19.1 Sequences and Series
INTRODUCTION
Much of the theory of complex sequences and series is analogous to that encountered in real calculus. In this section we explore the definitions of convergence and divergence for complex sequences and complex infinite series. In addition, we give some tests for convergence of infinite series. You are urged to pay special attention to what is said about geometric series since this type of series will be important in the later sections of this chapter.
Sequences
A sequence {zn} is a function whose domain is the set of positive integers; in other words, to each integer n = 1, 2, 3, . . ., we assign a complex number zn. For example, the sequence {1 + in} is
(1)
If limn→∞ zn = L we say the sequence {zn} is convergent. In other words, {zn} converges to the number L if, for each positive number ϵ, an N can be found such that |zn − L| < ϵ whenever n > N. As shown in FIGURE 19.1.1, when a sequence {zn} converges to L, all but a finite number of the terms of the sequence are within every ϵ-neighborhood of L. The sequence {1 + in} illustrated in (1) is divergent, since the general term zn = 1 + in does not approach a fixed complex number as n → ∞. Indeed, the first four terms of this sequence repeat endlessly as n increases.
EXAMPLE 1 A Convergent Sequence
The sequence converges, since
As we see from
and FIGURE 19.1.2, the terms of the sequence spiral toward the point z = 0. ≡
The following theorem should make intuitive sense.
Theorem 19.1.1 Criterion for Convergence
A sequence {zn} converges to a complex number L if and only if Re(zn) converges to Re(L) and Im(zn) converges to Im(L).
EXAMPLE 2 Illustrating Theorem 19.1.1
The sequence converges to i. Note that Re(i) = 0 and Im(i) = 1. Then from
we see that Re(zn) = 2n/(n2 + 4) → 0 and Im(zn) = n2/(n2 + 4) → 1 as n → ∞. ≡
Series
An infinite series of complex numbers
is convergent if the sequence of partial sums {Sn}, where
Sn = z1 + z2 + z3 + … + zn,
converges. If Sn → L as n → ∞, we say that the sum of the series is L.
Geometric Series
For the geometric series
(2)
the nth term of the sequence of partial sums is
Sn = a + az + az2 + … + azn−1. (3)
By multiplying Sn by z and subtracting this result from Sn, we obtain Sn − zSn = a − azn. Solving for Sn gives
(4)
Since zn → 0 as n → ∞, whenever |z| < 1 we conclude from (4) that (2) converges to
when |z| < 1; the series diverges when |z| ≥ 1. The special geometric series
(5)
(6)
valid for |z| < 1, will be of particular usefulness in the next two sections. In addition, we shall use
(7)
in the alternative form
(8)
in the proofs of the two principal theorems of this chapter.
EXAMPLE 3 Convergent Geometric Series
The series
is a geometric series with a = (1 + 2i)/5 and z = (1 + 2i)/5. Since |z| = , the series converges and we can write
≡
Theorem 19.1.2 Necessary Condition for Convergence
If zk converges, then limn→∞ zn = 0.
An equivalent form of Theorem 19.1.2 is the familiar nth term test for divergence of an infinite series.
Theorem 19.1.3 The nth Term Test for Divergence
If limn→∞ zn ≠ 0, then the series diverges.
For example, the series (k + 5i)/k diverges since zn = (n + 5i)/n → 1 as n → ∞. The geometric series (2) diverges when |z| ≥ 1, since, in this case, limn→∞ |zn| does not exist.
DEFINITION 19.1.1 Absolute Convergence
An infinite series is said to be absolutely convergent if converges.
EXAMPLE 4 Absolute Convergence
The series (ik/k2) is absolutely convergent since |ik/k2| = 1/k2 and the real series (1/k2) converges. Recall from calculus that a real series of the form (1/kp) is called a p-series, p a real number, and converges for p > 1 and diverges for p ≤ 1. ≡
As in real calculus,
Absolute convergence implies convergence.
Thus in Example 4, because the series
converges absolutely, it is also convergent.
