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MATH1113 Calculus SlidesSequencesThe numbera1is called the first term,a2is the second term, and ingeneralanis the n-th term. We will deal exclusively with infinitesequences and so each termanwill have a successoran-1.Remark:1.Since we will be dealing exclusively with infinite sequences we willsimply call them sequences.2.Notice that for every positive integernthere is a correspondingnumberanand so a sequence can be defined as a function whosedomain isN, the set of positive integers. But we usually write aninstead of the function notationf(n) for the value of the func tionat the numbern.3.The sequence{a1,a2,a3,· · ·,}is also denoted by{an}or{an}n=1.5,-425,5125,-6625,73125,· · ·.assuming that the pattern of the first few terms continues.Notice that the numerators of these fractions start with 3 and increase by1 whenever we go to the next term. In general, the nth term will havenumeratorn+ 2. The denominators are the powers of 5, so the n-th termhas denominator 5n. The signs of the terms are alternately positive andnegative, so we need to multiply by a power of-1. Therefore,an= (-1)n-1n+ 25n.A sequence{an}has thelimitL, writtenlimn→∞an=L, if we can makeanas close toLas we want by takingnsufficiently large. Iflimn→∞anexistswe say the sequenceconverges. Otherwise we say the sequencediverges.Remark:Note the similarity between the definition oflimn→∞anandlimx→∞f(x). In fact, given a functionfwhose domain includes everynon-negative integer we get a sequence by settingan=f(n), and iflimx→∞f(x) =Lthenlimn→∞an=Las well.DefinitionA sequence{an}isincreasingifan+1>anfor alln. Similarly, thesequence isdecreasingifan+1<anfor alln.Theorem (Monotonic sequence theorem)If{an}is increasing (or non-decreasing) and bounded above then{an}
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(v)Find a formula for the general term an of the sequence35,-425,5125,-6625,73125,· · ·.assuming that the pattern of the first few terms continues.Notice that the numerators of these fractions start with 3 and increase by1 whenever we go to the next term. In general, the nth term will havenumeratorn+ 2. The denominators are the powers of 5, so the n-th termhas denominator 5n. The signs of the terms are alternately positive andnegative, so we need to multiply by a power of-1. Therefore,an= (-1)n-1n+ 25n.A sequence{an}has thelimitL, writtenlimn→∞an=L, if we can makeanas close toLas we want by takingnsufficiently large. Iflimn→∞anexistswe say the sequenceconverges. Otherwise we say the sequencediverges.Remark:Note the similarity between the definition oflimn→∞anandlimx→∞f(x). In fact, given a functionfwhose domain includes everynon-negative integer we get a sequence by settingan=f(n), and iflimx→∞f(x) =Lthenlimn→∞an=Las well.DefinitionA sequence{an}isincreasingifan+1>anfor alln. Similarly, thesequence isdecreasingifan+1<anfor alln.Theorem (Monotonic sequence theorem)If{an}is increasing (or non-decreasing) and bounded above then{an}
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