Series
These notes review the most basic definitions and properties of series.
Definition 1
(a) A sequence, sℓ ℓ∈IN = s1 , s2 , s3 , · · · , is an (ordered) infinite list of numbers.
(b) A sequence sℓ ℓ∈IN is said to converge to S, denoted lim sℓ = S, if sℓ approaches S
as ℓ tends to infinity. The full technical definition is
∀ ε > 0 ∃ L > 0 such that
ℓ→∞
sℓ − S < ε whenever ℓ > L
A sequence which does not converge is said to diverge.
(c) A series is an infinite sum
∞
P
an .
n=0
∞
∞
P
P
(d) A series
an is said to converge to S, denoted
an = S, if the sequence of partial
n=0
n=0
n
o
ℓ
P
sums, sℓ =
an
, converges to S. A series which does not converge is said to
n=0
diverge.
(e) A series
number.
∞
P
ℓ∈IN
an is said to converge absolutely if the series
n=0
∞
P
|an | converges to a finite
n=0
(f) A power series is a series of the form
∞
P
an (z − z0 )n .
n=0
Example 2 (The Geometric Series) The geometric series is the power series
∞
P
az n .
n=0
If z 6= 1, then
(1 − z)(1 + z + z 2 + · · · + z ℓ ) = 1 − z ℓ+1
(just multiply out the left hand side) so that
ℓ
X
n=0
az n = a
1 − z ℓ+1
1−z
As ℓ tends to ∞, z ℓ+1 tends to zero if |z| < 1 and tends to infinity if |z| > 1. So,
1
if |z| < 1 and diverges if |z| > 1. If |z| = 1, the
the geometric series converges to a 1−z
geometric series diverges too, by Theorem 3.a, below.
c Joel Feldman.
2012. All rights reserved.
March 4, 2012
Series
1
Theorem 3 Let {an }n∈IN∪{0} , {cn }n∈IN∪{0} and {dn }n∈IN∪{0} be sequences of complex
numbers. Then
∞
P
(a) (Divergence Test or nth Term Test) If
an converges, then lim an = 0.
n→∞
n=0
(b) If
∞
P
an converges absolutely, then it converges too.
n=0
(c) (Comparison Test) Let N0 ∈ IN.
(i) If |an | ≤ cn for all n ≥ N0 and
∞
P
cn converges, then
n=0
(ii) If an ≥ dn ≥ 0 for all n ≥ N0 and
∞
P
an converges absolutely.
n=0
∞
P
dn diverges, then
n=0
∞
P
an diverges.
n=0
∞
P
(d) (Radius of Convergence) Each power series
an (z − z0 )n has a characteristic “num-
n=0
ber”, 0 ≤ R ≤ ∞, called its radius of convergence, with the property that
◦ the series converges absolutely for each z ∈ C obeying |z − z0 | < R and
◦ the series diverges for each z ∈ C obeying |z − z0 | > R.
Remark. The radius of convergence of a power series provides no information about
whether or not the series converges when |z − z0 | = R.
Proof:
Denote the partial sums sℓ =
Pℓ
n=0
an .
(a) Assume that the series converges to S. As n → ∞, n − 1 → ∞ too, so that
lim sn = lim sn−1 = S
n→∞
n→∞
and
lim an = lim (sn − sn−1 ) = lim sn − lim sn−1 = S − S = 0
n→∞
n→∞
n→∞
n→∞
P∞ P∞ Pℓ (b) Denote by tℓ = n=0 an the partial sums for n=0 an . By hypothesis, n=0 an converges to some finite number T . So
∞
X
an = lim T − tℓ = 0
lim
ℓ→∞
n=ℓ+1
ℓ→∞
and for all m > 0
ℓ+m
∞
X
X X
ℓ+m
ℓ→∞
sℓ+m − sℓ = an −→ 0
a
≤
a
≤
n
n
n=ℓ+1
c Joel Feldman.
2012. All rights reserved.
n=ℓ+1
n=ℓ+1
March 4, 2012
Series
2
This is saying that the partial sums sℓ are getting closer and closer together as ℓ increases.
