This section is devoted to a very quick discussion of sequences and series of complex numbers.

Transcription

1 1.3. Complex sequences and series: a snapshot This section is devoted to a very quick discussion of sequences and series of complex numbers. Formally a sequence of complex numbers is a function from the set N of positive integers into C. More informally, and perhaps more helpfully, we think of a sequence as a collection of complex numbers indexed by the positive integers, or, equivalently, as an enumeration of a collection of complex numbers. We use the notation {z n : n N} or more simply {z n } to denote a sequence (the choice of n to denote the (dummy) index is not unique; other letters may be used as well). Thus the first element in the sequence is z 1, the second is z 2, and so on. A sequence need not be a set, for we do not assume that the elements in a sequence be distinct. Thus {1, 1, 1,...,1,...} is a sequence, all of whose elements are 1. However, as a set this would have to be written as the singleton {1}. A subsequence of a given sequence {z n } is an enumeration {z n1,z n2,...,z nk,...} where {n 1,n 2,...,n k,...} is a set of positive integers satisfying the condition n 1 <n 2 < n k < n k+1 < ; in other words, {n k : k N} is a strictly increasing sequence of positive integers. Definition (i) A sequence {z n } of complex numbers is said to be bounded if there is a positive number such that z n for every positive integer n. (ii) We say that a sequence {z n } of complex numbers converges to a complex number z, and write lim z n = z, if lim z n z = 0; the latter expression is a familiar limit of real numbers. The number z is then called the limit of the sequence. A sequence is said to diverge if it does not converge to any complex number. Remark (i) Suppose that {a n } is a sequence of nonnegative real numbers. Recall that the statement lim a n = 0 means the following: given ɛ>0, there is a positive integer N, whichmay depend on ɛ, such that a n <ɛfor every n N. Thus the statement lim z n = z (vide supra) is equivalent to the following: given ɛ>0, there is a positive integer N, which may depend on ɛ, such that z n z <ɛfor every n N. Goemetrically, this means that for every positive number ɛ, there is a positive integer N (which may depend on ɛ) such that z n D(z; ɛ) for every n N. (ii) The reader should verify that a sequence {z n } is unbounded if and only if there is a subsequence {z nk : k N} of {z n } such that z nk >kfor every positive integer k. (iii) Suppose that {z n } is a sequence and z is a fixed complex number. The reader should prove that {z n } does not converge to z if and only if there exists a positive number ɛ 0 and a subsequence {z nk : k N} such that z nk z ɛ 0 for every positive integer k. Example (i) If z n := n/(n + i), then z n 1 = i n + i = 1 n + i = 1 n2 +1, and the last quantity tends to zero as n tends to infinity. Thus lim z n =1. (ii) The sequence {i n : n N} is bounded, whereas the sequence {1+ni : n N} is unbounded. (iii) It is an instructive exercise for the reader, especially for one who is a tyro in such matters, to show that the sequence {i n : n N} cannot converge to any complex number. As expected, the convergence of a complex sequence is intimately tied to the convergence of its real and imaginary parts; precisely one has the following. 1

2 Theorem Suppose that {z n } is a sequence of complex numbers, and let z n =: x n + iy n, n N. The following are equivalent: (i) lim z n = z = x + iy. (ii) lim x n = x and lim y n = y. Proof. Assumme that (i) holds, that is, lim z n z = 0. Proposition implies that x n x z n z and y n y z n z, n N, whence the Squeeze Theorem (Sandwich Principle) asserts that lim x n x = 0 = lim y n y. In other words lim x n = x and lim y n = y. Conversely, if lim x n = x and lim y n = y, then lim x n x = 0 = lim y n y. A second appeal to Proposition gives the inequality z n z x n x + y n y, n N, whence lim z n z = 0 (equivalently lim z n = z) via the Squeeze Theorem. The next result records some basic implications of convergence of complex sequences. Theorem Suppose that {z n } is a sequence converging to the complex number z. Thefollowing hold: (i) {z n } is bounded. (ii) lim z n = z. (iii) lim z n = z. Proof. (i) There is a positive number N such that z n z < 1 for every n>n. So the triangle inequality implies that z n = z n z + z z n z + z <1+ z, n > N. (1.3.1) Choosing := max{ z 1,..., z N,1+ z } we find, via (1.3.1), that z n for every positive integer n. Thus{z n }is a bounded sequence. (ii) lim z n z = lim z n z = lim z n z =0. (iii) Corollary implies the relation z n z z n z, n N, and the latter term tends to zero as n tends to infinity. Our next offering sets out the standard limit laws governing the basic algebra of convergent complex sequences. Theorem Suppose that {z n } and {w n } are sequences of complex numbers, and assume that lim z n = z and lim w n = w. Letαbe a fixed complex number. The following hold: 2

