Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan

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1 Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan 8. Sequences We start this section by introducing the concept of a sequence and study its convergence. Convergence of Sequences. An infinite sequence is a list of numbers a, a 2, a 3,. Example 8.. (), 2, 3,. (2) 0, 2, 0, 2, 0,. Formally, a sequence is a function f with domain the set of nonnegative integers IN. The image of n IN is denoted by f(n) = a n. We represent a sequence by the notation {a n } n= and we call a n the n th term of the sequence. For example, the n th term of the sequence () above is a n = n, n while the n th term of (2) is a n = + ( ) n. We say that a sequence {a n } n= converges to a number L if and only if a n approaches L as n gets larger and larger, i.e., if the difference a n L can be made as small as we wish by taking n large enough. Using the ɛ-δ argument, if ɛ > 0 is given then we can find a positive integer N such that We write a n L < ɛ for n N. a n = L. Figure 8.. illustrates the definition of convergence of a sequence. Figure 8.. If a sequence does not converge then we say that it diverges.

2 Example 8..2 () Since 3n 2 +5 n 2 +n+ = 3, the sequence { 3n2 +5 n 2 +n+ } n= converges to 3. (2) Since ( ) n alternates between and, the sequence { + ( ) n } n= alternates between 0 and 2 and so it diverges. The following theorem shows the connection between the convergence of a function and the convergence of a sequence.(see Figure 8..2). This basically allows us to replace its of sequences with its of functions. In particular, this is useful for using L Hôpital s rule in computing its of sequences. Theorem 8.. If x f(x) = L and a n = f(n) then a n = L. Figure 8..2 Let ɛ > 0 be given. Since f(x) = L, we can find a δ > 0 such that x f(x) L < ɛ if x δ. Let N = (integer part of δ) +. Then N is a positive integer greater than or equal to δ. Thus, if n N then This shows that a n = L Example 8..3 (a) Find (b) Find ln (n+) n+. sin n n+. a n L = f(n) L < ɛ. 2

3 Solution. (a) Applying L Hôpital s rule we have ln (n+) ln (x+) n+ = x x+ = x x+ = 0. (b) Since sin n we obtain n+ sin n n+ n+. But x x+ = 0. By the squeeze rule we have sin n n + = 0 We next discuss some of the important properties of sequences. n+ = Theorem 8..2 (Convergence Properties of Sequences) Let {a n } n= and {b n} n= be two sequences such as a n = L and b n = L where L and L are finite numbers. Then (a) (a n + b n ) = a n + b n = L + L. (b) ka n = k a n = kl, where k is a constant. (c) a nb n = ( a n)( b n) = LL. a (d) n = L bn bn L provided that b n 0. (e) Squeeze Principle: If a n b n c n for all n N 0 and if = an c n = L then b n = L. (f) If a n = 0 then a n = 0. a n = We will prove (a) and leave the rest as an exercise. Let ɛ > 0 be given. Then there exist positive integers N and N 2 such that a n L < ɛ 2 for n N and b n L < ɛ 2 for n N 2. Let N = N + N 2. Then for n N we have (a n + b n ) (L + L ) a n L + b n L < ɛ 2 + ɛ 2 = ɛ. 3

4 Example 8..4 Show that the sequence with n th term a n = ( )n n converges to 0. Solution. Since ( ) n n = By (f) of the above theorem, we have ( ) n n = 0 n = 0, In order to discuss the next result we need the following definitions: We say that a sequence {a n } n= is bounded from below if there is a number M such that M a n for all n. It is said to be bounded from above if a n M for all n. We say that a sequence is bounded if there are two numbers K and M such that K a n M for all n. The next result shows that convergent sequences are always bounded. Theorem 8..3 Suppose that a n = L. Then there exists a positive number M such that M a n M for all n. That is the sequence is bounded. Let ɛ =. Then there exists N > 0 such that a n L < whenever n N. This implies that L < a n < L + for all n N. As a result of that we have a n < L + for all n N. Let M = max{ a, a 2,, a N, L + }. Then a n M for all n, i.e. M a n M for all n. This establishes a proof of the theorem Now, we discuss a very useful theorem that establishes the convergence of a given sequence (without, however, revealing the it of the sequence). But first we introduce the following definitions: We say that a sequence {a n } n= is increasing if a n a m whenever n m. A sequence is said to be decreasing if a m a n whenever n m. Theorem 8..4 (Monotone Convergence Theorem) An increasing sequence that is bounded from above is always convergent. A decreasing sequence that is bounded from below is always convergent. 4

5 We will prove the first part and leave the second part for the reader. Since {a n } n= is bounded from above, there exists a C > 0 such that a n C for all n. Let c be the smallest positive constant such that a n c for all n. We will show that a n = c. Let ɛ > 0. Then there is an integer N such that (See Figure 8..3). a N > c ɛ. Figure 8..3 Since the sequence {a n } n= is increasing, we find for all n N. Thus, a n > c ɛ a n c = c a n < ɛ for all n N. This shows that the sequence converges to c The following is a list of useful sequences. Theorem 8..5 (a) If r = then rn =. (b) If r = then the sequence {r n } n= is divergent. (c) If r > then the sequence {r n } n= is divergent. (d) If r < then rn = 0. (e) For r > 0, r n =. (a) Trivial. (b) If r = then the sequence {( ) n } n= alternates between the two values and and so the sequence is divergent. (c) r > implies r > or r <. Suppose first that r >. Let ɛ > 0. Let 5

6 N be a positive integer greater than r n =( + (r )) n ɛ r. Then for n N we have + n(r )(by the binomial formula) > + N(r ) > + ɛ >ɛ. This shows that for any given positive number we can find a term in the sequence {r n } n= which is greater than the number. This means that rn as n. If r < then r n = ( ) n ( r) n with r >. Thus, as n becomes large, r n alternates between large positive numbers and negative numbers with large absolute value so that again the it r n as n does not exist. (d) If 0 < r < then r n = (r ) n with (r ) n as n. (See (c)). Hence, r n 0 as n. If < r < 0 then 0 < r <. In this case, r n = ( ) n ( r) n 0 as n. If r = 0 then r n = 0 and r n = 0. (e) Let d = r n. Then ln d = ln r n 0 as n. Hence, d = eln d e 0 = as n. 6