Sequences. We know that the functions can be defined on any subsets of R. As the set of positive integers

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1 Sequences We know that the functions can be defined on any subsets of R. As the set of positive integers Z + is a subset of R, we can define a function on it in the following manner. f: Z + R f(n) = a n The range of this function is the infinite set {a 1, a, a 3, }. The set may not necessarily contain infinitely many distinct numbers because some of a i s could be equal. That set {a 1, a, a 3, } is called a sequence. We call a r the r th term of the sequence. For our convenience, we shall denote the whole sequence {a 1, a, } by {a n }. Example 1 Consider the sequence {a n } defined as a n = 1 n. Here, a 1 = 1 a = 1 = 0.5 a 3 = 1 3 = a = 1 = 0.5 One can see that as n gets bigger a n is getting closer to 0. For example, a 15 = a 0 = Because of the behaviour we have seen with the sequence {a n }, we suspect that by taking n large enough, we can make a n as close to 0 as we wish. Because a 0 is within 0.05 units of 0 and a 1 is within units of 0, we can expect that a will be even closer. These observations prepare us for our first definition. Definition 1 Let {a n } be a sequence. A number L is said to be the limit of {a n } if ε > 0, n 0 Z + such that n > n 0 a n L < ε When this happens, we write lim a n = L Now, let us consider the sequence in example 1. Let ε > 0.Then 1 ε R. By Archimedean property, n 0 Z + such that n 0 > 1 ε n > n 0 n > 1 ε 1 n < ε a n 0 = 1 0 < ε n lim a n = 0 1

2 Definition A sequence {a n } is said to be convergent if L R such that lim a n = L. Because we don t consider as a real number, this L is understood to be finite. The sequence in the example 1 is convergent. Not all the sequences are convergent. When a sequence is not convergent, it is said to be divergent. Example a n = n Assume {a n } is convergent. Then L R such that lim a n = L Because 1 > 0, n 0 Z + such that n L < 1 n > n 0 (n 0 + 1) L < 1 and (n 0 + 3) L < 1 = (n L) (n L) n L + n L < 1 contradiction The sequence {a n } is divergent. Theorem 1 A convergent sequence has a unique limit. Proof: Suppose a convergent sequence {a n } which doesn t have a unique limit. There are L, M R such that L M, lim a n = L and lim a n = M. Then > 0. n 1 Z + such that n > n 1 a n L < n Z + such that n > n a n M < Let n 0 be such that n 0 > n 1 and n 0 > n. Then = L a n0 + a n0 M a n0 L + a n0 M < + The limit is unique. = - contradiction Theorem Let {a n }, {b n } be two sequences such that lim a n = L and lim b n = M. Then, lim (a n + b n ) = L + M

3 Proof: Let ε > 0. n 1 Z + such that n > n 1 a n L < ε n Z + such that n > n b n M < ε Let n 0 = Max {n 1, n }. n > n 0 (a n + b n ) (L + M) a n L + b n M lim (a n + b n ) = L + M < ε + ε = ε Theorem 3 Let {a n }, {b n } be sequences such that lim Then the following claims are true. i) lim (ra n ) = rl r R ii) lim (a n b n ) = LM a n = L and lim b n = M. a n iii) lim = L b n M provided that M 0 The proof is an exercise for the student. Definition 3 Let {a n } be a sequence. i) If a n a n+1 n, then the sequence is said to be increasing. ii) If a n a n+1 n, then the sequence is said to be decreasing. iii) If a sequence is either increasing or decreasing, then it is said to be a monotonic sequence. If the strict inequalities are being used in i) or ii), then we say that the sequence is strictly increasing or strictly decreasing. Definition Let {a n } be a sequence. i) If A R such that a n A n, then {a n } is said to be bounded above. ii) If B R such that a n B n, then {a n } is said to be bounded below. iii) If there are A, B R such that B a n A n, then {a n } is said to be bounded. Theorem Every convergent sequence is bounded. Proof: Let {a n } be a convergent sequence. Then L R such that lim a n = L n 0 Z + such that n > n 0 a n L < 1 a n < L + 1 Let M = Max { a 1, a,., a n0, L + 1} Then a n M n M a n M n {a n } is bounded. 3

4 The student should realize that this theorem provides us with another way to make the claim we made in the example. The converse of theorem is false. Let a n = ( 1) n Then < a n < n It will be easy proving that the sequence {a n } is divergent. Suppose a sequence {a n } is bounded above. Then from the completeness axiom (chapter ), we can say that the supremum of {a n n Z + } exists. In the same way, if {a n } is bounded below, we can say that the infimum of {a n n Z + } exists. Theorem 5 If {a n } is increasing and bounded above, then lim a n = L where L = sup{a n n Z + }. Proof is a tutorial exercise. Theorem 6 If {a n } is decreasing and bounded below, then lim a n = M where M = inf{a n n Z + }. Proof is a tutorial exercise. Example 3 b n = n Clearly, {b n } is decreasing. 1 b n n So {b n } is bounded below. The student should be able to prove that 1 = inf{b n } From the above theorem, lim b n = 1 Cauchy Sequences We know that in a convergent sequence the terms get closer and closer to its limit. It may also be possible to have a sequence whose terms get closer and closer together and, we should be able to explain the behaviour of those terms without even mentioning a limit. The following definition will explain such a behaviour. Definition 5 A sequence {a n } is said to be a Cauchy sequence if and only if ε > 0, n 0 Z + such that m, n > n 0 a m a n < ε If the terms are getting closer and closer together, one would expect those to get closer and closer to a limit. In other words, one would expect such a sequence to be convergent. That fact needs to be proven and we will devote the next two theorems for that purpose.

5 Theorem 7 Every Cauchy sequence is bounded. Proof: Let {a n } be a Cauchy sequence. n 0 Z + such that m, n > n 0 a m a n < 1 m > n 0, a m a n0 +1 < 1 m > n 0 a m a n0 +1 < 1 a m < a n Let M = Max{ a 1, a,., a n0, a n } Then a m < M m Z + {a m } is bounded. Theorem 8 sequence. Let {a n } be a sequence. Then {a n } is convergent if and only if {a n } is a Cauchy Proof: Suppose {a n } is convergent. Let ε > 0. Then n 0 Z + such that n > n 0 a n L < ε where L is the limit of {a n}. Take m, n such that m > n 0, n > n 0. a m a n = a m L + L a n a m L + a n L < ε {a n } is a Cauchy sequence. Proving the converse is not necessary for engineering students. Theorem 7 and some other material will be needed for that proof. Because of this theorem, we can establish the convergence of a sequence without knowing the limit. Example : Suppose there is a sequence with the property a n+ = a n + a n+1 n Z + For this sequence, a i+1 a i = a i 1 + a i a i = a i 1 a i = a i 1 ( a i + a i 1 ) = a i 1 a i It follows that a i+1 a i = a i a i 1 = a i 1 a i 5

6 = = a a 1 i 1 If m = n, then a m a n = 0 So, without loss of generality, take m > n. m 1 a m a n = (a i+1 a i ) i=n m 1 a m a n a i+1 a i i=n m 1 = a a 1 ( 1 i 1 ) < a a 1 ( 1 n ) i=n Let ε > 0. If a = a 1, then a m a n < ε So let a a 1 Choose n 0 such that n0 > a a 1 ε n > n 0 n > a a 1 ε a a 1 n < ε a m a n < ε {a n } is a Cauchy sequence. Hence, {a n } is convergent. Note that we were able to establish the convergence without knowing the limit. U. A. Senevirathne Dept. of Mathematics University of Moratuwa. 6

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