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1 THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each aswer book. Calculators are ot eeded ad are ot permitted i this examiatio. ANSWER 3 QUESTIONS. If you have aswered more tha the required 3 questios i this examiatio, you will oly be give credit for your 3 best aswers. The umbers i the margi idicate approximately how may marks are available for each part of a questio.. a) Give a example of a sequece which is ot bouded above but which does t ted to ifiity. b) State without proof whether the sequece cos()/(2 si()) is coverget, bouded or mootoe. c) Defie what it meas for ( ) to be ull sequece. d) If ( ) coverges to a, ad >, for all, the show that a e) State without proof the possible limits of the sequece (x ) for x > 0. f) Give a example of a sequece that satisfies the coditios of the Ratio Lemma, ad explai your aswer. g) Give a example of a sequece ( ) such that + <, for all, ad such that does ot ted to 0. [3] h) Cosider the recursive sequece + = +, a =. Assumig that the sequece coverges, what is the limit? i) Fid if{ π < x < π }. [3] ta x 4 2 j) Give a example of a Cauchy sequece ( ) such that is ratioal for all, ad such that its limit is ot ratioal. Questio cotiued overleaf

2 Questio cotiued k) Give a example of a set A that has o supremum. Explai why it does ot violate the Completeess Axiom. l) State without proof for which a R the series = a coverges. m) Does there exist a rearragemet (b ) of ( ) = ( cos(π) ) such that = b = π? ) Decide whether the followig is true or false, ad if false give a couter-example. Every coverget series is either absolutely coverget or coditioally coverget. o) Decide whether the followig is true or false, ad if false give a couter-example. Suppose b for all. If = b coverges the = coverges ad = = b. [3] p) Assume that oe kows that 0 e M =0 2! (M+)!, for M. Fid the smallest N such that the error i the approximatio e N =0 that /50. is less! 2. a) Defie what it meas for a sequece ( ) to coverge to a limit a. Give the limits of the followig sequeces: (i) ( ) / ; (ii) ; (iii) ( ) /. Justify your aswers. b) Defie what it meas for a sequece ( ) to evetually have some property. Show that if the limit b of a sequece (b ) satisfies b > 0 the evetually b > b/2. Prove the quotiet rule for sequeces, i.e., if ( ) coverges to a ad (b ) coverges to b, with b > 0 for all ad b > 0, the ( a b ) coverges to a b. [8] c) State the Ratio Lemma for sequeces ( ) for which there exists a costat 0 l < so that + l, for. Show how to prove the Ratio Lemma assumig the Sadwich Theorem. [4] 3. a) Oe of the followig is the Bolzao-Weierstrass theorem: (i) Every mootoic sequece has a coverget subsequece. (ii) Every bouded sequece has a coverget subsequece. (iii) Every bouded above sequece has a coverget subsequece. 2 Questio 3 cotiued overleaf

3 Questio 3 cotiued (iv) Every icreasig sequece has a coverget subsequece. Which oe is the correct claim? Give a couterexample for each of the icorrect claims. b) Cosider the sequece ad the set A = {a, a 2,...}. = ( ), (i) Write explicitly the first six elemets of the sequece. (ii) Are all the elemets of A ratioal? Are all the elemets of A irratioal? (iii) Show that A is bouded, ad fid the ifimum ad the supremum. (iv) Prove that ( ) is Cauchy. c) Suppose that ( ) is Cauchy, ad defie the sequece (b ) by [8] [8] b = a 2. Show that b is Cauchy ad fid its limit. [4] 4. a) Explai what is meat by the partial sums of a series =. State precisely what it meas for the series = to diverge to ifiity. b) State what it meas for a series = to be absolutely coverget. Give a example (without proof) of a series which is coverget but ot absolutely coverget. c) State the versio of the Ratio Test for series that does ot require every term to be positive. d) Determie whether each of the followig series coverges or ot. If you use a covergece test, state the ame of the test ad show that the appropriate coditios are satisfied. (i) π +e +3 = 4 +π (ii) ( ) = (iii) = / ! e) Prove that the harmoic series = diverges to ifiity. Give the full statemets of ay results you use i your proof. [5] [3] [3] [3] 3 CONTINUED

4 MATHEMATICS DEPARTMENT FIRST YEAR UNDERGRADUATE EXAMS Course Title: Aalysis I Model Solutio No: a) ( + ( ) ). b) ot coverget; bouded; ad ot mootoe. c) ( ) is ull sequece if ɛ > 0 N > 0 such that < ɛ wheever > N. d) Assume for a cotradictio that a < the lettig ɛ = a the for N sufficietly 2 large a N < a + ɛ = +a <, givig a cotradictio. 2 e) 0 ad (igore if give as a aswer). f) The sequece ( ) = (2 ), sice + = 2 <. g) The sequece ( ) = ( + ). We have + = < for all. But, it does ot ted to 0. h) Let a be the limit. It satisfies the equatio a = + a. Solvig, we fid a = i) The ifimum is 0. j) = ( + ). Or let d be the digits of 2; the = d k k=0. Those sequeces 0 k coverge i R, so they are Cauchy. k) Eg A = {, 2, 3,...}. It is ot bouded, so the Completeess Axiom does ot apply. l) a <. m) Yes. ) True. o) False. E.g. =, b = ( ). p) N = 4. 4 CONTINUED

