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## SEQUENCES & SERIES

by: Joanne Bergnaum

29

0

6

# SEQUENCES & SERIES MATH 3100

Joanne Bergnaum
UGA
GPA 3.56

Clark

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COURSE
PROF.
Clark
TYPE
Class Notes
PAGES
6
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 6 page Class Notes was uploaded by Joanne Bergnaum on Saturday September 12, 2015. The Class Notes belongs to MATH 3100 at University of Georgia taught by Clark in Fall. Since its upload, it has received 29 views. For similar materials see /class/202075/math-3100-university-of-georgia in Mathematics (M) at University of Georgia.

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Date Created: 09/12/15
PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM General Comments These are practice problems You should regard each indi vidual problem as being a plausible exam problemi Certainly there are too many problems here to make one hourlong exam roughly twice as many The exam will be closedbook and calculators will not be permitted 1 would urge you not to use calculators or software when doing the practice problems 7 that would not be such good practice When in doubt you should justify your answers by correct logical reasoning brief explanations are often suf cient 1 Let ab E R a Suppose a and b are both positive Show that a lt b ltgt a2 lt 122 b Show that a lt b ltgt a3 lt 123 c Suggest a generalization of the rst two parts to nth powers for an arbitrary positive integer n You need not prove it First Solution a Suppose ab gt 0 We want to show that a 7 b lt 0 ltgt a2 7 b2 lt 0 In general given nonzero numbers A and B to show that A and B have the same sign it is enough to show that g is positive As a2 7 b2 a 7 b a b we have a2 7 b2 a 7 b which is indeed positive since a and b are both positive b Now suppose a and b are arbitrary but both zero if a b 0 neither a 7 I nor a3 7 b3 is negative As above we have a37b3 7 a7ba2abb2 b 7 a7 ab a2abb2 so it s enough to show that a2 ab 2 gt 0 This is clear if a and b have the same sign all the terms are nonnegative and at least one is positive If a and b have opposite sign then a WW 7 w 7 W le 2 W 7 2w W 7 M 7 W 2 o and is strictly positive unless I 7a in which case a2 ab b2 a2 gt 0 c The desired generalization is that if I and n is even then for all ab gt 0 a lt b ltgt a lt 12 whereas ifn is odd then for all ab E R a lt b ltgt a lt bni Remark To prove this by the current factorization method involves the identity an 7 b a 7 ban 1 anTQb i i i abn 2 bnili When a and b are both positive it is clear that a 1 an Qb i i i abn Q b 1 gt 0 Pete L Clark 2011 2 PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM and this is enough to show the result when n is even However when n is odd we need to establish the above inequality for all ab which are not both zero and off the top of my head this does not seem so straightforward It is worth pointing out however that the result is actually equivalent to this inequality so the alternate solution below gives an indirect proof of it Second Solution We will solve all three parts simultaneously For n E N con sider the function f R 7gt R given by z l gt I The derivative is nzn l Suppose rst that n is odd then is positive for all 1 except that it is zero at z 0 Thus f is strictly increasing on 7000 and on 0 Moreover is negative for negative I and positive for positive I so indeed f is strictly increasing on 70000 Thus altb gt anfaltfbbn and by 7 bnfltbgtfltagta z ie a lt b ltgt fa lt Now suppose that n is even so f is positive for all z gt 0 and thus f is increasing on 000 Thus for a b gt 0 altb gt a faltfbbn and b a gt b fb fa a so a lt b ltgt fa lt 2 Find all real numbers I satisfying l2zl lz 7 3 5 Solution As usual we need to consider cases depending upon whether each of the quantities appearing inside an absolute value are positive or negative so here four cases in all Case 1 21 Z 0 and z 7 3 Z 0 These inequalities are equivalent to z 2 3 When they hold we have l2zllz73l 2zz733z735 so I But lt 3 so this solution must be thrown out Case 2 21 lt 0 and z 7 3 Z 0 These inequalities are inconsistent Case 3 21 Z 0 and z 7 3 lt 0 The inequalities are equivalent to z 6 03 When they hold we have l2zllz73l 2z37z z35 so I 2 This solution does lie in 03 so it is a solution of the equation Case 3 21 lt 0 and z 7 3 lt 0 These inequalities are equivalent to z lt 0 When they hold we have l2zllz73l 72z37z 73z35 so I This solution does lie in 700 0 so it is a solution of the equation Thus the two solutions are I T 2 We de ne a real sequence an il to be eventually positive if there exists N E N PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM 3 such that for all n 2 N an gt 0 3 a Show that an n3 5n2 7 lln 7 43 is eventually positivei b Show that an e 7 105n100 is eventually positive Solution In each case we in fact have limn 00 an 00 which certainly im plies that the sequence is eventually positivei As for part a this is routine even in calculus class As for part b one method is to factor out the term of largest growth so here 6 to get 105n100 an 6 lt1 7 7 i Repeated application of L Hopital s Rule shows that 105 100 um L n7gtoltgt e an7gtoo170ooi 4 Let an and 127 be two sequences a Show that if both sequences are eventually positive so is an 127 and anbn Solution There exist positive integers N1 and N2 such that n 2 N1 gt an gt 0 and n 2 N2 gt In gt 0 