MATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1

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1 MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The mai idea of this problem is to show that {s } is mootoic ad bouded. Claim. For every positive iteger, we have s < 2. Proof. We proceed by iductio. Base Step. For =, we have s = 2 < 2. Iductive Step. Suppose k is a positive iteger such that s k < 2. The, s k+ = 2 + s k < < 2. Hece, by Mathematical Iductio, we have s < 2 for all. Claim 2. The sequece {s } is mootoically icreasig. Proof. To show that s < s 2 < s 3 <... is equivalet to provig that > 0, s < s +. Agai, we proceed by iductio. Base Step. s = 2 ad s 2 = > 2 = s. Iductive Step. Suppose k is a positive iteger such that s k < s k+. The, sk < s k+ 2 + s k < 2 + s k+ 2 + s k < 2 + s k+ which meas s k+ < s k+2. This completes he iductio. Thus, by Mathematical Iductio, we have s < s + for all. Claim 3. For every positive iteger, we have s [ 2, 2]. Proof. From Claim 2, we kow that {s } is mootoically icreasig, ad so, s s = 2. Claim shows that, s < 2. Hece,, s [ 2, 2]. Claims 2 ad 3 imply that {s } is mootoic ad bouded. Theorem 3.4, we deduce that {s } coverges. Therefore, from Date: Sprig 206. Theorem 3.4 (Rudi). Suppose a sequece of real umbers {s} is mootoic. The {s } coverges if ad oly if it is bouded.

2 2 CA PRO JIRADILOK Problem 2: Rudi, Chapter 3, Problem 6 (a) - (c). Problem. Ivestigate the behavior (covergece or divergece) of a if (a) a = + ; + ; (b) a = (c) a = ( ). Proof. For parts (a) ad (b), the mai tool we are goig to use is Theorem To show that a diverges, we shall fid a sequece {d } for which d diverges ad a d 0. O the other had, to show that a coverges, we shall fid a sequece {c } for which c coverges ad a c. (a) We claim that a diverges. Solutio. Note that a = + = + + > 2,. + We kow that the series = 2 + = diverges. Therefore, 4 a diverges. Solutio 2. Alteratively, we ca evaluate N = a explicitly. Note that N a = N + N + = N +. = Therefore, N N = a = N N + = +. (b) We claim that a coverges. Note that + a = = ( + + ) < =,. 3/2 We kow that = coverges. Therefore, a 3/2 coverges. For part (c), we use the root test (Theorem ). (c) We claim that a coverges. Set α := sup a. We have α = sup = 0 by Theorem 3.20(c) 4. Therefore, a coverges. 2 Theorem 3.25 (Rudi). (a) If a c for N 0, where N 0 is some fixed iteger, ad if c coverges, the a coverges. (b) If a d 0 for N 0, ad if d diverges, the a diverges. 3 Theorem 3.33 (Rudi). (Root Test) Give a, put α := sup a. The, (a) if α <, a coverges; (b) if α >, a diverges; (c) if α =, the test gives o iformatio. 4 Theorem 3.20(c) (Rudi). =.

3 MATH 2: HOMEWORK 6 SOLUTIONS 3 Problem 3: Rudi, Chapter 3, Problem 7. Problem. Prove that the covergece of a implies the covergece of a, if a 0. Proof. First, we show the followig hady lemma. Lemma 4. (AM-GM Iequality) For o-egative real umbers x, y 0, we have x + y xy. 2 Proof. Observe that ( x y) 2 0 for all x, y 0. Therefore, x + y 2 xy 0. This gives xy x+y 2. Suppose that a coverges. For each, usig Lemma 4 with (x, y) = ( a, ), 2 we obtai a = a a Sice a 2 ad both coverge, we have ( a 2 + ) coverges. Therefore, 2 a coverges. If you like this problem, you ca try the followig problem for practice. Problem. Is there a sequece {a } of positive real umbers such that both ad = a coverge? = a Problem 4: (2) o the problem set Problem. Suppose that {a } is a sequece of poits i a metric space (X, d), such that d(a, a + ) < c 2 d(a, a ) for some c (0, ). Show that {a } is a Cauchy sequece. (Fu Fact: This little problem is a key step i the proof of the implicit fuctio theorem). Proof. To show that {a } is a Cauchy sequece, we have to prove that sup M, 2 M Let r := d(a, a 2 ). First, we show the followig. d(a, a 2 ) = 0. Claim 5. For all 0, we have d(a +, a +2 ) c 2 r. Proof. We proceed by iductio. Base Step. d(a, a 2 ) = r c 0 r is true by defiitio of r. Iductive Step. Suppose k is a o-egative iteger such that d(a k+, a k+2 ) c 2k r. From the iequality give i the problem, we have d(a k+2, a k+3 ) < c 2 d(a k+, a k+2 ) c 2k+2 r. This fiishes the iductive step. Thus, by Mathematical Iductio, we have 0, d(a +, a +2 ) c 2 r.