The following two tests are the complex versions of the ratio and root tests that are encountered in calculus:
Theorem 19.1.4 Ratio Test
Suppose zk is a series of nonzero complex terms such that
(9)
(i) If L < 1, then the series converges absolutely.
(ii) If L > 1 or L = ∞, then the series diverges.
(iii) If L = 1, the test is inconclusive.
Theorem 19.1.5 Root Test
Suppose zk is a series of complex terms such that
(10)
(i) If L < 1, then the series converges absolutely.
(ii) If L > 1 or L = ∞, then the series diverges.
(iii) If L = 1, the test is inconclusive.
We are interested primarily in applying these tests to power series.
Power Series
The notion of a power series is important in the study of analytic functions. An infinite series of the form
(11)
where the coefficients ak are complex constants, is called a power series in z − z0. The power series (11) is said to be centered at z0, and the complex point z0 is referred to as the center of the series. In (11), it is also convenient to define (z − z0)0 = 1 even when z = z0.
Circle of Convergence
Every complex power series has radius of convergence R, where R is a real number. Analogous to the concept of an interval of convergence in real calculus, when 0 < R < ∞, a complex power series (11) has a circle of convergence defined by |z − z0| = R. The power series converges absolutely for all z satisfying |z − z0| < R and diverges for |z − z0| > R. See FIGURE 19.1.3. The radius R of convergence can be
- zero (in which case (11) converges at only z = z0),
- a finite number (in which case (11) converges at all interior points of the circle |z − z0| = R), or
- ∞ (in which case (11) converges for all z).
A power series may converge at some, all, or none of the points on the actual circle of convergence.
EXAMPLE 5 Circle of Convergence
Consider the power series . By the ratio test (9),
Thus the series converges absolutely for |z| < 1. The circle of convergence is |z| = 1 and the radius of convergence is R = 1. Note that on the circle of convergence, the series does not converge absolutely, since the series of absolute values is the well-known divergent harmonic series (1/k). Bear in mind this does not say, however, that the series diverges on the circle of convergence. In fact, at z = −1, ((−1)k+1/k) is the convergent alternating harmonic series. Indeed, it can be shown that the series converges at all points on the circle |z| = 1 except at z = 1. ≡
It should be clear from Theorem 19.1.4 and Example 5 that for a power series ak(z − z0)k, the limit (9) depends on only the coefficients ak. Thus, if
(i) = L ≠ 0, the radius of convergence is R = 1/L;
(ii) = 0, the radius of convergence is ∞;
(iii) = ∞, the radius of convergence is R = 0.
Similar remarks can be made for the root test (10) by utilizing .
EXAMPLE 6 Radius of Convergence
Consider the power series . Identifying an = (−1)n + 1/n!, we have
Thus the radius of convergence is ∞; the power series with center 1 + i converges absolutely for all z. ≡
EXAMPLE 7 Radius of Convergence
Consider the power series . With an = , the root test in the form
shows that the radius of convergence of the series is R = . The circle of convergence is |z − 2i| = ; the series converges absolutely for |z − 2i| < . ≡
19.1 Exercises Answers to selected odd-numbered problems begin on page ANS-47.
In Problems 1–4, write out the first five terms of the given sequence.
- {5in}
- {2 + (−i)n}
- {1 + enπi}
- {(1 + i)n} [Hint: Write in polar form.]
In Problems 5–10, determine whether the given sequence converges or diverges.
- {e1/n + 2(tan−1n)i}
In Problems 11 and 12, show that the given sequence {zn} converges to a complex number L by computing limn→∞ Re(zn) and limn→∞ Im(zn).
In Problems 13 and 14, use the sequence of partial sums to show that the given series is convergent.
In Problems 15–20, determine whether the given geometric series is convergent or divergent. If convergent, find its sum.
In Problems 21–28, find the circle and radius of convergence of the given power series.
- Show that the power series is not absolutely convergent on its circle of convergence. Determine at least one point on the circle of convergence at which the power series converges.
-
- Show that the power series converges at every point on its circle of convergence.
- Show that the power series diverges at every point on its circle of convergence.