SOLUTIONS OF HW5
MINGFENG ZHAO
March 18, 2011
1. [Problem 1, in Page 106] If the series u1 +u2 +· · · is convergent, then there exists a positive number
M such that the inequality
|uν + uν+1 + · · · + uν+p | < M.
holds for all ν and p.
Proof. Let sn =
Pn
k=1
uk . By the assumption, we assume sn is convergent to s ∈ C. Then {sn }∞
n=1
is Cauchy, that is, for any > 0, there exists N ∈ N such that whenever n, m ≥ N , we have
|sn − sm | < .
Take = 1, then there exists N ∈ N such that whenever n, m ≥ N , we have
|sn − sm | < 1.
(1)
Let
MN = max {|si − sj | : 1 ≤ i, j ≤ N.}
and M = MN + 1.
Claim: |sn − sm | ≤ M for all n, m ∈ N.
In fact, for any n, m ≥ 1, without loss of generality, we can assume n ≥ m ≥ 1. We have three
cases:
Case I: n ≥ m ≥ N .
By (1), we have |sn − sm | < 1 ≤ M .
Case II: N ≥ N ≥ m ≥ 1.
By the definition of MN , we have |sn − sm | ≤ Mn < M .
Case III: n ≥ N ≥ m ≥ 1.
Then we have |sn − sm | ≤ |sn − sN | + |sN − sm | < 1 + MN = M .
So anyway, we have |sn − sm | ≤ M for all n, m ≥ 1. In particular, we take n = ν + p, m = ν − 1,
we have
ν+p
ν−1
X
X |sν+p − sν−1 | = uk −
uk k=1
1
k=1
2
MINGFENG ZHAO
=
|uν + uν+1 + · · · + uν+p |
<
M.
2. [Problem 2, in Page 106] Prove the Cauchy product rule in Section 7.3: If the series
U=
∞
X
uν ,
and,
V =
ν=1
∞
X
vν .
ν=1
converges, and if at least one of the them converges absolutely, then
UV =
∞
X
(u1 vn + u2 vn−1 + · · · + un v1 ).
n=1
P∞
P∞
Proof. Without loss of generality, we assume i=1 ui absolutely converges, and j=1 vj converges.
P∞
P∞
P∞
And U = i=1 ui , V = j=1 , and W = i=1 |ui |. Let U0 = V0 = 0, W0 = 0, and
Un =
n
X
ui ,
Wn =
i=1
n
X
|ui |,
i=1
Vn =
n
X
vj ,
and Tn =
j=1
n
X
(u1 vk + u2 vk−1 + · · · + uk v1 ).
k=1
Then we have
lim an
=
0
lim Un
=
U
lim Wn
=
W
lim Vn
=
V
n→∞
n→∞
n→∞
n→∞
We also can define Rn =
P∞
k=n+1
vk , then we have Vn + Rn = V , and
lim Rn = 0.
n→∞
And we have
Tn
=
n
X
(u1 vk + u2 vk−1 + · · · + uk v1 )
k=1
= u1 v1 + (u1 v2 + u2 v1 ) + · · · + (u1 vn + u2 vn−1 + · · · + un v1 )
= u1 (v1 + v2 + · · · + vn ) + u2 (v1 + · · · + vn−1 ) + · · · + un v1
= u1 Vn + u2 Vn−1 + · · · + un V1
= u1 (V − Rn ) + u2 (V − Rn−1 ) + · · · + un (V − R1
SOLUTIONS OF HW5
=
3
(u1 + u2 + · · · + un )V − u1 Rn − u2 Rn−2 − · · · − un R1
= Un V − u1 Rn − u2 Rn−2 − · · · − un R1 .
Let Zn = u1 Rn + u2 Rn−2 + · · · + un R1 . Since limn→∞ Rn = 0, then for any > 0, there exists
N ∈ N such that whenever n ≥ N , we have
|Rn | < .
