30.4 Working with Series and Power Series 961 f(x)=C 0 +C 1 x − C 0 2! x 2 − C 1 3! x 3 + C 0 4! x 4 + C 1 5! x 5 − C 0 6! x 6 − C 1 7! x 7 +··· f(x)=C 0 1− x 2 2! + x 4 4! −···+(−1) n x 2n (2n)! +··· cos x + C 1 x − x 3 3! + x 5 5! − x 7 7! +···+(−1) n x 2n+1 (2n + 1)! +··· sin x f(x)=C 0 cos x + C 1 sin x ◆ EXERCISE 30.6 Verify that f(x)=C 0 cos x + C 1 sin x is a solution to the differential equation y =−y. Wehave shown that if a solution to y =−y has a power series representation, then that solution must be of the form C 0 cos x + C 1 sin x, where C 0 and C 1 are constants. In the example just completed, we recognized the Maclaurin series for sin x and cos x. It is entirely possible that we can solve for all the coefficients of a power series and simply have the solution expressed as and defined by the power series expansion. There are well-known functions defined by power series that arise in physics, astronomy, and other applied sciences. An example of such functions are the Bessel functions, named after the astronomer Bessel who came up with them in the early 1800s while working with Kepler’s laws of planetary motion. The Bessel function J 0 (x) is defined by J o (x) = ∞ k=0 (−1) k x 2k (k!) 2 2 2k . As is often the case in mathematics, while Bessel functions arose in a particular astronomical problem they are now used in a wide array of situations. One such example is in studying the vibrations of a drumhead. A graph of the partial sum J 0 (x) ≈ 13 k=0 (−1) k x 2k (k!) 2 2 2k is given in Figure 30.9. .5 –.5 –1 1 2 –2 x y Graph of ∑ (–1) k k = 0 13 x 2k (k!) 2 2 2k Figure 30.9 962 CHAPTER 30 Series Transition to Convergence Tests Because this chapter began with Taylor polynomials, it was natural to move on to Taylor series directly, without the traditional lead-in of convergence tests for infinite series. Taylor’s Theorem enables us to deal with some convergence issues quite efficiently. Not only are we able to show that the series for e x , sin x, and cos x converge, but we can determine that each converges to its generating function. Our previous work with geometric series allows us to conclude that the series for 1 1−x converges to its generating function on (−1, 1). When we find a Taylor series by manipulating a known Taylor series, whether by substitution, differentiation, or integration, we can calculate the radius of convergence. But, faced with a generic power series, we have few tools at our disposal with which to determine convergence and divergence. More fundamentally, we have no systematic way of determining the convergence or divergence of an infinite series of the form a k . The next section will remedy this situation. PROBLEMS FOR SECTION 30.4 For each series in Problems 1 through 9, determine whether the series converges absolutely, converges conditionally, or diverges. 1. ∞ k=1 (−1) k k! (k−1)! 2. ∞ k=1 (−1) k+1 k! (k+1)! 3. ∞ k=1 (−1) k 1 3k 4. ∞ k=2 (−1) n k ln k 5. ∞ k=10 cos(kn) 10k 6. ∞ k=0 − 11 12 k 7. ∞ k=1 1 100 sin kπ 2 8. ∞ k=1 (−1) k 2 k k 9. ∞ k=0 (−1) n k 2 −10 2k 2 +5k 10. Is it possible for a geometric series to converge conditionally? If it is possible, produce an example. 11. How many nonzero terms of the Maclaurin series for ln(1 + x) are needed to approx- imate ln 3 2 with an error of less than 10 −4 ? 12. Approximate 1 e with error less than 10 −5 . 30.4 Working with Series and Power Series 963 13. Arrive at the series for cos x by differentiating the Maclaurin series for sin x. 14. Find the Maclaurin series for arcsin x using the fact that 1 √ 1−x 2 dx =sin −1 x + C. What is the radius of convergence of the series? In Problems 15 through 17, write the given integral as a power series. 15. cos(x 2 )dx 16. e x 3 dx 17. 1 1+x 5 dx 18. Approximate 0.5 0 sin(x 2 )dx with error less than 10 −8 . Is your approximation an overestimate, or an underestimate? 19. Approximate 0.1 0 x 1+x 3 dx with error less than 10 −10 . 20. Find the Maclaurin series for ln(2 + x) along with its radius of convergence. 21. (a) Find the Maclaurin series for ln 1+x 1−x by subtracting the Maclaurin series for ln(1 − x) from that for ln(1 + x). (b) Show that when x = 1 3 , 1+x 1−x = 2. (c) Use the first four nonzero terms of the series in part (a) to approximate ln 2. Compare your answer with the approximation given by the first four terms of the series for ln(1 + x) evaluated at x =1, and the value of ln 2 given by a calculator or computer. 22. Show that ∞ k=0 (2x) k k! is a solution to the differential equation f (x) = 2f(x).What familiar function does this series represent? 23. Show that if f(x)= ∞ k=0 a k x k is a power series solution to f (x) =−f(x), then f(x)= ∞ k=0 (−1) k x k k! . What function does this series represent? 24. Use power series to solve the differential equation f (x) = 9f(x). What familiar function(s) does this series represent? 25. The Bessel function J 0 (x) is given by J 0 (x) = ∞ k=0 (−1) k x 2k (k!) 2 2 2k . It converges for all x. (a) If the first three nonzero terms of the series are used to approximate J 0 (0.1), will the approximation be too large, or too small? Give an upper bound for the magnitude of the error. (b) How many nonzero terms of the series for J 0 (1) must be used to approximate J 0 (1) with error less than 10 −4 ? 964 CHAPTER 30 Series 30.5 CONVERGENCE TESTS In this section we focus on ways of determining whether or not a given series converges. We begin by looking at series of constants; in the last subsection we apply our results to the convergence of power series. The Basic Principles A series ∞ k=1 a k converges to a sum S if the sequence of its partial sums converges to S, where S is a finite number. In other words, if lim n→∞ S n = S, where S n = n k=1 a k , then the infinite series converges to S. Otherwise, the series diverges. Note that if lim x→∞ f(x)=L, then lim n→∞ f (n) = L. The converse is not true. Our first case study was geometric series. (Refer to Chapter 18.) For a geometric series we are able to express S n in closed form and directly compute lim n→∞ S n .Wefind that ∞ k=0 ar k converges to a 1 − r if |r| < 1 and diverges if |r|≥1. Once we leave the realm of geometric series it can be difficult or impossible to express S n in closed form, so we generally can’t compute lim n→∞ S n directly. Instead, we might determine convergence or divergence by comparing the series in question to a geometric series or an improper integral. We have already established one test for divergence; if the terms of the series don’t tend toward zero then the series diverges. nth Term Test for Divergence. If lim n→∞ a n = 0, then ∞ n=1 a n diverges. If lim n→∞ a n = 0, we have no information and must turn our attention back to the sequence of partial sums. Definition A sequence {s n } is increasing if s n ≤ s n+1 for all n ≥ 1. It is decreasing if s n ≥ s n+1 for all n ≥ 1. If a sequence is either increasing or it is decreasing it is said to be monotonic. A sequence {s n } is bounded above if there is a constant M such that s n ≤ M for all n ≥ 1. It is bounded below if there is a constant m such that m ≤ s n for all n ≥ 1. A sequence is said to be bounded if it is bounded both above and below. A bounded sequence may or may not converge. It could oscillate between the bounds like {(−1) n }. However, if the sequence is bounded and increasing, then its terms must cluster about some number L ≤ M. A similar statement can be made for a decreasing sequence. The following theorem will prove very useful. 30.5 Convergence Tests 965 Bounded Monotonic Convergence Theorem 15 A monotonic sequence converges if it is bounded and diverges otherwise. Suppose the terms of the series ∞ k=1 a k are all positive. Then the sequence of partial sums is increasing: s n ≤s n+1 =s n +a n+1 . Because the terms are positive, the sequence of partial sums is bounded below by zero. Therefore, if {S n }is bounded above, then {S n }converges and consequently ∞ k=1 a k converges; otherwise they diverge. We will use this line of reasoning repeatedly. We’ll refer to it as the Bounded Increasing Partial Sums Theorem. The Bounded Increasing Partial Sums Theorem A series ∞ k=1 a k , where a k ≥ 0, converges if and only if its sequence of partial sums is bounded above. Our focus in this section is on the question of convergence versus divergence and not on the sum of a convergent series. Therefore, the starting point of the series is not important; the first hundred or thousand terms of the infinite series can be chopped off without impacting convergence issues. Keep this in mind when applying the results of this section. For example, if the sequence of partial sums is eventually monotonic, then the Bounded Increasing Partial Sums Theorem can be applied. In the next few subsections we will discuss convergence tests with the specification that the terms of the series are positive. From the observation made above, you can see that what is really required is that the terms a k are positive for all k greater than some fixed number, or, more generally, have any of the required specifications in the long run. The Integral Test We revisit the idea of comparing an infinite series and an improper integral in the next example. 16 ◆ EXAMPLE 30.26 Determine whether the following series converge or diverge. (a) ∞ k=1 1 k 3 = 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 +··· (b) ∞ k=1 1 √ k = 1 √ 1 + 1 √ 2 + 1 √ 3 +··· SOLUTION In both of these series the terms are positive, decreasing, and going toward zero, but the terms of the series in part (b) are heading toward zero much more slowly than those in part (a). The values of some partial sums are given in the table on page 966. The information in the table is inconclusive, but it leads us to guess that ∞ k=1 1 k 3 might converge and ∞ k=1 1 √ k might diverge. 15 A formal proof of this theorem rests on the Completeness Axiom for real numbers, which says that if a nonempty set of real numbers has an upper bound it must have a least upper bound. 16 This was first introduced in Section 29.4. 966 CHAPTER 30 Series 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1.0 1.125 1.162037 1.177662 1.185662 1.190291 1.193207 1.195160 1.196531 1.197531 1.198283 1.198862 1.199317 1.199681 1.199977 1.204607 1.204811 1.204982 1.205128 1.205253 1 1.707106 2.284457 2.784457 3.231670 3.639918 4.017883 4.371436 4.704770 5.020997 5.322509 5.611184 5.888534 6.155795 6.413994 6.663994 6.906530 7.142232 7.371648 7.595255 n S n = ∑ n k =1 1 k 3 S n = ∑ n k =1 1 √k (a) (b) Partial sums are recorded up to six decimal planes. (a) To prove that ∞ k=1 1 k 3 converges it is enough to show that the increasing sequence of partial sums is bounded. We do this by comparing the partial sums to ∞ 1 1 x 3 dx,as shown in Figure 30.10. y x 102345 y = (1, 1) (4, ) The areas of the shaded rectangles correspond to the terms of the series 1 1 3 1 x 3 1 4 3 (3, ) 1 3 3 (2, ) 1 2 3 1 2 3 + 1 3 3 + . . . This figure is not drawn to scale. Figure 30.10 Each of the shaded rectangles has a base of length 1. The area = (base) · (height), so the areas of the rectangles, from left to right, are 1 1 3 , 1 2 3 , 1 3 3 , ···.The sum of the areas of the rectangles is ∞ k=1 1 k 3 . Chop off the first rectangle. n k=2 1 k 3 < n 1 1 x 3 dx; the rectangles lie under the graph of 1 x 3 . Consequently, lim n→∞ n k=2 1 k 3 ≤ ∞ 1 1 x 3 dx. But ∞ 1 1 x 3 dx converges. ∞ 1 1 x 3 dx = lim b→∞ b 1 x −3 dx = lim b→∞ x −2 −2 b 1 = lim b→∞ −1 2b 2 + 1 2 = 1 2 Therefore, the partial sums of ∞ k=1 1 k 3 are bounded by 1 + 1 2 , and the series converges by the Bounded Increasing Partial Sums Theorem. The sum of the series is greater than 1 and less than 1.5. (b) To prove that ∞ k=1 1 √ k diverges it is enough to show that the sequence of partial sums is unbounded. We do this by comparing the partial sums to ∞ 1 1 √ x dx as shown in Figure 30.11. y x 12345 y = 1 √x (2, 1 √2 (1, 1) ) (3, 1 √3 ) (4, 1 √4 ) The areas of the shaded rectangles correspond to the terms of the series 1 √1 1 √2 + 1 √3 ++ . . . Figure 30.11 30.5 Convergence Tests 967 ∞ 1 1 √ x dx = lim b→∞ b 1 x − 1 2 dx = lim b→∞ 2x 1 2 b 1 = lim b→∞ 2 √ b − 2 =∞ Because we want to show that the partial sums are unbounded, we draw rectangles that lie above the graph of 1 √ x . The areas of the shaded rectangles are, from left to right, 1 √ 1 , 1 √ 2 , 1 √ 3 , ···,sothe sum of the areas of the shaded rectangles is ∞ k=1 1 √ k .Wesee that n+1 1 1 √ x dx < S n ,so lim n→∞ n+1 1 1 √ x dx ≤ lim n→∞ S n . ∞≤ lim n→∞ S n . The series diverges. ◆ REMARKS To show that a series with positive terms converges we show that the increasing sequence of partial sums is bounded, that is, is less than some constant M. To show that it diverges, we show that g(n)<S n where lim n→∞ g(n) =∞. Suppose we compare the series ∞ k=1 f(k)with the improper integral ∞ 1 f(x)dx.If f(x)is positive, continuous, and decreasing on [1, ∞), then by including or omitting the first term of the series, we can depict the area corresponding to the sum as lying above or below the area corresponding to the improper integral. (See Figures 30.12 and 30.13.) y x 123456 The sum of the area of the shaded rectangles is a 1 + a 2 + a 3 + a 4 + a 5 a 1 a 2 a 3 a 4 a 5 y = f(x) (2, f(2)) Figure 30.12 y x 123456 The sum of the area of the shaded rectangles is a 2 + a 3 + a 4 + a 5 + a 6 a 2 a 3 a 4 a 5 a 6 y = f(x) (2, f(2)) Figure 30.13 968 CHAPTER 30 Series Using the reasoning given we can obtain the Integral Test. The Integral Test Let ∞ k=1 a k be a series such that a k =f(k)for k =1, 2, 3 ,where the function f is positive, continuous, and decreasing on [1, ∞). Then ∞ k=1 a k and ∞ 1 f(x)dx either both converge or both diverge. The proof of the Integral Test is constructed along the lines of Example 30.26 and makes a nice exercise for the reader. Construct the proof for yourself. If you have difficulty, consult the proof in Appendix H. EXERCISE 30.7 Use the Integral Test to show that the harmonic series diverges. The harmonic series, ∞ k=1 1 k , and the series from Example 30.26, ∞ k=1 1 k 3 and ∞ k=1 1 √ k , are all examples of a class of series referred to as p-series because they can be written in the form ∞ k=1 1 k p , where p is a constant. ◆ EXAMPLE 30.27 Show that the p-series ∞ k=1 1 k p = 1 1 p + 1 2 p + 1 3 p + 1 4 p +···converges if p>1and diverges if p ≤1. SOLUTION For p ≤ 0 the nth term doesn’t tend toward zero: For p = 0, lim n→∞ 1 n p = lim n→∞ 1 = 1 = 0 For p<0, lim n→∞ 1 n p =∞=0 So for p ≤0, ∞ k=1 1 k p diverges by the nth Term Test for Divergence. For p>0, f(x)= 1 x p = x −p is positive, continuous, and decreasing on [1, ∞). For p = 1, we know ∞ k=1 1 k diverges. Let’s consider p = 1 and apply the Integral Test. For p>0, p = 1, ∞ 1 1 x p dx = lim b→∞ b 1 x −p dx = lim b→∞ x −p+1 −p +1 b 1 = lim b→∞ b −p+1 − 1 −p +1 , so ∞ 1 1 x p dx converges to 1 p−1 if p>1 diverges if p ∈ (0, 1). Therefore ∞ k=1 1 k p converges for p>1and diverges for p ≤1. ◆ The conclusion that ∞ k=1 1 k p converges for p>1and diverges otherwise will be useful to keep in mind; we’ll compare other series to p-series. Essentially, we’ve shown that for p>1the terms 1 k p tend toward zero rapidly enough to make the series converge. If we run 30.5 Convergence Tests 969 into an unfamiliar series whose terms go to zero more rapidly than those of a convergent p-series, we will find that this series converges as well. EXERCISE 30.8 Which of the following series converge? (a) ∞ n=1 n −1.1 (b) ∞ k=1 1 5 √ k 2 (c) ∞ k=1 3 k √ k (d) ∞ n=1 1 e n Answers (a), (c), and (d) converge. (a), (b), and (c) are p-series; (d) is not. (d) is a geometric series. Comparison Tests In Example 30.26(b) we showed ∞ k=1 1 √ k diverges by comparing it to the divergent integral ∞ 1 1 √ x dx. This approach was taken to illustrate a tool that enabled us to treat all p-series easily. However, if we had been concerned only with ∞ k=1 1 √ k it would have been simpler to compare this series with the harmonic series. k ≥ √ k for k ≥ 1, so 1 k ≤ 1 √ k for k ≥ 1. Comparing ∞ k=1 1 √ k = 1 + 1 √ 2 + 1 √ 3 + 1 √ 4 +···with ∞ k=1 1 k = 1 + 1 2 + 1 3 + 1 4 +···, we see that each term of the former series is greater than or equal to the corresponding term of the harmonic series. The harmonic series diverges because the partial sums increase without bound. Therefore, ∞ k=1 1 √ k must diverge as well. Having shown that ∞ k=1 1 √ k diverges because the partial sums increase without bound, we can reason that ∞ k=4 1 √ k−1 diverges as well. For k ≥4 the latter is, term for term, larger. 0 < 1 √ k < 1 √ k − 1 for k ≥ 4. Similarly, knowing that the geometric series ∞ k=1 1 2 k converges, we can conclude that ∞ k=1 1 2 k +1 converges as well. 0 < 1 2 k +1 < 1 2 k , so the partial sums of ∞ k=1 1 2 k +1 are smaller than the corresponding partial sums of the geometric series. The partial sums of ∞ 2 k =1 1 2 k +1 are increasing and bounded. Therefore ∞ k=1 1 2 k +1 converges. We can generalize this line of reasoning to compare pairs of series with positive terms. The Comparison Test Also known as the Direct Comparison Test i. Let 0 ≤ a k ≤ b k for all k (or all k ≥ N for some constant N). If ∞ k=1 b k converges, then ∞ k=1 a k converges as well. ii. Let 0 ≤ c k ≤ a k for all k (or all k ≥ N for some constant N). If ∞ k=1 c k diverges, then ∞ k=1 a k diverges as well. 970 CHAPTER 30 Series y x y = x y = lnx Figure 30.14 Less formally, this says that we can compare series whose terms are positive as follows. We can show that a series is convergent by showing that its terms are (eventually) smaller than the terms of a series known to be convergent. We can show that a series is divergent by showing that its terms are (eventually) larger than the terms of a series known to be divergent. Proof i. Let S n = n k=1 a k and ˆ S n = n k=1 b k and suppose 0 ≤ a k ≤ b k for all k. Then {S n } and { ˆ S n } and both increasing sequences, bounded below by 0. S n ≤ ˆ S n for all n. ∞ k=1 b k converges, so { ˆ S n } is bounded by some M. Then {S n } is likewise bounded by M, and hence ∞ k=1 a k converges by the Bounded Increasing Partial Sums Theorem. Because the first N terms of a series don’t affect whether or not it converges, we can draw the same conclusion if 0 ≤ a k ≤ b k for all k ≥ N. ii. Let ˜ S n = n k=1 c k and suppose 0 ≤ c k ≤ a k for all k. Then {S n } and { ˜ S n } are both increasing sequences and ˜ S n ≤ S n for all n. But ∞ k=1 c k diverges. By the Bounded Increasing Partial Sums Theorem { ˜ S n } is unbounded, and therefore {S n } is unbounded and ∞ k=1 a k diverges. ◆ EXAMPLE 30.28 Does ∞ k=2 1 ln k converge or diverge? SOLUTION Compare with the harmonic series. k>ln k for all k ≥ 2 (See Figure 30.14.) Therefore 0 ≤ 1 k < 1 ln k for all k ≥ 2. The series ∞ k=1 1 k diverges; therefore ∞ k=2 1 ln k diverges by the Comparison Test. ◆ To use the Comparison Test effectively we need to have some familiar series on call. The series most commonly used for comparison are the following: The geometric series ∞ k=1 ar k converges for |r| < 1 diverges for |r|≥1. The p-series ∞ k=1 1 k p converges for p>1 diverges for p ≤ 1. Suppose we’re working with the series ∞ k=1 1 10 k −2 . Basically this “looks like” ∞ k=1 1 10 k = ∞ k=1 1 10 k , which converges. Our series should behave similarly. But 10 k > 10 k − 2, so 1 10 k < 1 10 k −2 .Wecan’tapply the Direct Comparison Test. This is frustrating, because we know that the tail of the series is what really matters, and for large k,10 k ≈10 k − 2. The Limit Comparison Test gets us out of this bind. . lim b→∞ −1 2b 2 + 1 2 = 1 2 Therefore, the partial sums of ∞ k=1 1 k 3 are bounded by 1 + 1 2 , and the series converges by the Bounded Increasing Partial Sums Theorem. The sum of the series is greater than 1 and less than 1.5. (b). question of convergence versus divergence and not on the sum of a convergent series. Therefore, the starting point of the series is not important; the first hundred or thousand terms of the infinite. diverge. (a) ∞ k=1 1 k 3 = 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 +··· (b) ∞ k=1 1 √ k = 1 √ 1 + 1 √ 2 + 1 √ 3 +··· SOLUTION In both of these series the terms are positive, decreasing, and going toward zero, but the terms of the series in part (b) are heading toward zero much more slowly than those in part (a). The