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(7/19/09) Math 20B. Lecture Examples. Section 10.3. Convergence of series with positive terms† Theorem 1 (Convergence of series with positive terms) An infinite series with positive terms either converges or diverges to ∞. The series converges if its partial sums are bounded and diverges if its partial sums are not bounded. Theorem 2 (The Integral Test for series with positive terms) ∞ X Suppose that the series an (1) n=n0 with positive terms is such that an = f (n) for integers n ≥ c with some c, where y = f (x) is continuous on [c, ∞) and decreasing for x ≥ c. Then the infinite series (1) converges if and only if the improper integral Z ∞ f (x) dx (2) c converges. Figures 1 and 2 show why the series converges (1) if the integral (2) converges. The area of the rectangles in Figure 1 is less than the area of the region in Figure 2. If the integral converges, then the area of the region in Figure 2 is bounded as N → ∞, so that the partial sums of the series are bounded and the series converges. Area = N X Area = an n=c+1 FIGURE 1 † Lecture Z N f (x) dx c FIGURE 2 notes to accompany Section 10.3 of Calculus, Early Transcendentals by Rogawski. 1 Math 20B. Lecture Examples. (7/19/09) Section 10.3, p. 2 Figures 3 and 4 show why the series (1) diverges if the integral (2) diverges. The area of the rectangles in Figure 4 is greater than the area of the region in Figure 3. If the integral diverges, then the area of the region in Figure 3 tends to ∞ as N → ∞, so that the partial sums of the series tend to ∞ and the series diverges. Area = Z N +1 f (x) dx Area = c Does an n=c FIGURE 3 Example 1 N X ∞ X FIGURE 4 2 ne−n converge? n=1 Answer: Z ∞ 2 xe−x dx = 1 series are plotted in Figure A1.) 1 −1 e 2 • ∞ X 2 ne−n converges by the Integral Test. (Some partial sums of the n=1 y y = sN 1.5 sN = ∞ X n=1 1 −n2 ne 0.5 Figure A1 10 20 N Section 10.3, p. 3 Example 2 Math 20B. Lecture Examples. (7/19/09) Does ∞ X n=2 Answer: Z ∞ 1 converge or diverge? n ln n 1 x ln x 2 dx = ∞ • plotted in Figure A2.) ∞ X n=2 1 diverges by the Integral Test. (Some partial sums of the series are n ln n y y = sN 2 sN = N X n=2 1 1 n ln n Figure A2 10 20 N The Integral Test is used to establish the following result. Theorem 3 (Convergence of the p-series) The infinite series ∞ X 1 1 1 1 p = 1 + p + p + p + ··· n 2 3 4 n=1 converges if p > 1 and diverges if p ≤ 1. Example 3 Does ∞ X 6 1.75 n n=1 Answer: ∞ X n=1 6 =6 n1.75 with p = 1.75 > 1. Example 4 Does Answer: ∞ X n=1 1 converges because it is a constant multiplied by the p-series n1.75 n=1 1 1.75 n 1 √ converge or diverge and why? 9 n n=2 ∞ X 1 √ 9 n 1 < 1. with p = 2 ∞ X Does n=1 1 = 9 ∞ X n=2 1 n1/2 diverges because it is a constant multiplied by the p-series ∞ X n=2 1 n1/2 7n converge or diverge and why? 6n + 1 Answer: The series diverges because an = Example 6 ∞ X ∞ X n=2 Example 5 converge or diverge and why? 7n 6n + 1 = 7 6 + 1/n → 7 6 as n → ∞; the terms do not tend to 0. ∞ X 1 is called the harmonic series. Does it converge or diverge? The series n n=1 Answer: ∞ X n=1 1 n diverges because it is the p-series with p = 1. Math 20B. Lecture Examples. (7/19/09) Section 10.3, p. 4 Theorem 4 (The Comparison Test with positive terms) Suppose that ∞ X an and n=n0 (a) If ∞ X n=n0 ∞ X ∞ X bn are series with positive terms. n=n0 bn converges and there are constants M and N such that an ≤ M bn for n ≥ N, an also converges. then n=n0 ∞ X (b) If n=n0 bn dinverges and there are constants M > 0 and N such that an ≥ M bn for n ≥ N, then ∞ X an also diverges. n=n0 Example 7 Does the series ∞ X (0.6)n converge or diverge and why? n+1 n=0 Answer: ∞ X (0.6)n n=0 (0.6)n n+1 ≤ converges by the Comparison Test with the convergent geometric series n+1 (0.6)n ∞ X (0.6)n because n=0 for n ≥ 0. (The partial sums of the first series in Figure A7a are less than the partial sums of the second series in Figure A7b.) y y 2 2 1 1 10 20 y= N N X (0.6)n n=0 n+1 Figure A7a Example 8 Does Answer: n2 + 1 3 5n 20 y= N X N (0.6)n n=0 Figure A7b ∞ X n2 + 1 converge or diverge and why? 3 5n n=1 ∞ X n2 + 1 n=1 2 > 10 n 3 5n 5n3 = 1 5 diverges by the Comparison Test with the divergent harmonic series ∞ X n=1 1 n is a positive constant. for n ≥ 1, and 1 5 1 n because Section 10.3, p. 5 Math 20B. Lecture Examples. (7/19/09) Theorem 5 (The Limit-Comparison Test with positive terms) Suppose that an =L b n→∞ n lim (3) where the an ’s and bn ’s are positive and L is either a nonnegative number or ∞. (a) If L is finite and positive, then ∞ X an converges if and only if (c) If L = ∞ and ∞ X bn converges, then n=n0 ∞ X bn converges. n=n0 n=j0 (b) If L = 0 and ∞ X an converges. n=n0 ∞ X bn diverges, then n=n0 ∞ X an also diverges j=n0 ∞ X 10(2n ) Does converge or diverge and why? 5n − 1 Example 9 n=1 Answer: ∞ X n=1 n 10(2n ) n converges by the Limit Comparison Test with the convergent geometric series = 5 −1 ∞ X n=0 2 n 5 10(2 ) because 5n − 1 2 n 5n 2 n 5n 10(2n ) n 5 −1 = 5n 2 n 10(2n ) n 5 (1 − 1/5n ) = 10 1 − 1/ 5n → 10 as n → ∞, and 10 is a positive number. Example 10 Does the series ∞ X n=1 Answer: ∞ X n=1 n 6n2 + 1 n 2 6n + 1 converge or diverge and why? diverges by the Limit Comparison Test with the divergent harmonic series ∞ X n=1 1 n because n 6n2 + 1 1 = n2 2 6n + 1 = n2 2 n (6 + 1/n2 ) = 1 6 + 1/n2 → 1 6 as n → ∞, and 1 is a positive number. 6 n Interactive Examples Work the following Interactive Examples on Shenk’s web page, http//www.math.ucsd.edu/˜ashenk/:‡ Section 10.3: Examples 1–4 Section 10.4: Examples 1–4 ‡ The chapter and section numbers on Shenk’s web site refer to his calculus manuscript and not to the chapters and sections of the textbook for the course.