So we have
|Zn | = |u1 Rn + u2 Rn−2 + · · · + un R1 |
≤
|u1 ||Rn | + |u2 ||Rn−2 | + · · · + |un−N ||RN | + |un−N +1 ||RN −1 | · · · + |un ||R1 |
≤
|u1 | + |u2 | + · · · + |un−N | + |un−N +1 ||RN −1 | + · · · + |un ||R1 |
=
(|u1 | + |u2 | + · · · + |un−N |) + |un−N +1 ||RN −1 | + · · · + |un ||R1 |
= Wn−N + |un−N +1 ||RN −1 | + · · · + |un ||R1 |
≤ W + |un−N +1 ||RN −1 | + · · · + |un ||R1 |. Since Wn increasingly converges to W
Fix N , take n → ∞, we have
lim sup |Zn |
≤ W + lim sup(|un−N +1 ||RN −1 | + · · · + |un ||R1 |)
n→∞
n→∞
= W + lim sup |un−N +1 ||RN −1 | + · · · + lim sup |un ||R1 |
n→∞
n→∞
= W . Notice N is fixed.
By the arbitrary of , we get
lim sup |Zn | = 0.
n→∞
So we have
lim Tn
n→∞
=
=
lim (Un V − u1 Rn − u2 Rn−2 − · · · − un R1 )
n→∞
lim (Un V − Zn )
n→∞
=
V lim Un − lim Zn
=
U V.
n→∞
n→∞
4
MINGFENG ZHAO
3. [Problem 5, in Page 106] Expand
1
z+a
into increasing powers of (z − b) and of
1
z−b ,
and determine
the domain of convergence of each of the resulting series.
Proof. a. If b = −a, then
z−b a+b < 1, we have
1
z+a
1
z+a
=
=
=
=
1
z−(−a) .
So without loss of generality, we assume b 6= −a. If
1
z−b+b+a
1
1
Since a + b 6= 0
·
z−b
a + b 1 + a+b
"∞
k #
X
1
z−b
k
if
·
(−1)
a+b
a+b
k=0
=
∞
X
k=0
z − b
a + b < 1
(−1)k
(z − b)k .
(a + b)k+1
Figure 1. |z − b| < |a + b|
The domain of convergence is: {z ∈ C : |z − b| < |a + b|}
z−b And if a+b
> 1, we have
1
z+a
=
=
=
1
z−b+b+a
1
1
·
Since |z − b| > 0
z − b 1 + a+b
z−b
"∞
k #
X
a + b
a+b
1
k
<1
·
(−1)
if z−b
z−b
z − b
k=0
=
∞
X
k=0
(−1)k (a + b)k
.
(z − b)k+1
The domain of convergence is: {z ∈ C : |z − b| > |a + b|}
SOLUTIONS OF HW5
5
Figure 2. |z − b| > |a + b|
4. [Problem 6, in Page 106] Expand the rational function
1
z 2 +1
into increasing powers of the difference
z − 1 and determine the domain of convergence of the series.
Proof. Since
1
z 2 +i
=
1
(z+i)(z−i)
=
i 1
2 z+i
−
i 1
2 z−i .
And when |z − 1| < |1 − i| = |1 + i| =
Problem 5, in Page 106, we have
1
z+i
=
1
z−i
=
∞
X
k=0
∞
X
k=0
(−1)k
(z − 1)k
(i + 1)k+1
(−1)k
(z − 1)k
(−i + 1)k+1
Figure 3. |z − 1| < |1 − i| = |1 + i| =
√
2
√
2, by the
6
MINGFENG ZHAO
And the domain of convergence is: {z ∈ C : |z − 1| < |1 − i| = |1 + i| =
1
2
z +i
=
=
=
=
√
2}. Hence we can get
∞
∞
i X (−1)k
iX
(−1)k
k
(z
−
1)
(z − 1)k
−
2
(i + 1)k+1
2
(−i + 1)k+1
k=0
k=0
∞ k
k
X
(−1)
i
(−1)
−
(z − 1)k
2
(i + 1)k+1
(−i + 1)k+1
k=0
∞ iX
(−1)k ik+1
(−1)k
− k+1
(z − 1)k
2
(i + 1)k+1
i
(−i + 1)k+1
k=0
∞ (−1)k ik+1
iX
(−1)k
−
(z − 1)k
2
(i + 1)k+1
(i + 1)k+1
k=0
∞
=
i X (−1)k + (−i)k+1
(z − 1)k
2
(i + 1)k+1
k=0
5. [Problem 8, in Page 107] Set z = r(cos φ + i sin φ), in the series
∞
X
zn,
n=1
and split each term into its real and imaginary parts. Form the series of real parts and the series of
imaginary parts, and find the sum of each when it converges.
