Power Series
3.1. Complex power series
Let A be a subset of C, and let {fn} = {fn}∞n=0 be a sequence of functions defined on A (in the previous chapter, sequences were indexed by Z>0; for convenience, in this chapter, they will be indexed by Z≥0). We form the new sequence (known as a series) {SN}, where SN(z) =
N n=0
fn(z), and the formal infinite series ∞ n=0
fn(z) .
The sequence of complex numbers {SN(z)}∞n=0 is also known as the sequence of partial sums associated with the infinite series
∞ n=0
fn(z) at the point z ∈ A. When the indices of summation are clear from the context, we often omit them. For example,
fn(ζ) usually means the infinite sum
∞ n=0
fn(ζ). Other similar abbreviations are used.
Definition 3.1. i) We say that the infinite series ∞ n=0
fn(z) con- verges at a pointζ ∈Aif{SN(ζ)}converges. In this case, we write ∞
n=0
fn(ζ) = lim
n→∞SN(ζ).
ii) We say that the infinite series ∞ n=0
fn(z)converges pointwise inA if {SN(ζ)} converges for everyζ ∈A.
iii) We say that the infinite series ∞ n=0
fn(z) converges absolutely at a point ζ ∈A if the infinite series
∞ n=0
|fn(ζ)|converges.
iv) We say that the infinite series ∞ n=0
fn(z) converges uniformly in A if the sequence of partial sums {SN(z)} converges uniformly in A.
3.1. COMPLEX POWER SERIES 25
v) We say that the infinite series ∞ n=0
fn(z)converges normally on a set B ⊂ A if there exists a sequence of positive constants {Mn} such that
(1) |fn(z)| ≤Mn for allz ∈B and all n, and
(2)
Mn<∞.
vi) We say that the infinite series ∞ n=0
fn(z) diverges at a point of A if it does not converge at that point.
We speak of the pointwise, uniform, absolute, or normal conver- gence of a series as well as of thedivergence of a series.
Remark 3.2. The fact that the sequence {SN(ζ)}converges if and only if {SN(ζ)} is Cauchy allows us to rephrase conditions (i–iv) as if and only if statements when useful: (i)
fn(z) converges at a point ζ ∈Aif and only if{SN(ζ)}is Cauchy; (ii)
fn(z)converges pointwise in A if and only if {SN(ζ)} is Cauchy for every ζ ∈ A; (iii)
fn(z) converges absolutely at a point ζ ∈ A if and only if the infinite series |fn(ζ)| converges; and (iv)
fn(z) converges uniformly inA if and only if the sequence of partial sums {SN(z)}converges uniformly inA.
Remark3.3. Many questions on convergence of complex sequences are reduced to the real case by the trivial but important observation that absolute convergence at a point implies convergence at that point.
Remark 3.4. Two other such observations, both trivial but impor- tant, are that if an infinite series
∞ n=0
fn(z) converges at a point ζ, then
nlim→∞fn(ζ) = 0 and lim
n→∞
∞ k=n
fk(ζ) = 0.
Some relationships between some different types of convergence of a series are given in the following result.
Weierstrass M-test. Normal convergence implies uniform and absolute convergence.
Proof. With notation as in the last definition, if N1 < N are positive integers, then
|SN(z)−SN1(z)| ≤ N n=N1+1
Mn for all z ∈A (needed for the uniform convergence argument), and
| |SN(z)| − |SN1(z)| | ≤ N n=N1+1
Mn for allz ∈A (needed for the absolute convergence argument).
Given any > 0, we can find a positive integer N0 such that N >
N1 > N0 implies N n=N1+1
Mn< .
We shall be mostly interested in series of the form ∞
n=0anzn with an∈C(these are known aspower series) and theassociated real-valued series ∞
n=0|an|rn where r =|z|. We define, for each N ∈Z≥0, SN∗ (r) =
N n=0
|an|rn for r∈R≥0, and we observe that
SN∗+1(r)≥ SN∗(r) for all N ∈Z≥0 and for all r ∈R≥0 . We refer to SN∗(r) as thereal partial sum at r.
An elementary but most important example is provided by the geo- metric series with r ∈R≥0:
1 +r+r2+. . . .
Note thatSN(1) =N + 1, and SN(r) = 1−rN+1
1−r for 0≤r <1, as well as for r >1.
Thus ∞ n=0
rn = 1
1−r if 0≤r <1 and
rn diverges if r≥1.
We now introduce two special cases of divergence of a sequence of real numbers. It will be useful to regard these sequences as convergent sequences with infinite limits.