Sequences sℓ with this property are called “Cauchy sequences”. There is a property of C
(and also a property of IR) called “completeness” which says that every Cauchy sequence
converges. (Roughly speaking if the numbers in a sequence get closer and closer to each
other then they also get closer and closer to some fixed number.)
n
o
Pℓ
(c.i) The sequence sℓ = n=0 |an | is monotonically increasing (i.e. if ℓ′ > ℓ, then sℓ′ ≥
PN0 −1
P∞
sℓ ) and bounded above by n=0
|an | + n=N0 cn and hence converges (by completeness
P∞
once again). So n=0 an converges absolutely and hence also converges by part (b).
P∞
P∞
(c.ii) If n=0 an were to converge, then n=0 dn would also converge by part (c.i). This
is a contradiction.
(d) Step 1:
We first observe that
◦ if ρ > 0 has the property that there is an N0 ∈ IN such that |an | ≤ ρ1n for all n ≥ N0 ,
◦ then the series converges absolutely for all z ∈ C obeying |z − z0 | < ρ
by the comparison test (part (c.i) of this theorem) with an replaced by an (z − z0 )n and
0| n
. Here we used that if |an | ≤ ρ1n then |an (z − z0 )n | ≤ cn and that, when
cn = |z−z
ρ
P∞
0|
<
1
so
that
|z − z0 | < ρ, we have that |z−z
n=0 cn is a convergent geometric series.
ρ
Step 2:
G=
Denote
ρ > 0 there is an N0 ∈ IN such that |an | ≤
1
ρn
for all n ≥ N0
Observe that if ρ ∈ G and 0 < ρ′ < ρ, then ρ1n < ρ1′ n so that ρ′ ∈ G too. So there are
three possibilities.
◦ G is the empty set. In this case, set R = 0.
◦ G is the entire positive half, (0, ∞), of the real line. In this case, set R = ∞.
◦ G is an interval of the form (0, r) or of the form (0, r] for some 0 < r < ∞. In this
case, set R = r.
In all of these cases, Step 1 says that the series converges absolutely for each z obeying
|z − z0 | < R.
Step 3:
It only remains to prove that if |z − z0 | > R (with the R chosen in Step 2),
then the series diverges. So fix any z ∈ C with |z − z0 | > R and set ρ = |z − z0 |. Then ρ
is not in G. So there does not exist an N0 ∈ IN with |an | ≤ ρ1n for all n ≥ N0 . So there
must exist a sequence n1 , n2 , n3 , · · · of natural numbers, that tend to infinity, such that
P∞
|anℓ | > ρn1ℓ , and hence anℓ (z − z0 )nℓ > 1, for each ℓ. But this forces n=0 an (z − z0 )n
to diverge by the nth term test (part (a) of this theorem).
c Joel Feldman.
2012. All rights reserved.
March 4, 2012
Series
3
P∞
P∞
Theorem 4 If n=0 an z n and n=0 bn z n both converge for |z| < R, then
P∞
(a) so does n=0 (an + bn )z n and
∞
X
(an + bn )z n =
n=0
∞
X
n=0
for all |z| < R and
P∞
(b) so does n=0 can z n and
∞
X
n
can z = c
n=0
for all |z| < R and
P∞
(c) so does n=1 nan z n−1 and
an z n +
∞
X
∞
X
bn z n
n=0
∞
X
an z n
n=0
nan z
n−1
=
d
dz
n=1
∞
X
an z n
n=0
for all |z| < R and
P∞
Pn
(d) so does n=0 can z n where cn = ℓ=0 aℓ bn−ℓ and
∞
X
n
cn z =
n=0
X
∞
an z
n=0
n
X
∞
for all |z| < R and
(e) if Γ is a contour in z ∈ C |z| < R , then
Z hX
∞
Γ
c Joel Feldman.
n=0
2012. All rights reserved.
an z
n
i
bn z
n=0
dz =
∞
X
n=0
an
Z
n
z n dz
Γ
March 4, 2012
Series
4