Questions with Solutions in Final Exam - Fundamental Analysis | MATH 424, Exams of Mathematics

Download Questions with Solutions in Final Exam - Fundamental Analysis | MATH 424 and more Exams Mathematics in PDF only on Docsity! Mathematics 424/574 Final Exam Name: Answers December 12, 2007 Instructions: This is a closed book exam, no notes or calculators allowed. Justify all of your answers, unless the problem says otherwise. Unless otherwise specified, you may refer to and use any result from the book, homework, or in-class problems. This is a timed exam, so you may use abbreviations and symbols (such as “∀”): as long as I can make sense of what you write without struggling too much, it’s okay. Notation: R is the set of real numbers. 1. (10 points) Let {an} be a sequence of real numbers, let {pk} be a sequence of positive real numbers so that the partial sums ∑nk=1 pk → ∞, and let cn = p1a1 + p2a2 + · · ·+ pnan p1 + p2 + · · ·+ pn . Prove that if the sequence {an} converges to some number a, then {cn} converges to a also. Solution: Fix ε > 0, and consider cn−a: cn−a = p1a1 + p2a2 + · · ·+ pnan p1 + p2 + · · ·+ pn −a = p1(a1−a)+ · · ·+ pn(an−a) p1 + · · ·+ pn . There is an integer N so that if n > N, then |an − a| < ε/2. There is also an integer M so that if n > M, then ∣∣∣∣ p1(a1−a)+ · · ·+ pN(an−a)p1 + · · ·+ pn ∣∣∣∣ < ε/2 : the numerator here is fixed, and the denominator gets larger as n → ∞, so the fraction can be made arbitrarily close to zero. Therefore for any n > max(N,M), we have cn−a = p1(a1−a)+ · · ·+ pn(an−a) p1 + · · ·+ pn = p1(a1−a)+ · · ·+ pN(aN −a)+ pN+1(aN+1−a)+ . . . pn(an−a) p1 + · · ·+ pn = p1(a1−a)+ · · ·+ pN(aN −a) p1 + · · ·+ pn = p1(a1−a)+ · · ·+ pN(aN −a)+ pN+1(aN+1−a)+ . . . pn(an−a) p1 + · · ·+ pn = p1(a1−a)+ · · ·+ pN(aN −a) p1 + · · ·+ pn + pN+1(aN+1−a)+ · · ·+ pn(an−a) p1 + · · ·+ pn . Now take absolute values: the first fraction is less than ε/2 since n > M. For the second fraction, Mathematics 424/574 Palmieri we have ∣∣∣∣ pN+1(aN+1−a)+ · · ·+ pn(an−a)p1 + · · ·+ pn ∣∣∣∣≤ pN+1|aN+1−a|+ · · ·+ pn|an−a|p1 + · · ·+ pn < pN+1ε/2+ · · ·+ pnε/2 p1 + · · ·+ pn < p1ε/2+ · · ·+ pnε/2 p1 + · · ·+ pn = ε/2. Therefore for n > max(M,N), we have |cn−a|< ε . Since ε was arbitrary, we conclude that cn → a. 2. Investigate the behavior (convergence or divergence) of ∑an if (a) (5 points) an = nn3+1 Solution: We rewrite an as 1n2+ 1n . Since n2 + 1n > n 2, we have 1 n2+ 1n < 1n2 . Everything here is positive, so we have |an| < 1n2 . The series ∑ 1 n2 converges, so by the comparison test, so does ∑an. (b) (5 points) an =  (−1)n/2 23n+4 if n is even, 1 32n if n is odd. Solution: There are various ways to do this. For example, we can use the root test. We have n √ |an| → { 1 8 if n is even, 1 9 if n is odd. (For the case when n is even, the sign is of course irrelevant. The nth root of the denominator 23n+4 = 23n24 is n √ 23n n √ 24 = 23 n √ 24. As n goes to infinity, n √ 24 = 24/n goes to 1, so the limit is 8.) Therefore limsup n→∞ n √ |an| ≤ 1 8 , so the root test tells us that ∑an converges. Alternatively, we can break this into two sums, the even terms and the odd terms. If we can show that each of these converges, then the whole thing will. Each of these is a geometric series with ratio less than one, so they converge.