Istructor: Marius Ioescu 1. Let a =. (5pts) (a) Prove that for every ε > 0 there is N 1 such that a +1 a < ε if N. Solutio: Let ε > 0. The a +1 a < ε is equivalet with + 1 < ε. Simplifyig, this iequality is equivalet with 1 + 1 + < ε. Note that the previous expressio is positive so we do ot eed the absolute value bars. Sice 1 < 1 + 1 + 2 if we choose N = (1/(2ε)) 2, it follows that a +1 a < ε if N. (5pts) Is a a Cauchy sequece? Why? Solutio: a is ot a Cauchy sequece. To prove this, suppose by cotradictio that a is Cauchy. The a must be coverget by Theorem 6.4. The sequece a is, however, diverget. Thus it ca ot be Cauchy.
(7pts) 2. (a) Show that if a coverges absolutely, the so does a 2. Is this true without the hypothesis of absolute covergece (prove or give a couterexample)? Solutio: If the series a coverges absolutely it coverges. The lim a = 0. The there is N 1 such that a < 1 for all N. It follows that a 2 a for all N. The compariso test for positive series implies that =N a2 coverges. From the tail theorem we coclude that a 2 coverges. The coclusio fails without the hypothesis of absolute covergece. Cosider ( 1). This series is coverget by the Cauchy test. The series 1, however, diverges (it is a p-series with p = 1). (3pts) If a coverges ad a 0, does it follow that a coverges? Prove or give a couterexample. Solutio: If a coverges ad a 0, it does ot follow that a coverges. For example, 1 coverges (p-series with p = 2) but 1 2 diverges. Page 2 of 8 Please go o to the ext page...
(20pts) 3. Fid the radius of covergece of =0 (si )x (with proof). Solutio: The mai poit about this problem is that we ca ot use the root or the ratio test to determie the radius of covergece of this series (because the sequeces (si()) (1/) ad si(+1) si() are diverget). Notice that si()x x for all 1. If x < 1 we kow that x coverges absolutely. The compariso theorem implies that si()x coverges absolutely. For x = 1, si() diverges sice { si() } does ot coverge to 0. Thus R = by Theorem-Defiitio 8.1. Page 3 of 8 Please go o to the ext page...
(10pts) 4. Determie if the series is coverget. If yes, fid it s value. l + 2 =1 Solutio: Sice l +2 = l() l(+2) we see that the series is a telescopig series. We compute the th partial sum as follows: The s = ( ) l(k) l(k + 2) = l(1) + l(2) l( + 1) l( + 2). k=1 Thus the series is diverget. lim s ( ) = l(2) lim l( + 1) + l( + 2) =. Page 4 of 8 Please go o to the ext page...
(5pts) 5. Let b be a decreasig sequece with lim b = 0. (a) Prove that =1 (b b +1 ) coverges. Solutio: We otice this is a telescopig series. The partial sum equals s = (b 1 b 2 ) + (b 2 b 3 ) + + (b b +1 ) = b 1 b +1. Sice lim b = 0 it follows that the sequece of partial sums is coverget ad equals b 1. Thus (b b +1 ) coverges. (10pts) Let a be a bouded sequece. Prove that =1 a (b b +1 ) coverges. Solutio: By hypothesis, there exists M > 0 such that a M for all 1. Sice b b +1 0 we have that a (b b +1 ) M(b b +1 ). By the liearity theorem M(b b +1 ) coverges. By the compariso theorem, a (b b +1 ) coverges. Thus =1 a (b b +1 ) coverges. Page 5 of 8 Please go o to the ext page...
(10pts) 6. Prove the followig two statemets: (a) Every real umber is a cluster poit of some sequece of ratioal umbers. Solutio: Let r be a real umber. For ay 1, Theorem 2.5 o page 25 implies that there is a ratioal umber a such that r < a < r + 1. By the squeeze theorem, lim a = r. Theorem 6.2 implies that r is a cluster poit for the sequece a. (5pts) Every real umber is a cluster poit of some sequece of irratioal umbers. Solutio: The proof is similar with the previous part. If r is ay real umber, Theorem 2.5 implies that there is a irratioal umber a with r < a < r + 1 for all 1. Thus r = lim a ad r is a cluster poit of a sequece of irratioal umbers. Page 6 of 8 Please go o to the ext page...
(3pts) 7. Give examples for the followig or explai why o example exists. (a) A series that has bouded partial sums but does ot coverge. Solutio: Cosider the sequece a = ( 1) for all 1. The s 2k+1 = 1 ad s 2k = 0 for all k 1. Thus { s } is a bouded sequece which diverges. (4pts) A sequece which has a ifiite umber of cluster poits. Solutio: Example 6.2A a) i the textbook says that for the sequece 1; 1, 2; 1, 2, 3;... every iteger is a cluster poit. (3pts) (c) A power series whose radius of covergece is 2. Solutio: Cosider the power series x 2. We compute the radius of covergece usig the ratio test: lim a +1 a = x 2. Thus the series is absolutely coverget if x < 2 ad diverget for x > 2. Thus R = 2. Page 7 of 8 Please go o to the ext page...
(10pts) 8. Prove, usig the defiitio, that if { a } ad { b } are Cauchy sequeces the { a b } is Cauchy. Solutio: We proved i class that a Cauchy sequece is bouded (see also part (A) of the proof of Theorem 6.4). Thus there are M 1 > 0 ad M 2 > 0 such that a M 1 ad b M 2 for all 1. Let M = max{m 1, M 2 }. Thus a M ad b M for all 1. Let ε > 0. Sice the sequece a is Cauchy there exists N 1 1 such that a a m < 2M for all m, N 1. Similarly there exists N 2 1 such that b b m < for all m, N 2. Let N = max{n 1, N 2 }. If, m N we have ε 2M a b a m b m = a b a b m + a b m a m b m Sice ε was arbitrary, {a b } is a Cauchy sequece. ε a b a b m + a b m a m b m = a b b m + b m a a m ε < M 2M + M ε 2M = ε. Page 8 of 8 Ed of exam.