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

Series and Convergence Copyright © Cengage Learning. All rights reserved. 9.2

3 Understand the definition of a convergent infinite series. Use properties of infinite geometric series. Use the nth-Term Test for Divergence of an infinite series. Objectives

4 One important application of infinite sequences is in representing “infinite summations.” Informally, if {a n } is an infinite sequence, then is an infinite series (or simply a series). The numbers a 1, a 2, a 3, are the terms of the series. For some series it is convenient to begin the index at n = 0 (or some other integer). As a typesetting convention, it is common to represent an infinite series as simply Infinite Series

5 In such cases, the starting value for the index must be taken from the context of the statement. To find the sum of an infinite series, consider the following sequence of partial sums. If this sequence of partial sums converges, the series is said to converge. Infinite Series

6

7 Example 1(a) – Convergent and Divergent Series The series has the following partial sums.

8 Example 1(a) – Convergent and Divergent Series Because it follows that the series converges and its sum is 1. cont’d

9 Example 1(b) – Convergent and Divergent Series The nth partial sum of the series is given by Because the limit of S n is 1, the series converges and its sum is 1. cont’d

10 Example 1(c) – Convergent and Divergent Series The series diverges because S n = n and the sequence of partial sums diverges. cont’d

11 Infinite Series The series is a telescoping series of the form Note that b 2 is canceled by the second term, b 3 is canceled by the third term, and so on. The nth partial sum of this series is S n = b 1 – b n + 1

12 Infinite Series Because the nth partial sum of this series is S n = b 1 – b n + 1 it follows that a telescoping series will converge if and only if b n approaches a finite number as Moreover, if the series converges, its sum is Because the limit of S n = 1, the series converges and its sum is 1.

13 Example 2: Writing a series in Telescoping Form.  Find the sum of the series:  Using partial fractions:  From this form, the nth partial sum is:  So the series converges and its sum is 1.

14 Ex: Find the sum: All the fractions cancel except the first ones in the 2 leading terms and the last 2 in the last two terms.

15 Geometric Series

16 The series is a geometric series. In general, the series given by is a geometric series with ratio r. Geometric Series

17 Geometric Series

18 Example 3(a) – Convergent and Divergent Geometric Series The geometric series has a ratio of with a = 3. Because 0 < |r| < 1, the series converges and its sum is

19 Example 3(b) – Convergent and Divergent Geometric Series The geometric series has a ratio of Because |r| ≥ 1, the series diverges. cont’d

20 Ex. #4  Use a geometric series to find the sum of the repeating decimal.  …= …  Infinite geometric sum =

21 Ex: Use a geometric series to find the sum of the repeating decimal … … = … This is a geometric series where r =.0001 < 1.

22 Ex: Find all values of x Write the sum of the for which the series converges. series as a function of x. #81

23 Geometric Series As we progress through this chapter, we will become more interested in whether or not a given series converges than in the actual sum itself. Actual sums can be left to computers.

24 nth-Term Test for Divergence

25 nth-Term Test for Divergence If a series converges, the limit of its nth term must be 0. Careful: The converse is not true, just because the nth term is zero, doesn’t mean the series converged! The contrapositive of Theorem 9.8 provides a useful test for divergence. This nth-Term Test for Divergence states that if the limit of the nth term of a series does not converge to 0, the series must diverge.

26 Example 5 – Using the nth-Term Test for Divergence a. For the series you have So, the limit of the nth term is not 0, and the series diverges. b. For the series you have So, the limit of the nth term is not 0, and the series diverges.

27 c. For the series you have Because the limit of the nth term is 0, the nth-Term Test for Divergence does not apply and you can draw no conclusions about convergence or divergence. cont’d Example 5 – Using the nth-Term Test for Divergence

28 Two Theorems As we progress through this chapter, we will become more interested in whether or not a given series converges than in the actual sum itself. Actual sums can be left to computers.

29 Ex: # Assume NOT… (proof by contradiction) Q.E.D.

30 However, surprisingly, the converse is NOT true. Example: Be verrrry careful when you mess with infinity! I’m so scared that I think I’ll do ALL my homework tonight! However,

31 Σ1/n is called the Harmonic Series. Terms after the first two are the harmonic means of those on either side. (Harmonic mean is the reciprocal of the mean of the reciprocals).

32 About the divergence of Σ1/n. Don’t confuse the sequence {1/n} with the series Σ1/n. The sequence {1/n} does converge. As n approaches infinity, then 1/n approaches 0. The series Σ1/n means something different. It means 1 + 1/2 + 1/3 + 1/4 + 1/ So in the series Σ1/n, intuitively you are adding up things which eventually get small, but you are adding up infinitely many of them, which can yield something large. The following is one way to see how Σ1/n diverges.

33 If you group the series Σ1/n like this: (1) + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16) +... Then the first parenthesized expression has value 1. The second parenthesized expression has value 1/2. The third parenthesized expression has value greater than 2/4, because there are 2 fractions and all are at least as large as 1/4. The fourth parenthesized expression has value greater than 4/8, because there are 4 fractions and all are at least as large as 1/8. The fifth parenthesized expression has value greater than 8/16, because there are 8 fractions and all are at least as large as 1/16. By continuing in this manner, you can view the series Σ1/n as the sum of infinitely many "groupings," all with value greater than 1/2. So the series diverges, because if you add up 1/2 enough times, the sum will eventually get as large as you like. Therefore, Σ1/n diverges.

34 Writing Assignment: Explain the difference between what it means for a sequence to converge or a series to converge.

35 Homework Section 9.2 Day 1: pg. 612 #1-33 odd Day 2: pg. 612 #35-71 odd

36 Integral test for divergence of the harmonic series.