Math 409-502Harold P. [email protected] on radius of convergenceWe can find the radius of convergence of a power series by either the ratio test or the root test,but some other test is needed to determine the endpoint behavior.Useful tests for endpoint behavior are:• nth-term test• comparison tests• alternating series testMath 409-502 October 11, 2004 — slide #2Follow-up on endpoint convergenceLast time we saw (by the ratio test) that∞∑n=1(n!)2(2n)!xnhas radius of convergence equal to 4, and∞∑n=1n! xn1 ·3 ·5 ···(2n −1)has radius of convergence equal to 2. At the right-hand endpoint, bothseries become∞∑n=1(n!)24n(2n)!. That series diverges by the nth-term test. Indeed,4n= (1 + 1)2n= 1 +¡2n1¢+¡2n2¢+ ··· +¡2nn¢+ ··· +¡2n1¢+ 1, so 4n>¡2nn¢=(2n)!(n!)2. Thus(n!)24n(2n)!> 1, so the series cannot converge. For the same reason, divergence occurs at the left-hand endpoint in this example.Math 409-502 October 11, 2004 — slide #3Operations on power seriesAddition, subtraction, multiplication, and division of power series work the way you expect.Examplecos(x) = 1 −x22!+x44!−x66!+ ···sin(x) = x −x33!+x55!−x77!+ ···so the coefficient of x5in the product cos(x) sin(x) equals15!+12! 3!+14!=215.Math 409-502 October 11, 2004 — slide #4Remark on the multiplication theoremTheorem (page 121): If∞∑n=0anand∞∑n=0bnboth converge absolutely, then the product of the twoseries equals the absolutely convergent series∞∑n=0cn, where cn=n∑k=0akbn−k.Counterexample in case of conditional convergence: Set an= bn= (−1)n/√n + 1. Then∑anand∑bnare conditionally convergent by the alternating series test, but the series∑cnis divergent. Indeed, cn=n∑k=0(−1)k(−1)n−k√k + 1√n − k + 1. All the terms in this sum have the samesign (−1)n, so |cn| ≥n∑k=01n + 1= 1. Hence∑cndiverges.Math 409-502 October 11, 2004 — slide #5Homework1. Read Chapter 9, pages 125–134.2. Do Exercises 9.2/3 and 9.3/1, pages 134–135.Math 409-502 October 11, 2004 — slide