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Infinite Series 9 Copyright © Cengage Learning. All rights reserved.

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1 Infinite Series 9 Copyright © Cengage Learning. All rights reserved.

2 Sequences Copyright © Cengage Learning. All rights reserved. 9.1

3 Warm-up: Ex #5 Section 8.7: Evaluate the limit.

4 L’Hopital’s Rule

5

6 In mathematics, a sequence is an ordered list of objects (or events). Like a set, it contains members (also called elements or terms), and the number of terms (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and exactly the same elements can appear multiple times at different positions in the sequence. A sequence is a discrete function whose domain is the set of positive integers.mathematicssetmembersdiscrete function In this course we will only be interested in infinite sequences with a rule for generating the terms (range values) and using the set of whole numbers or counting numbers as the domain.

7 Don’t let the new notation confuse you… Other notation:

8 Example 1 – Listing the Terms of a Sequence a.The terms of the sequence {a n } = {3 + (–1) n } are 3 + (–1) 1, 3 + (–1) 2, 3 + (–1) 3, 3 + (–1) 4,... 2, 4, 2, 4,.... b. The terms of the sequence {b n } are

9 Example 1 – Listing the Terms of a Sequence c. The terms of the sequence {c n } are d. The terms of the recursively defined sequence {d n }, where d 1 = 25 and d n + 1 = d n – 5, are 25, 20, 15, 10,..... cont’d

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11 Limit of a Sequence Sequences whose terms approach to limiting values, are said to converge. For instance, the sequence {1/2 n } converges to 0.

12 The car converged to the tree. Some oscillating sequences never settle down to approach a single number.

13 Limit of a Sequence

14 Example 5 – Using the Squeeze Theorem Show that the sequence converges, and find its limit. Solution: To apply the Squeeze Theorem, you must find two convergent sequences that can be related to the given sequence. Two possibilities are a n = –1/2 n and b n = 1/2 n, both of which converge to 0. By comparing the term n! with 2 n, you can see that, n! = n = and 2 n = =

15 Example 5 – Solution This implies that for n ≥ 4, 2 n < n!, and you have as shown in Figure 9.2. So, by the Squeeze Theorem it follows that Figure 9.2 cont’d

16 Step 1: line up the most reasonable domain #’s … The numerators are all 1’s. The denominators are all powers of “2”. If we started with n=1…

17 Example – Finding the nth Term of a Sequence Find a sequence {a n } whose first five terms are and then determine whether the particular sequence you have chosen converges or diverges.

18 Example – Solution First, note that the numerators are successive powers of 2, and the denominators form the sequence of positive odd integers. By comparing a n with n, you have the following pattern. Using L’Hôpital’s Rule to evaluate the limit of f(x) = 2 x /(2x – 1), you obtain So, the sequence diverges.

19 Ex: Find the limit (if possible): Final answer: this sequence converges to zero by the Squeeze Theorem.

20 Ex: Find the limit (if possible): Use L’Hopital’s Rule L’Hopital again… One more time! This sequence converges to zero.

21 Ex: Don’t forget to anti-

22 Ex: Find the limit of the sequence: Sequence mode is way too slow at creating tables…so let’s switch back to function mode!

23 Ex: Find the limit of the sequence:

24 Monotonic Sequences and Bounded Sequences

25

26 Example 8– Determining Whether a Sequence Is Monotonic Determine whether each sequence having the given nth term is monotonic. Solution: a. This sequence alternates between 2 and 4. So, it is not monotonic. Figure 9.3(a)

27 Example – Solution b. This sequence is monotonic because each successive term is larger than its predecessor. Figure 9.3(b) cont’d

28 Example – Solution b. This sequence is monotonic because each successive term is larger than its predecessor. To see this, compare the terms b n and b n + 1. [Note that, because n is positive, you can multiply each side of the inequality by (1 + n) and (2 + n) without reversing the inequality sign.] Figure 9.3(b) cont’d

29 Example – Solution Starting with the final inequality, which is valid, you can reverse the steps to conclude that the original inequality is also valid. cont’d

30 Example – Solution c. This sequence is not monotonic, because the second term is larger than the first term, and larger than the third. (Note that if you drop the first term, the remaining sequence c 2, c 3, c 4,... is monotonic.) cont’d Figure 9.3(c)

31 Monotonic Sequences and Bounded Sequences

32 One important property of the real numbers is that they are complete. This means that there are no holes or gaps on the real number line. (The set of rational numbers does not have the completeness property.) The completeness axiom for real numbers can be used to conclude that if a sequence has an upper bound, it must have a least upper bound (an upper bound that is smaller than all other upper bounds for the sequence).

33 Monotonic Sequences and Bounded Sequences For example, the least upper bound of the sequence {a n } = {n/(n + 1)}, is 1.

34 Example – Bounded and Monotonic Sequences a.The sequence {a n } = {1/n} is both bounded and monotonic and so, by Theorem 9.5, must converge. b.The divergent sequence {b n } = {n 2 /(n + 1)} is monotonic, but not bounded. (It is bounded below.) c.The divergent sequence {c n } = {(–1) n } is bounded, but not monotonic.

35 9.1 Homework Pg. 602: EOO MMM pg (includes notes)


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