Taking N maXN1N2 for all n 2 N an and 12 are both positive so an 127 and anbn are positive Thus these latter two sequences are eventually positivei b Give an example where neither sequence is eventually positive but anbn is eventually positivei Solutionzz We may take for instance an bn 71 for all n Remark In the Review Session we discussed the analogous question Are there sequences an bn neither of which is eventually positive but for which an In is eventually positive It turns out that the answer is yes but counterexamples are slightly tricky to construct For instance we could take an 271271Hi and 12 712712Hi so that anbn1111ui 5 Let an be a sequence converging to a real number Li a Show that if L gt 0 then an is eventually positivei 4 PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM Solution In the de nition of limit take 6 then there exists N E N such that for all n 2 N we have law 7 Ll lt ie 5 lt lt a 2 a 2 Thus for all suf ciently large n an gt g gt 0 so the sequence is eventually positive b Show that if L lt 0 then an is not eventually positivei Solution Arguing as above we get 3L lt lt L i a it 2 n 2 Thus for all suf ciently large n an lt I lt 0 so the sequence is eventually negative hence certainly not eventually positive c Exhibit a sequence which converges to 0 and is eventually positive Solution We may take an d Exhibit a sequence which converges to 0 and is not eventually positivei Solution We may take an 6 In each part either exhibit a sequence or in the last three parts a pair of sequences satisfying the given conditions or prove that no such sequence or pair of sequences exists a A bounded sequence which does not convergei Solution We may take an 71 It has subsequences converging to l and also to fl so it cannot convergei b A convergent sequence which is not boundedi Solution As was proved in the text and in class every convergent sequence is bounded So there are no such sequencesi c A positive sequence an which is convergent and for which I is not convergenti Solution We may take an d A sequence an with an 2 0 for all n such that limn 00 an 73 Solution There is no such sequence In the de ninition of limit take any 6 lt 3 Then all suf ciently large terms of the sequence must be negative contradicting an 2 0 for all n e Sequences an and 127 which both converge such that an 127 divergesi Solution There is no such sequencer As was proved in class and in the text PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM 5 ifanHLand bnaMso anbn4gtLMl f Sequences an and Inn exactly one of which converges such that an bn converges Solution There is no such sequence Suppose without loss of generality that an converges and b7 diverges Put on an 12 Suppose that on convergesl Then 12 on 7 an is a difference of two convergent sequences hence convergentl g Sequences an and 127 both of which diverge such that an bn convergesl Solution We may take an n In in In fact we may take for an any divergent sequence and take 12 ianl For the next two problems we x a real number a Z 1 7 Consider the function 000 a R de ned by z o gt z Show that f has a global minimum at z a and thus 2 2a for all z gt 0 Solution It s calculus as usual 0 2 f I 1 i 17 The derivative is 0 precisely when I a note that z 7a is not in the domain of Moreover as I approaches zero from the right approaches 700 so f is negative on 0a whereas lirngcn00 f z 1 so f is positive on a Thus f has a global minimum at z a so for any I 6 000 mo 2 fa 2a 8 Let a Z 5 be a real number We de ne a real sequence as follows a1 a and VnZ 1 an1 ltanigtl an a Show that an gt 0 for all n Solution a1 a 2 W gt 0 Moreover the recursive equation for an1 shows that if an gt 0 so is an1 so an easy induction argument gives an gt 0 for al n b Show that a3 2 a for all n Suggestion go by induction on n The result of problem 7 may come in handy Solution By induction on n n l a a2 Z a by hypothesisl Now let n E N and assume that a3 2 at Since 2 a2 2 a2 an2agg angz72a 4 a 4 2 7 anlia7 5 PRACTICE PROBLEMS FOR FIRST MATH 3100 MIDTERM 2 Applying the result of Problem 7 With 1 a we nd a3 37 2 2a so ai 2a 2a72a70 4 i 4 7 We are nished by induction c Show that an is decreasing Solution For n E N l a a an a 7 a3 anTlian7 ltanangtian72ani 2 7 Zuni By part a the denominator is positive and by part b a 7 a3 g 0 so an1 an S 0 d Show that the sequence converges it converges to r Put L lian00 ani Then taking limits in the equation 1 a an1 E an din L L gives mp 2 Clearing denominators gives 2L2 L2 a so L i i Since every term of the sequence is positive the sequence certainly cannot converge to 7 clfi Problem 6d abovel so if it converges it must con verge to 1 e Extra Credit Use a fundamental result from ll6 to deduce lian00 an r Solution To sum up what we ve shown so far an is decreasing nonnegative 7 hence bounded below by 0 and if it converges to anything it converges to r But by the Monotone Sequence Lemma the sequence is convergent so indeed lian00 an 1 Remark This gives a 39 quoti 39 of the n 1 39 q algorithmi 9 Let an be an increasing unbounded sequencer Show that lian00 an 00 Solution Let M E El Since an is unbounded there exists N E N with aN gt Mi Since the sequence is increasing for all n 2 N we have an 2 aN gt Mi Thus the sequence diverges to in nity 10 Let an be a sequence converging to a real number L and for n E N put 12 71 ani a lfL 0 show that In 7gt 0

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