Jim Lambers MAT 169 Fall Semester 2009-10 Lecture 6 Notes These notes correspond to Section 8.3 in the text. The Integral Test Previously, we have defined the sum of a convergent infinite series 1 X an n=1 1 to be the limit of the sequence of partial sums fsngn=1, defined by k X sk = a1 + a2 + ··· + ak = an: n=1 In other words, if S is the sum of the series, then k X S = lim sk = lim an: k!1 k!1 n=1 Figure 1 illustrates the partial sum s8 for the series 1 X n−3=2: (1) n=1 It can be seen from the figure that this partial sum can be viewed as the sum of the areas of rectangles, each with width 1 and height n−3=2, for n = 1; 2;:::; 8. If we exclude the leftmost rectangle, then the sum of the areas of the remaining rectangles can be viewed as an approximation of the area of the region bounded by the curves y = x−3=2, y = 0, x = 1, and x = 8. The exact value of this area is given by the integral Z 8 x−3=2 dx; (2) 1 1 which is defined to be the limit of a sequence of Riemann sums fRmgm=1. Each Riemann sum approximates the area of this region by the sum of the areas of m rectangles, each of width ∆x = (8 − 1)=m. Specifically, we have m X 7 R = n−3=2 : (3) m m n=1 1 Figure 1: Partial sum s8 of the series (1). The leftmost rectangle, shown with dotted edges, is excluded from the approximation of the area given by the integral (2). All other rectangles, shown with solid edges, represent the partial sums ~7 of the modified series (4). From the definitions of the original series (1) and the Riemann sum (3), we can see that the partial sum shown in Figure 1, s8, is equal to R7 + 1, since 1 is the area of the leftmost, excluded rectangle. Furthermore, the partial sums ~8 of the modified series 1 X n−3=2; (4) n=2 defined by 8 −3=2 −3=2 X −3=2 s~8 = 2 + ··· + 8 = n ; n=2 is a lower bound for the exact area given by the integral. This last point suggests that if we extend the interval of integration to [1; 1), and find that the improper integral Z 1 Z k x−3=2 dx = lim x−3=2 dx 1 k!1 1 exists and is finite, then the modified series (4) must converge, because each partial sum of the modified series is bounded above by the \partial integral" on the interval [1; k]. If this sequence 2 of integrals converges, it follows that the sequence of modified partial sums must converge, and therefore the modified series must be convergent. Because the original series (1) and modified series (4) only differ by the inclusion of the first term 1−3=2 = 1, we conclude that the original series must be convergent as well. Evaluating the integral (2), we obtain Z 1 Z k x−3=2 dx = lim x−3=2 dx 1 k!1 1 k = lim −2x−1=2 k!1 1 = lim 2 − 2k−1=2 k!1 = 2: It follows that the sum of the modified series (4) is less than 2, and therefore the sum of the original series (1) must be less than 3. Unfortunately, it is not possible to analytically compute the exact value of the sum, although it can be approximated numerically. Nonetheless, we at least know that in some circumstances, we can use integrals to determine whether a series converges, as this example illustrates. Integrals can also be used to determine that a series is divergent. Suppose that the integral Z 1 f(x) dx 1 is divergent, meaning that it is not equal to any finite value. By examining Figure 2, we can see that just as the integral from 1 to k of f(x) is an upper bound on the partial sums ~k, that excludes the first term, this integral is also a lower bound on the partial sum sk, that includes the first term. Therefore, if the sequence of integrals from 1 to k diverges as k ! 1, it follows that the sequence 1 of partial sums fskgk=1, consisting of values that are greater than these integrals, must diverge as well. 1 We summarize this discussion by stating the Integral Test: Let the sequence fangn=1 be defined by an = f(n), for n ≥ 1; where f(x) is a continuous, positive, decreasing function defined on [1; 1). If the improper integral Z 1 Z k f(x) dx = lim f(x) dx 1 k!1 1 converges to a finite value F , then the series 1 X an n=1 is convergent. Furthermore, if S is the sum of this series, then S < F + a1. On the other hand, if the integral diverges, then the series diverges. 3 Figure 2: Partial sum s7 of the series (1). The reason why it is important that f(x) is positive and decreasing is that it must be ensured that the partial sums of the modified series (with the first term excluded) are a lower bound for the value of the integral, because otherwise, no conclusive statement can be made about the convergence of the series based on the existence of the integral. It should be mentioned that it is not absolutely necessary for f(x) to be positive and decreasing on the entire interval [1; 1). It is sufficient that there exist some c ≥ 1 such that f(x) is positive decreasing for all x > c. This is because the convergence of a series is not affected by the behavior of a finite number of terms. Example Consider the p-series 1 X 1 ; np n=1 where p > 0. The function f(x) = 1=xp, for x ≥ 1, is a continuous, positive, decreasing function, so we can apply the Integral Test to determine whether this series converges. We have Z 1 k 1 1 1−p 1 1−p p dx = lim x = lim (k − 1): 1 x k!1 1 − p 1 k!1 1 − p In order for the integral to converge, we must have p > 1. We conclude that the p-series converges if p > 1, and diverges if p ≤ 1. 2 4 Example Consider the series 1 X 1 : 1 + n2 n=0 We attempt to determine whether this series converges using the Integral Test, by evaluating the integral Z 1 1 2 dx: 0 1 + x Note that while the Integral Test, as stated previously, uses 1 for both the initial index and the lower limit of integration, it can also be applied with different values for these quantities, as long as they are consistent. We have Z 1 1 −1 k −1 −1 π 2 dx = lim tan x 0 = lim (tan k − tan 0) = : 0 1 + x k!1 k!1 2 It follows from the convergence of this integral that the series is convergent. 2 5 Summary 1 • The Integral Test can be used to determine whether a series converges. If the terms fangn=1 of a series are defined to be the values of a continuous, positive, decreasing function f(n) where n is each positive integer, and if f(x) is integrable on the interval [1; 1), then the series converges, and the sum of the series, excluding the first term a1, is bounded above by the value of the integral. • The Integral Test can also be used to test for the divergence of a series. If the integral of f(x) from 1 to 1 diverges, then the series whose terms are defined by an = f(n), for n ≥ 1, also diverges. • The function f(x) need not be positive and decreasing for the entire domain of integration, but it needs to be positive and decreasing for all x > c, for some c in the domain. • The initial index of the series and the lower limit of integration do not have to be equal to 1, but the same value should be used for both. −p • Using the Integral Test, it can be shown that the p-series, whose terms are given by an = n , converges if p > 1 and diverges if p ≤ 1. 6.