1.10. INDEPENDENT PROJECTS 29 From this exercise, it follows that a complex power series around z0 that converges absolutely at any point other then z0 will have a disk of convergence of the form {z ∈ C | |z − z0| r}. The supremum of all such r is called the radius of convergence of the power series. If the power series converges only at the point z = z0, we say that the series has radius of convergence equal to 0. To determine the radius of convergence for a complex power series, we use the convergence tests developed above, in particular the root test. Theorem 1.10.32. Suppose that lim sup n→∞ |an|1/n = r. If 0 r ∞, then the power series ∑ ∞ n=0 an(z − z0)n has radius of convergence 1/r. If r = 0, then the radius of convergence is infinity. If r = ∞, then the radius of convergence is 0. Examples 1.10.33. (i) Consider the series ∑ ∞ n=0 n(z − z0)n. Then limn→∞ n1/n = 1, and the power series converges absolutely for |z − z0| 1, that is, the radius of convergence is 1. (ii) Consider the series ∑ ∞ n=1 nn(z − z0)n. Then limn→∞(nn)1/n = ∞, so the radius of convergence is 0 and the series converges only for z = z0. Exercise 1.10.34. Determine the radius of convergence of the following power series: (i) ∑ ∞ zn n! (ii) ∑n=1 ∞ zn ln(n) (iii) ∑n=2 ∞ n=1 nn n! zn. 1.10.3. Decimal Expansions of Real Numbers. In Appendix A, we used a decimal representation of the real numbers to show that the real numbers between 0 and 1 form an uncountable set. In this project, we actually prove that every real number between 0 and 1 has a decimal expansion which is unique with the condition that no expansion can terminate in all 9’s. In addition, we discuss the fact that rational numbers have decimal expansions of three different types. The first is terminating decimals, the second is rational numbers whose denominators are relatively prime to 10, and the third is a combination of the first two. Since we know that every real number lies between two consecutive integers, we start with a real number x so that 0 x 1. Let S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Assume first that x is irrational. The construction proceeds as follows. Let a1 be the largest element of S which is less than 10x. Then 0 x−a1/10 1/10. Let a2 be the largest integer in S less than 100x − 10a1. Proceeding as before, we get 0 x − a1/10 − a2/102 1/102. Continuing this process, we obtain a monotonic increasing sequence Sn = a1/10 + a2/102 + · · · + an/10n,

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