexample we let A 01 and Then f is continuous on 000 see Example 438 and therefore f is uniformly continuous on the compact set 01i1 However we have that f17f0 fem Thus given any M gt 0 ifwe let y 0 and z 6 01 satisfy 1 lt M17 then 1 gt Mi Therefore for no M gt 0 can satisfy f17 y lt M I 7 y 7 I for all z y 6 01 with z y y We conclude that is not Lipschitz on 0 1i D 1 One can use problem 1 to show that uniformly continuous on the entire halfiline 0 00 but we won t need that here 3 Exeri 4517 Let f 01 4gt R be continuous on 01 and assume that the range of f is contained in 01i Prove that there exists an I 6 01 satisfying 1 Proof De ne a function 9 z 0 1 a R by gltzgt fltzgt 7 z Since f is assumed to be continuous and the function hz z is continuous it follows from Theorem 41314 that 9 is continuous on 01 Finding a point z 6 01 satisfying z is equivalent to nd a point z 6 01 where 91 0 To nd a z 6 01 satisfying 91 0 we rst note that the assumption that the range of f is contained in 0 1 is equivalent to 0 g 1 for all z 6 01 We therefore nd that 90 7 0 2 0 and 91 7 1 g 0 If either 90 0 or 91 0 there is nothing left to prove so we assume that 90 0 and 91 y 0 Then the preceding observation tells us that 90 gt 0 gt 90 Since 9 is continuous on 0 1 the Intermediate Value Theorem Theorem 4151 allows us to conclude that there is an I 6 01 satisfying 91 0 as desired Math 3207 Homework 4 Solutions For many of the problems in this assignment the following adjusted version of the comparison test Theorem 274 is useful The proof of this theorem is a straightforward modi cation of the one given in the book for Theorem 274 see the discussion on page 64 Theorem Generalized Comparison Test Let an and 127 be sequences and assume that there exists an N 6 N so that Ugan bnforalanNi Then 0 If 221 12 converges then 221 an converges 0 If 221 an diverges then 221 12 diverges 1 Let an and 127 be sequences with In gt 0 for all n 6 Ni Assume that liran L with L gt 0 Prove that the series 221 an converges if and only if the series 221 12 convergesi Proof Since we assume that lim 1 L gt 0 ngtoltgt In there exists an N 6 N so that for n 2 N a L in 7 L 7 bn lt 2 Therefore for n 2 N we will have that 0 lt L lt a lt 3L 2 bn 2 and using 12 gt 0 we conclude 1 0lt bnltanltbnforalln2Ni Assume that 221 an convergesl Then we know from 1 and the generalized comparison test that 221 127 converges and by Theorem 271 this implies that 221 12 converges as well Conversely if we assume that 221 12 converges then we can conclude from Theorem 271 that 1 bn converges alsoi Using this with l and the generalized comparison test then implies that 221 an convergesl Therefore 221 an converges if and only if 221 12 convergesl D 2 2 a Assume that the series Ezlan is convergent and that an 2 0 for all n 6 Ni Prove that 00 nl a3 converges Proof Since we assume that 221 an is convergent Theorem 273 allows us to conclude that lim an 0 By the de nition of convergence to 0 there exists an N 6 N so that if n 2 N then M wan 7 0 lt1 Since we assume that an 2 0 for all n we can multiply this inequality through by an lanl to nd that Ogai anforalanNi Since the series 221 an is assumed to converge it follows from the generalized comparison test that 200 a2 conver es as we nl n g A Alternate proof This result follows from the special case of problem 3 when 12 an and an 2 0 for all n 6 N If we give a proof to problem 3 that doesn t assume this result is true then there is nothing more to do Such a proof is given belowi See alternate proof to problem 3 Find an example that shows that the assumption an 2 0 for all n E N cannot be removed from part a i Proof We rst observe that for any real number I y if y 2 z gt 0 then 2 To see this we assume to the contrary that y 2 z gt 0 and lt Since and are both positive we can square both sides of this last inequality to conclude that y lt I which contradicts our assumption y 2 If Next consider the sequence In 1 El N gt 31 so that TIN lt 5 Then for n2 N we will have that l lxN so We claim that In converges to 0 Given 5 gt 0 choose 1quot n W7 previously shown that the series 201 221 2 divergesi The alternating series test then tells us that the series 221 convergesi However we have D 3 Assume that the series an and 220112 are absolutely convergenti Prove that the series En1 anbn is absolutely convergent Prob em 2 and the fact that A 7B2 Z 0 for any A B 6 R might be useful Proof We rst claim that for any two real numbers A B E R that 1 AB g 5A2 B2 To see this we write 0g AiB2A272ABB2i Adding 2AB to both sides and multiplying by gives us AB 3142 B2 for any A B E R Applying this inequality with A lanl and B lbnl we get that 1 3 lanbnl E a3 123 for all n 6 N The series 221 an is assumed to be absolutely convergent which by de nition means the series 201lanl convergesi Using problem 2 we can conclude that the series 221 a3 convergesi An identical argument then lets us conclude that the series 2113 convergesi Applying Theorem 271 we then conclude that the series 201 a3 12 converges as well Using this with the comparison test Theorem 274 we can conclude that the series 201lanbnl converges as well which by de nition means that the series 201anbn is absolutely convergenti Alternate proof We will use the Cauchy criterion for series Theorem 2 72 rst observe that for any n gt m we lam1l 39 39 39 lanl lbm1l 39 39 39 lbnl lam1l lbm1l 39 39 lbnl lam2l lbmll 4 i i i lanl lbm1l 39 39 W0 2 lam1bm1l lam2bm2l lanbnl 7 where the 2 follows because we have discarded nonnegative terms from our sumi Now since 201an and 21 7 are assumed to be absolutely convergent this means that 221 lanl and 221 lbnl are convergenti Given 5 gt 0 we apply the Cauchy criterion for series to 201lanl and 221 lbnl to nd an N so that if n gt m 2 N then1 lam1l w lt v5 and mm lbnl lt Therefore for n gt m 2 N we can use 4 to conclude that lam1bm1llam2bm2lgtlanbnl S lam1l lanllbm1llbnl lt WW a The Cauchy criterion for series then lets us conclude that 201lanbnl converges or equivalently that 221 anbn is absolutely convergenti 1 Find an N1 for that works for the series 2201 lanl and an N2 that works for the series 2201 lbnl and choose N to be the larger of the two 5 4 The Ratio Test Let an be a sequence with an y 0 for all n 6 N and assume that lim ngtoltgt Tltli an1 an Prove that the series 221 an converges absolutely See Exercise 279 for some hints on how to break this down into smaller steps i Proof Choose an s satisfying 7 lt s lt 1 Letting 5 s 7 7 gt 0 choose an N so that for n 2 N we will have lan1l lt537Ti lanl 77 This inequality implies that a wirltsirforalanN lanl which in turn implies that lan1l lt s lanl for all n 2 Ni We next claim that laNkl 8k laNl for all k 6 Ni We prove this by induction on kl The base case is 1 follows immediately from 5 applied with n Ni We assume then that for some k 6 N that laNkl lt 8k laNlA From 5 applied with n N k we get that laNk1l lt SlaNkl which together with our inductive assumption laNkl lt 8k MM and the fact that s gt 0 implies that 61 laNlA laNk1l lt 8 We conclude that laNkl g 8k laNl for all k 6 N or equivalently that lanl g sniN laNl for all n gt Ni Now since 8 satis es 0 lt s lt l we know that the series 221 as converges for any constant a E R Example 275 In particular ifwe choose a S NlaNl we see that the series 2 3 lale 201lalen N convergesi It then follows from 6 and the Generalized Comparison Test that 00 00 n1lanl converges wh1ch means that Enlan 1s absolutely convergenti D 1 a A 0quot V Math 3207 Homework 3 Solutions Exeri 233 The Squeeze Theorem Assume that and are sequences both of which converge to L Assume further that is a sequence satisfying In S 2n S yn for all n 6 Ni Prove that also converges to L See part b for the wrong answer Proof Given 5 gt 0 choose N1 so that yn 7 L lt 5 and for n 2 N1 and choose N2 so that In 7 L lt 5 for n 2 N2 Choosing N max N17N2 then n 2 N implies that lyn 7Ll lt 5 and lzn7Ll lt a Subtracting L from In lt 2n lt yn gives zn7L 2n7L yn7Ll Thus for n 2 N we have that 2n7L yn7L lyn7Ll lt8 and 2n7LZzn7LZ7 zn7L gt 75 The preceding two inequalities let us conclude that for n 2 N