3 (i) lim (z n + w n )=z+w. (ii) lim (αz n)=αz. (iii) lim (z nw n )=zw. (iv) If z 0,then lim (1/z n)=1/z. (v) If z 0,then lim (w n/z n )=w/z. Proof. (i) z n + w n z w z n z + w n w, and both summands on the right converge to zero as n tends to infinity. (ii) lim αz n αz = lim α z n z =0. (iii) Theorem 1.3.5(i) provides a positive number such that z n for every positive integer n. Now z n w n zw = z n w n z n w + z n w zw z n w n z n w + z n w zw = z n w n w + w z n z w n w + w z n z, and the last two summands converge to zero as n tends to infinity. (iv) As z > 0, Remark 1.3.2(i) provides a positive integer N such that z n z < z /2 for every n N. Therefore Corollary shows that z n = z (z z n ) z z z n > z /2, n N ; (1.3.2) in particular z n 0foreveryn N and the expression lim (1/z n) is indeed meaningful. Furthermore, (1.3.2) also leads to the relations 1 1 z n z = z z n z n z < 2 z z n z 2, n N, and the last quantity on the right converges to zero as n tends to infinity. (v) Combine (iii) and (iv). We now turn our attention to infinite series of complex numbers. Definition (i) Suppose that {z n } is a complex sequence, and define s n := z k, n N. We say that the infinite series z k converges to a complex number z if lim s n = z. (ii) We say that the infinite series z k converges absolutely if the (real) series z k is convergent. Remark (i) Writing z n = x n + iy n, u n := x k,andv n := y k, we find, thanks to Theorem 1.3.4, that z k = z = x + iy if and only if lim u n = x and lim v n = y; equivalently, z k = z = x + iy if and only if x k = x and y k = y. 3 n n n

4 (ii) Suppose that z k = z, sothatz= lim s n,wheres n = n z k.asz= lim s n 1 as well, one finds that lim z n = lim s n s n 1 =0,i.e., lim z n =0. Thusthen-th term of a convergent series of complex numbers tends to zero as n tends to infinity. The converse of this statement is false in general. (iii) Suppose that z k is absolutely convergent, and let z k = x k + iy k, k N. As x k z k and y k z k, the familiar Comparison Theorem for series of nonnegative terms implies that the series x k and y k are both convergent. So the theory of real series asserts that x k and y k are convergent series. Hence Remark 1.3.8(i) ensures that z k is convergent. Thus an absolutely convergent series of complex numbers is convergent. Once again the converse of this statement is false in general. The complex version of the well-known geometric series is the subject of the next result. Theorem Suppose that z is a fixed complex number. The following hold: (i) lim zn =0if and only if z < 1. (ii) If n is any positive integer, then n n +1, if z =1; z k = 1 z n+1 1 z, if z 1. (iii) If z < 1, then the geometric series z k converges and z k = 1 1 z. (iv) If z 1, then the geometric series z k is divergent. Proof. (i)wenotethat z n 0 = z n which converges to zero (as n tends to infinity) if z < 1. On the other hand, if z 1, then z n = z n 1 for every n, so the sequence {z n : n N} cannot converge to zero. (ii) The result being obvious for z = 1, we assume that z 1. Lets n := n z k,sothat zs n = n z k+1.then(1 z)s n =s n zs n =1 z n+1, whence the result. (iii) If z < 1, then, lim zn+1 = 0 by (i). Therefore (ii) implies that lim n z k 1 z n+1 = lim 1 z = 1 1 z. (iv) If z 1, then lim zn 0 by (i), so the series is divergent (Remark 1.3.8(ii)). 4

5 As the absolute convergence of complex series is really a statement about series of nonnegative real numbers, one is at liberty to use the entire panoply of methods available to analyze the latter. One such result, namely the Comparison Theorem, has been used already; we now discuss yet another. Before stating it, however, we recall that if {x n } is a sequence of nonnegative real numbers, then the statement lim x n =+ means that for every positive number T, there is a positive integer N, which may depend on T, such that x n >T for every n N. Theorem (Ratio Test) Suppose that {z n } is a sequence of nonzero complex numbers. (i) If lim z n+1 z n = L and 0 L<1, then the infinite series z k is absolutely convergent, hence convergent. (ii) If lim z n+1 z n = L with L>1,orif lim z n+1 z n =+,then lim z n =+. In particular the series z k is divergent. Proof. (i) Choose a positive number ɛ such that L < L + ɛ =: α < 1. By assumption there is a positive integer N such that z n+1 z n <αfor every n N. Thus zn+1 <α z N, z N+2 < α z N+1 <α 2 z N, and so on. Inductively one then gets z N+k <α k z N for every positive integer k. As the geometric series α k is convergent (because 0 <α<1) and z N is a fixed number, we conclude, by comparison, that the series z N+k is also convergent. As N is a fixed number, it follows that the same is true of the series z k. (ii) Either assumption guarantees a number β>1andapositive integer M such that z n+1 z n > β for every n M, and this, in turn, implies the estimate z M+k >β k z M for every positive integer k. As lim k βk =+ (because β>1) and z M > 0, we conclude that lim z M+k =+, k and hence that lim z n =+. The divergence of z k now obtains via Remark 1.3.8(ii). The alert reader will have noticed that the previous result is silent about the case L =1. Thisis because no general conclusion can be drawn in this case; the situation is entirely dependent on the specific series at hand. Example Let z be a fixed complex number and let z n := z n /n 2, n N. Theseries is manifestly convergent for z = 0 (the sum being zero), so let us assume that z 0. As lim z n+1 z n = lim n 2 z (n +1) 2 = z, z k we deduce from Theorem that the series (z k /k 2 ) is absolutely convergent for z < 1and divergent for z > 1. Furthermore, when z =1, z k =1/k 2, so the said series also converges absolutely for all such values of z. 5