4 4 CA PRO JIRADILOK Claim 6. For ay positive iteger M ad for ay, 2 M, we have d(a, a 2 ) c2m 2 c 2 r. Proof. If = 2, the d(a, a 2 ) = 0 ad the claim is trivial. So let s assume 2. Without loss of geerality, suppose 2 >. By Triagle Iequality 5, we have d(a, a 2 ) d(a, a +) + d(a +, a +2) + + d(a 2, a 2 ). Usig Claim 5, we obtai as desired. d(a, a 2 ) c 2( ) r + c 2 r + + c 2(2 2) r = c 2( ) r ( ) + c c 2( 2 ) = c 2( ) r c2( 2 ) c 2 c2( ) c 2 r c2m 2 c 2 r Claim 6 shows that sup, 2 M d(a, a 2 ) c2m 2 r. Therefore, c 2 sup M, 2 M d(a, a 2 ) M We coclude that {a } is a Cauchy sequece. r c 2 ( c 2 ) (c2 ) M = 0. Problem 5: (3) o the problem set Problem. Let {a } ad {b } be two Cauchy sequeces i (X, d). Show that d(a, b ) is a Cauchy sequece i R. (Fu Fact: This little problem is used i oe method for costructig the real umbers). Proof. Suppose {a } ad {b } are two Cauchy sequeces i a metric space (X, d). Fix ɛ > 0. We wat to show that there exists N such that (), 2 N, d(a, b ) d(a 2, b 2 ) < ɛ. Because {a } is Cauchy, there exists M such that for all r, r M, we have d(a r, a r ) < ɛ/2. Similarly, because {b } is Cauchy, there exists M 2 such that for all r, r M 2, we have d(b r, b r ) < ɛ/2. Choose N := max(m, M 2 ). We claim that for this N, the equatio () is satisfied. For ay, 2 N, we have, by Triagle Iequality, d(a, b ) d(a 2, b 2 ) d(a, a 2 ) + d(a 2, b 2 ) + d(b 2, b ) d(a 2, b 2 ) = d(a, a 2 ) + d(b 2, b ) < ɛ 2 + ɛ 2 = ɛ. 5 Recall that Triagle Iequality i a metric space (X, d) says that for ay three poits x, x, x X, we have d(x, x ) d(x, x ) + d(x, x ). By iteratig, we deduce that for ay 3 poits x, x 2,..., x X, we have d(x, x ) d(x, x 2) + d(x 2, x 3) + d(x 3, x 4) + + d(x, x ).

5 MATH 2: HOMEWORK 6 SOLUTIONS 5 O the other had, we ca switch the roles 6 of ad 2 to obtai d(a 2, b 2 ) d(a, b ) < ɛ. Thus from the two iequalities above, we have d(a, b ) d(a 2, b 2 ) < ɛ for all, 2 N. Therefore, the equatio () is satisfied, ad so {d(a, b )} is a Cauchy sequece i R. Problem 6: (4) o the problem set Problem. Suppose that {a } is a Cauchy sequece i (X, d), ad suppose that some subsequece {a k } coverges to a poit p X. Show that {a } coverges to p. Proof. We wat to show that, for ay give ɛ > 0, there exists N such that N, d(a, p) < ɛ. Because the subsequece {a k } coverges to p, there exists M such that for every k M, d(a k, p) < ɛ/2. Because {a } is Cauchy, there exists M such that for every r, s M, d(a r, a s ) < ɛ/2. Take N := max( M, M). Also, cosider a idex κ for which κ > N. Note that we have κ > N M, so κ > M. Therefore, d(a κ, p) < ɛ/2. For ay N, we the have, by Triagle Iequality, d(a, p) d(a, a κ ) + d(a κ, p) < ɛ 2 + ɛ 2 = ɛ. Thus, {a } coverges to p. Good luck o the midterm! 6 This switch of roles betwee ad 2 is importat. Some studets correctly obtaied the iequality d(a, b ) d(a 2, b 2 ) < ɛ, ad immediately cocluded from it that d(a, b ) d(a 2, b 2 ) < ɛ. Although, for this problem, we see that it is true, oe caot i geeral take the absolute value both sides of a iequality ad expect the resultig iequality to still hold.

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