we have that 75 lt 2n 7 L lt 5 which is equivalent to lzn 7 Li lt a Thus 2n converges to L Explain what s wrong with the following proposed sketch of a proof for part a Proof Let L be the limit of zni Then the assumption In 2n used with Theorem 2 34 implies that L Ll Similarly the assumption 2n g yn used with Theorem 2 34 implies that L g L We thus conclude that L L so converges to L as claimed D Explanation The problem is that this proposed proof assumes that the limit of the sequence 2 exists In a correct proof of this fact it is necessary to prove that has a limit since this fact is not included in the hypotheses of the theoreml D 1 2 Use the squeeze theorem to determine the limit of the sequence In Zsinn3 7 2n you may use facts about sinz that you ve learned in other classes Proof We W111 need to use the fact that the function sinz satis es 71 sinz 1 for all z 6 R Thus 1 1 3 1 77 g 7s1nn 7 2n 7 n n n for all n 6 Ni Since and 7 both converge to 0 the squeeze theorem implies that isinm3 7 2n converges to 0 as wel i D 3 Exeri 2 3quot Assume that an is a bounded but not necessarily convergent sequence and that the sequence bu converges to 0 Prove that the sequence anbn converges to zero Proof Since an is assumed to be a bounded sequence there exists an M Z 0 so that an g M for all n 6 Ni Let 5 gt 0 Since we assume that In converges to 0 we can nd an N 6 N so that w meow lt em for n 2 Ni Therefore for n 2 N we will have that wanbn e oi angn M M lt M a We thus conclude that anbn converges to 0 1 4 Exeri 244 Show that the sequence de ned by Vim mm w converges and nd the limit Hint Find a recursive de nition for In ie a formula for In1 in terms of In and try to use Theorem 242 Proof The sequence de ned by the pattern above can be written I1 2 In1 x2Ini Note that each term of the sequence is positive e rst claim that the sequence satis es In lt 2 for all n 6 Ni We prove this by induction For the base case we observe that I1 2 lt 2 Next we assume that for some k that 1k lt 2 and attempt to prove that Ik1 lt 2 Assume to the contrary that 1k lt 2 and that Ik1 Z 2 Then we nd Ikl Z 2 j xZTk Z 2 and since both sides of this inequality are nonnegative we can square the inequality and continue gt 21k 2 4 gt 1k 2 2 This contradicts our assumption that 1k lt 2 Therefore we have shown that 1k lt 2 implies that Ik1 lt 2 and by induction we conclude that In lt 2 for al n 6 We next claim that the sequence In is increasing To see this we write In1 7 2In 7 2In 7 2 In 7 In 7 1 7 In Since we have shown that In lt 2 for all nit follows that 141 gt 1 and thus In1 gt In Indeed if we had 741 1 the above computation would let us conclude that 4 g 1 5 0 12 l gt In 2 2 which contradicts In lt 2 We have thus shown that In is an increasing sequencer Since we have shown that In is increasing and bounded it follows from the Monotone Convergence Theorem Theorem 242 that In convergesi Moreover it is a straightforward consequence of the de nition of convergence that the sequence de ned by yn In1 also converges and has the same limit as In Prove this See Exeri 2i4i2bi We write L lim In limIn1i To nd L we square the equation In1 2In to nd Ii1 2Ini Using Theorem 2i3i3iii we have that limIn12 lim In12 L2 and using Theorem 2i3i3i we have lim 2In 2lim In 2Li Thus taking the limit on both sides of 1 leads to L2 2L which tells us that either L 2 or L 0 However since the sequence is an increasing sequence we know that In 2 I1 2 for all n 6 Ni Therefore limIn Z 2 by Theorem 2i3i4iii so we must have L 2 D Further comments You might wonder Why the rst several paragraphs of the above proof are nec essary Why can t we just take the limit on each side of 1 and argue as in the nal paragraph For some insight into Why it s not ok to just take the limit in 1 Without rst proving the limit exists try this argument on the sequence de ned by 11 1 and zn1 731 16 What happens if you assume the limit exists and take the limit on both sides of the equation zn1 731 16 Does the sequence In convergei D

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