5 MATHEMATICS DEPARTMENT FIRST YEAR UNDERGRADUATE EXAMS Course Title: Aalysis I Model Solutio No: 2 a) The sequece ( ) coverges to a limit a if: ɛ > 0 the N > 0 such that a < ɛ for all > N ; (i) 4. We ca write ( ) / = 4( + (3/4) ) / 4 sice (3/4) 0. (ii) 2. We ca write = 4 (/2) (/2) + (/2) 0. 2 sice (/2) 0 ad (iii). We ca write ( ) / / 4/ sice / ad ( / ) 4 ad by use of the Sadwich Theorem. b) We say that a property evetually holds for (b ) if it there exists some N such that the property holds for the shifted sequece (b +N ). We kow that (b ) b ad therefore (bb ) b 2 by the Sum Rule. By defiitio, there exists N such that bb b 2 < b2 whe > N 2. It follows that b2 < bb 2 b 2 ad therefore bb > b2 whe > N 2. Hece, 0 b b = b b bb = b b bb = b b bb < 2 b b b 2 whe > N. Let ε > 0 ad choose N 2 so that > N 2 implies ( ) that b b < ε b2. The > max{n 2, N 2 } implies b < ε. Therefore b b. b ( ) ( ( ) a Fially, sice a b = (a) b ), we see that b a by the Product Rule. b c) Ratio Lemma Let ( ) be a sequece of positive umbers. Suppose 0 < l < ad + l for all. The ( ) 0. There is a simple proof by iductio. Whe = 0 we have that a la 0. For the iductive step, assume that l a 0. The sice + l it follows that + l l + a 0. This completes the proof by iductio. Now (l ) is ull sequece ad so a 0 l is ull sequece ad sice 0 l a 0 the result follows by the Sadwich theorem. 5 CONTINUED

6 MATHEMATICS DEPARTMENT FIRST YEAR UNDERGRADUATE EXAMS Course Title: Aalysis I Model Solutio No: 3 a) The secod claim is correct. Here are couterexamples for the other claims: ( ) = (). ( ) = ( ). ( ) = (). (Note: 2 marks should allocated for each of the 4 parts.) b) (i) ( ) =, 2/2, /3, 2/4, /5, 2/6,... (ii) The odd elemets are ratioal, but the eve elemets are irratioal. The A is either. (iii) We have 0. Also, =. The A is bouded. a = is the supremum sice < for all >. The ifimum is 0 sice 0. (iv) Let m > be arbitrary atural umbers. We have a m a m + 2( m + ) 2 2. For every ε > 0 we ca choose N = 4/ε, ad we ideed have for all m, > N. a m ε < ε c) Sice ( ) is Cauchy, for every ε > 0, there exists N such that a m < 2 ε for all m, > N. The b m b a 2m a m + a 2 < ε, for all m, > N, so (b ) is Cauchy. Clearly, (b ) 0. 6 CONTINUED

7 MATHEMATICS DEPARTMENT FIRST YEAR UNDERGRADUATE EXAMS Course Title: Aalysis I Model Solutio No: 4 a) Cosider the series =. The partial sums (s ), are give by s = i= a i. We say = diverges to ifiity if (s ) teds to ifiity. b) The series = is absolutely coverget if = is coverget. E.g. = ( ). c) Suppose 0 for all ad + l. The = coverges absolutely (ad hece coverges) if 0 l < ad diverges if l >. d) i) We have 0 π + e + 3 π ). 3( 4 + π 4 Sice π/4 <, the series series test. Hece, by the compariso test, = = b, with b = ( π 4 ) coverges by the geometric π +e +3 coverges. 4 +π ii) Settig = /3 /2, we wat to determie the covergece of = ( ). Sice /3 for all, ad ( /3 ) diverges to ifiity, ( ) is decreasig ad ull ad hece = ( ) coverges by the alteratig series test. iii) Sice = 2! + = > 0 for all, we may apply the ratio test. We have ( + )+ 2! 2 + ( + )! Hece the series = diverges. = + ( ) e 2 >. e) Let s = k=. The s k 2 = s + 2 k=+ s k + = s 2 +. By defiitio 2 s 2 = +. Suppose that s k. The s k 2 2 s k k + = + k+. k Therefore s 2 + for all by iductio. It follows that the sequece (s 2 2 ) teds to ifiity. Sice s 2 is a subsequece of (s ), it follows that (s ) caot coverge. Hece = or diverges, by defiitio. We require the itegral test for divergece of a series: Suppose the fuctio f(x) is o-egative ad decreasig for x. The = f() diverges if the icreasig sequece ( f(x)dx) is ubouded. 7 CONTINUED

8 Let f(x) =, which is o-egative ad decreasig whe x. Observe that x f(x)dx = dx = log x = log. Therefore x = diverges to ifiity. 8 END

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