3.1. COMPLEX POWER SERIES 27
Definition3.5. A sequence of real numbers{bn}converges to+∞ if and only if for all M >0 there exists an N ∈Z>0 such thatbn > M for all n > N. In this case we shall write limn→∞bn = +∞. A similar definition applies to real sequences converging to −∞.
With this notation either:
(a) lim
N→∞SN∗(r) exists and is finite: That is,∞
n=0|an|rnconverges.
In this case we write ∞
n=0|an|rn<+∞; or
(b) lim
N→∞SN∗(r) = +∞: That is,∞
n=0|an|rn diverges (in the pre- vious sense). In this case we write ∞
n=0|an|rn= +∞. For real sequences, we have thecomparison test. Let 0≤an ≤bn.
(a) If
an= +∞, then
bn = +∞. (b) If
bn<+∞, then
an <+∞.
Abel’s lemma. Let 0< r < r0. Assume there exists an M ∈R>0
such that
|anrn0| ≤M for all n∈Z>0 . Then the series
anznconverges normally for all zwith |z| ≤r. In particular, it converges absolutely and uniformly for all z with |z| ≤r.
Proof. For|z| ≤r we have
|anzn|=|an| |z|n ≤ |an| |r|n=|an| r
r0
n
rn0 ≤M r
r0
n
. Comparison with the geometric series [or an application of the Weierstrass M-test with Mn = M(rr
0)n] shows the normal conver-
gence.
IfSis any nonempty set of real numbers, then the least upper bound or supremum of S is denoted by supS and the greatest lower bound or infimum is denoted by infS. The possibilities that supS = +∞ or infS=−∞ are allowed.
Definition 3.6. The radius of convergence ρ of the power series anzn is given by
ρ= sup{r ≥0;
|an|rn <+∞} .
Note that 0 ≤ ρ ≤+∞. As a result of the next theorem it makes sense to define{z ∈C; |z|< ρ}as thedisk of convergenceof the power series
anzn.
Theorem 3.7. Let
anznbe a power series with radius of conver- gence ρ.
(a) If 0< r < ρ, then
anzn converges normally, absolutely, and uniformly for |z| ≤r.
(b) The series
anzn diverges for |z|> ρ.
Proof. (a) Choose r0 with r < r0 < ρsuch that
|an|r0n<+∞. Thus there exists an M > 0 with |an|rn0 ≤ M for all n in Z>0. Now apply Abel’s lemma.
(b) We claim that for |z| > ρ, the sequence {|an| |z|n} is not even bounded. Otherwise Abel’s lemma (with r0 = |z|) would guarantee the existence of an r with ρ < r < |z| and
|an|rn < +∞. This
contradicts the definition of ρ.
Corollary 3.8. Let
anzn be a power series with radius of con- vergence ρ. Then the function defined by S(z) =
anzn is continuous for |z|< ρ.
Proof. It follows immediately from Theorems 2.23 and 3.7.
We now turn to the obvious and important question: How do we compute ρ?
To answer this question, we introduce the concepts of lim sup and lim inf.
Definition 3.9. Let {un} be a real sequence. We use ≡ to indi- cate equivalent names [we also use this same notation with a different meaning in other places, such as f ≡ 0 or f ≡ g, to emphasize that these functions are (identically) equal] and define
limn un≡lim sup
n
un≡ upper limit of {un}
≡ limit superior of {un}= lim
p→∞sup
n≥p{un}= inf
p sup
n≥p{un}, and
lim
n
un≡lim inf
n un≡ lower limit of {un}
≡ limit inferior of {un}= lim
p→∞inf
n≥p{un}= sup
p
ninf≥p{un}. Note that every real sequence has a limit superior as well as a limit inferior, which are either real numbers or +∞or −∞.
Properties of limits superior and inferior. Let{un}and {vn} be real sequences. Then:
3.1. COMPLEX POWER SERIES 29
(a) lim inf
n un≤lim sup
n
un. (b) lim inf
n (−un) =−lim sup
n
un. (c) If r >0, then
lim inf
n (run) =rlim inf
n un, and lim sup
n
(run) =rlim sup
n
un. (d) If un≤vn for all n, then
lim inf
n un≤lim inf
n vn, and lim sup
n
un≤lim sup
n
vn. (e) lim
n un =L if and only if lim inf
n un=L= lim sup
n
un (L is allowed to be±∞ in this context).
(f) lim inf
n (un+vn)≥lim inf
n un+ lim inf
n vn, and lim sup
n
(un+vn)≤lim sup
n
un+ lim sup
n
vn.
Remark and exercise. Iff is a real-valued function of a complex variable, defined on a set S in C and ifζ is a limit point of S, then it is possible to define lim
z→ζf(z) and lim
z→ζ
f(z).