Proof. We know z n = (r cos θ + ir sin θ)n = rn cos nθ + irn sin nθ. We have
∞
X
zn
=
n=1
∞
X
(rn cos nθ + irn sin nθ)
n=1
=
∞
X
rn cos nθ + i
n=1
So the series of real parts is:
P∞
n=1
∞
X
rn sin nθ.
n=1
rn cos nθ, and the series of imaginary parts is:
When |z| < 1, we know
∞
X
zn
=
z
1−z
=
z(1 + z
|1 − z|2
=
z + |z|2
|1 − z|2
=
Re z + |z|2
Im z
+i
2
|1 − z|
|1 − z|2
n=1
P∞
n=1
rn sin nθ.
SOLUTIONS OF HW5
7
Since z = r cos θ + ir sin θ, then |z| = r, and |1 − z|2 = |1 − r cos θ + ir sin θ|2 = (1 − r cos θ)2 +
r2 sin2 θ = 1 + r2 − 2r cos θ. Hence we have
∞
X
rn cos nθ
=
r cos θ + r2
1 + r2 − 2r cos θ
rn sin nθ
=
r sin θ
.
1 + − 2r cos θ
n=1
∞
X
n=1
r2
6. [Problem 11, in Page 107] Determine the radii of convergence of the following power series:
a)
X
p n
n z ,
Proof. Recall for the series
X zn
b)
,
nn
P∞
n=1 cn z
n
c)
X
n2 n
q z (|q| < 1),
1
.
R = lim inf p
n→∞ n |c |
n
a. We have cn = np , then
R
1
lim inf √
n→∞ n np
1 −p
lim n n
=
=
1
nn ,
z
2 + (−1)n
n
.
, the radii of convergence formuale (Theorem 4, in Page 103)
is given by
b. We have cn =
d)
X
n→∞
=
1.
=
1
lim inf q
then
R
=
n→∞
n
1
nn
lim n
n→∞
= ∞.
2
c. We have cn = q n , |q| < 1, then
R
=
=
1
lim inf q n→∞ n n2 q
lim
n→∞
1
|q|n
= ∞. Since |q| < 1
8
MINGFENG ZHAO
d. We have cn =
1
(2+(−1))n ,
|q| < 1, then
R
1
lim inf r
n→∞
1
n (2+(−1))
n
=
=
lim inf q
n→∞
n
1
1
(2+(−1))n
=
lim inf 2 + (−1)n
=
1.
n→∞
7. [Problem 15, in Page 107] Prove that is
|an |
|an+1 |
ρ is the radius of convergence of the series
P
Proof. Assume limn→∞
|an |
|an+1 |
tends to a finite limit ρ, as n tends to infinity, then
an z n .
= ρ, and 0 ≤ ρ < ∞. Recall for the series
P∞
n=1
an z n , the radii of
convergence formuale (Theorem 4, in Page 103) is given by
1
.
R = lim inf p
n→∞ n |a |
n
Since limn→∞
|an |
|an+1 |
1
n|
= ρ, then we have limn→∞ ln |a|an−1
| = ln ρ . Hence, we have
|an−1 |
|a1 |
n|
ln |a|an−1
ln |an |
1
| + ln |an−2 | + · · · + ln |1|
= lim
= ln .
n→∞
n→∞
n
n
ρ
lim
1
So limn→∞ |an | n = ρ1 . Hence we can get
1
R = lim inf p
= ρ.
n→∞ n |a |
n
Remark 1. If limn→∞ an = a, −∞ ≤ a ≤ ∞, then
lim
n→∞
a1 + a2 + · · · + an
= a.
n
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu