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CHAPTER 5 Power Series and Laurent series 5.1. Infinite Sequences of Constants In Chapter 2, we developed properties of analytic functions. In Chapter 4, we developed many more properties. Here, in Chapter 5, infinite series yield even more properties of analytic functions. An infinite sequence of constants is a complex-valued function f whose domain is the set of positive integers Z+. It assigns a complex number f(n) to each n Z+. We often list the sequence in the form 2 f(1), f(2), . , f(n), . The elements of the sequence are called its terms: first term, second term, nth term, etc. We usually replace f(1) by c1, f(2) by c2, etc. to get c1, c2, . , cn, . For a more compact form we use braces: cn = c1, c2, . , cn, . Example. { } ( 1)n + in 1 i 1 1 + i 1 i 1 − = 1 + i, 0, − − , , − , 0, − − , , . n − 3 2 5 7 4 ⇢ Definition (5.1). A sequence cn has limit L if given any ✏ > 0 there exists an integer N such that c {L }< ✏ for all n > N. | n − | When L is the limit of a sequence c , we write { n} lim cn = L, n !1 and we say the sequence converges to L. 115 116 5. POWER SERIES AND LAURENT SERIES Theorem (5.1). If sequences c and d have limits C and D, then { n} { n} (i) kc has limit kC; { n} (ii) c d have limits C D; { n} ± { n} ± (iii) c d has limit CD; { n n} (iv) c /d has limit C/D provided D = 0 and none of the d = 0. { n n} 6 n Theorem (5.2). Suppose cn = xn + yni and L = X + Y i. Then lim cn = L lim xn = X and lim yn = Y. n n n !1 () !1 !1 ✓ni ⇥i Theorem (5.3). Suppose cn = rne and L = Re , where ✓n and ⇥ are principal values of arguments. Then lim cn = L if lim rn = R and lim ✓n = ⇥. n n n !1 !1 !1 Theorem (5.4). lim cn = 0 lim cn = 0. n n !1 () !1| | ( 1)n + in Problem (Page 204 # 3). Find lim − . n n !1 ⇢ Soution. For any n, ( 1)n + in has one of 4 values: 0, 2, 1 i. Then − − ± ( 1)n + in ( 1)n + in lim − = 0 = lim − = 0 n n ) n n !1 !1 ⇢ by using Theorem5.4. ⇤ n2 + i Problem (Page 204 # 5). Find lim sinh . n 1 n2i !1 ⇢ ✓ − ◆ Solution. n2 + i 1 lim sinh = sinh = sinh i = (sin 1)i. n 1 n2i − i !1 ⇢ ✓ − ◆ ✓ ◆ ⇤ 5.2. INFINITE SERIES OF CONSTANTS 117 Example (5.3). Show lim zn = 0 if z < 1, and that the limit does not n | | exist when z > 1. What can!1you say if z = 1? | | | | solution. If z < 1, | | lim zn = lim z n = 0 = lim zn = 0. n n Th.5.4 n !1| | !1| | ) !1 If z > 1, then | | lim zn = lim z n = = lim zn DNE. n n n !1| | !1| | 1 ) !1 If z = 1, then lim zn = 1. n !1 Now suppose z = 1, but z = 1. By part (ii) of Theorem 5.1, if a sequence | | 6 cn converges, then lim (cn cn 1) = 0. By Theorem 5.4, we then get n − { } !1 − lim cn cn 1 = 0. It follows that if lim cn cn 1 = 0, then the sequence n | − − | n | − − | 6 !1c cannot converge. That is the case here:!1 { n} n n 1 n 1 lim z z − = lim z − z 1 = lim z 1 = z 1 = 0. n n n ⇤ !1| − | !1| || − | !1| − | | − | 6 5.2. Infinite Series of Constants When the terms of an infinite sequence are added together, we obtain an infi- nite series 1 c = c + c + + c + . n 1 2 · · · n · · · n=1 X If the sequence of partial sums S , where { n} S = c + c + + c , n 1 2 · · · n has limit S, we say the series has sum S and that the series converges to S, and write 1 cn = S. n=1 X If lim Sn does not exist, we say the series does not have a sum or that it n diverges.!1 118 5. POWER SERIES AND LAURENT SERIES Example. The nth partial sum of the geometric series 2 n 1 a + az + az + + az − + · · · · · · is 2 n 1 S = a + az + az + + az − . n · · · Multiplying the equation by z, we get zS = az + az2 + az3 + + azn. n · · · Subtracting these equations, we get S zS = a azn n − n − which can be solved for a(1 zn) S = − , n 1 z provided z = 1. Using the results of Example− 5.3, 6 a , z < 1 lim Sn = 1 z | | = n − !1 (diverges, z 1 ) | | ≥ a 1 n 1 2 , z < 1 az − = a + az + az + = 1 z | | . · · · − n=1 (diverges, z 1 X | | ≥ Theorem (5.5). A complex series converges the series of real and () imaginary parts converge. Precisely, if cn = xn +yni and S = X +Y i, then 1 1 1 c = S x = X and y = Y. n () n n n=1 n=1 n=1 X X X 5.2. INFINITE SERIES OF CONSTANTS 119 1 ( 1)n Example. Does z converge where z = 2 + i − ? n n − n2 n=1 X n Solution. xn = 2 converges since it is a constant sequence. ( 1) takes − ( 1)n − on the values 1. Thus y = − converges to 0, so by Theorem 5.5, ± n n2 1 ( 1)n 2 + i − = 2. − n2 − ⇤ n=1 X ✓ ◆ 1 1 Theorem (5.6). If cn = C and dn = D, then n=1 n=1 X X 1 (i) (c d ) = C D, n ± n ± n=1 X 1 (ii) kcn = kC (where k is a constant). n=1 X 120 5. POWER SERIES AND LAURENT SERIES 1 Theorem (5.7). If cn converges, then lim cn = 0. n n=1 !1 X 1 Proof. If cn converges to S, its sequence of partial sums n=1 X S1, S2, . , Sn . has limit S, But then the sequence ⇤ 0, S1, S2, . , Sn 1, . − also converges to S. When we subtract these sequences term-by-term, the sequence S1 0, S2 S1, . , Sn Sn 1, . − − − − has limit S S = 0 by Theorem 5.1. Since cn = Sn Sn 1, lim cn = 0. − n − − !1 Corollary (1 – nth term test). If lim cn = 0 or does not exist, then n !1 6 1 cn diverges. n=1 X 1 Definition (5.2). A complex series cn is said to be absolutely con- n=1 X 1 vergent (or converge absolutely) if c converges. | n| n=1 X 5.2. INFINITE SERIES OF CONSTANTS 121 Theorem 1 (5.8). If a complex series converges absolutely, then it con- verges. 1 Proof. Suppose the series cn converges absolutely where cn = an +bni. n=1 X 1 1 This means c = a2 + b2 has a sum, call it S. | n| n n n=1 n=1 X X p 1 Now each term in the series a is nonnegative and is less than or equal to | n| n=1 X 1 2 2 the corresponding term in an + bn . n=1 X p 1 1 1 It follows that a converges = the real series a of real parts of c | n| ) n n n=1 n=1 n=1 converges absolutelyX . X X 1 Similarly, the series of imaginary parts bn converges absolutely. n=1 X 1 By Theorem 5.5, cn converges. ⇤ n=1 X 1 By this theorem, a complex series cn converges when its series of moduli n=1 X 1 c converges. Since this latter series is a series of nonnegative reals, we | n| n=1 canX use the tests for such series in real variable theory. However, we will phrase them in the context of complex series and absolute convergence. 122 5. POWER SERIES AND LAURENT SERIES Recall. (1) For c = 1, 6 1 1 cn+1 1 cn = lim − = if c < 1, n 1 c 1 c | | n=0 !1 X − − 1 and cn diverges if c 1. | | ≥ n=0 X 1 1 (2) The harmonic series diverges. n n=1 X 1 1 (3) The p-series converges for p > 1. np n=1 X 1 1 (4) The p-series diverges for p 1. np n=1 X 1 Theorem (5.9 – Comparison Test). If c a for all n N and a | n| n ≥ n n=1 X 1 converges, then cn converges absolutely. n=1 X 5.2. INFINITE SERIES OF CONSTANTS 123 1 Theorem (5.10 – Limit Comparison Test). If bn is a convergent series n=1 of positive real numbers, and X c lim | n| = l, 0 l < , n !1 bn 1 1 then cn converges absolutely. n=1 X Theorem (5.11 – Integral Test). If cn = f(n) and f(x) is a continuous, decreasing function for x N such that| | ≥ 1 f(x) dx < , 1 ZN 1 then cn converges absolutely. n=1 X cn+1 Theorem (5.12 –Limit Ratio Test). Suppose cn = 0 and lim = L. 6 n cn Then !1 1 (i) cn converges absolutely if L < 1; n=1 X 1 cn+1 (ii) cn diverges if L > 1 or if lim = ; n cn 1 n=1 !1 X 1 (iii) cn may converge or diverge if L = 1. n=1 X 124 5. POWER SERIES AND LAURENT SERIES n Theorem (5.13 – Limit Root Test). Suppose lim cn = L. Then n !1 | | 1 p (i) cn converges absolutely if L < 1; n=1 X 1 n (ii) cn diverges if L > 1 or if lim cn = ; n | | 1 n=1 !1 X p 1 (iii) cn may converge or diverge if L = 1.
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