Example 3.10. Let un = sinnπ
2
, n = 0,1,2, . . . . This is the sequence {0,1,0,−1, . . .}. Hence for all p∈Z≥0,
ap = sup
n≥p
un= 1 and thus lim sup
n
un= lim
p→∞ap = 1, and bp = inf
n≥p un=−1 and thus lim inf
n un= lim
p→∞bp =−1.
The ratio test will be well known to most readers (see Exercise 3.4). Suppose that vn>0 for all non-negative integers n.
(a) If lim
n→∞
vn+1
vn =L <1, then
vn converges.
(b) If lim
n→∞
vn+1 vn
=L >1, then
vn diverges.
Perhaps less familiar is the root test. Suppose that vn≥0 all n.
(a) If lim
n (vn)n1 =L <1, then
vn converges.
(b) If lim
n (vn)n1 =L >1, then
vn diverges.
Proof. (a) Choose > 0 such that 0< L+ <1. There exists a P ∈Z>0 such that
sup
n≥p
vn1n
< L+ for all p≥P . Thus
vn<(L+)n for all n≥P,
and comparison with the geometric series yields convergence.
(b) Suppose that
vn<+∞. Then lim
n vn = 0. Thus there exists a P in Z>0 such that n ≥P implies that vn <1 for all n ≥P. Hence (vn)n1 <1 for all n≥P and therefore lim
n (vn)n1 ≤1.
Remark 3.11. The root test applies (with the same value of L) whenever the ratio test applies. However, the converse is not true.
To see this take a sequence where the ratios are alternately 12 and 18. Then the root test will apply with L = 14, but the sequence of ratios, obviously, will not converge.
Example 3.12. Consider the sequence n−s with s a positive real number. Both the ratio and the root test end up with L = 1. The series diverges (converges to +∞) for 0 < s ≤ 1 and converges for s >1.
We now return to the study of the complex power series ∞
n=0anzn and to the problem of computing the radius of convergence.
Theorem3.13. Let
anzn be a power series. Suppose thatan = 0 for all n and that the limit lim
n
an+1 an
=L exists, with 0≤L≤+∞. Then the radius of convergence ρ of the power series ∞
n=0anzn is
1
L; in other words,
1
ρ = lim
n→∞
an+1
an , where 1
+∞ is to be understood as being = 0.
The hypotheses required for this result to hold are strong. As pointed out in Remark 3.11, the ratio test is stronger than the root test. The next result provides a way of computing the radius of con- vergence for any power series.
3.1. COMPLEX POWER SERIES 31
Theorem 3.14 (Hadamard). The radius of convergence ρ of the power series ∞
n=0anzn is given by 1 ρ = lim
n |an|n1 . Proof. Let L = lim
n |an|1n. Thus lim
n |anrn|1n = rL for all r ≥ 0 and we conclude by the root test that the associated series
|an|rn converges for 0 ≤r < L1 and diverges forr > L1. Thus ρ= L1.
Lemma 3.15. Let
un and
vn be two absolutely convergent se- ries. Define
wn= n
p=0
upvn−p.
Then
wn is absolutely convergent and
wn= (
un) ( vn).
Proof. Letαp =
n≥p
|un| and βp =
n≥p
|vn|. Then limp αp = 0 = lim
p βp . Also
N n=0
|wn| ≤∞
n=0
|un|∞
n=0
|vn|=α0β0 <+∞,
and therefore, ∞ n=0
|wn| < +∞. Thus we have proven the absolute convergence of the new series.
To show the required equality, choose m and n with m ≥ 2n and consider
m k=0
wk− n
k=0
uk
n k=0
vk
=L.
We have to show that L →0 as n → ∞ (we already know each of the above series converges). We rewrite
L=
m k=0
k i=0
uivk−i− n
j=0
n i=0
uivj
.
5
4
3
2
1
-1
k
2 4 i
(a) The first sum
5
4
3
2
1
-1 k
2 4 i
(b) The second sum
Figure 3.1. The (i, k) plane
By looking at the diagrams in the (i, k) plane shown in Figure 3.1, we see that
m k=0
k i=0
uivk−i= m
i=0
m k=i
uivk−i
= m
i=0
ui m
k=i
vk−i = m
i=0
ui
m−i
j=0
vj = m
i=0 m−i
j=0
uivj . Thus we can estimate
L=
n i=0
m−i
j=0
uivj− n
j=0
uivj
+ m i=n+1
m−i
j=0
uivj
≤
n i=0
m−i
j=n+1
uivj +
m i=n+1
m−i
j=0
uivj
≤ ∞
i=0
∞ j=n+1
|ui| |vj|+ ∞ i=n+1
∞ j=0
|ui| |vj|= (α0βn+1+β0αn+1), and the last expression approaches 0 as n goes to∞.