Advanced Calculus (I)
W EN -C HING L IEN
Department of Mathematics
National Cheng Kung University
W EN -C HING L IEN
Advanced Calculus (I)
6.2 Series with nonnegative terms
Theorem
Suppose that ak ≥ 0 for k ≥ N. Then
∞
P
ak converges if
k=1
and only if its sequence of partial sums {sn } is bounded,
i.e., if and only if there exists a finite number M > 0 such
that
n
X
ak ≤ M for all n ∈ N.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
6.2 Series with nonnegative terms
Theorem
Suppose that ak ≥ 0 for k ≥ N. Then
∞
P
ak converges if
k=1
and only if its sequence of partial sums {sn } is bounded,
i.e., if and only if there exists a finite number M > 0 such
that
n
X
ak ≤ M for all n ∈ N.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
ak converges, then sn
k=1
k=1
converges as n → ∞. Since every convergent sequence
∞
P
ak has bounded partial
is bounded (Theorem 2.8),
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
ak has bounded partial
is bounded (Theorem 2.8),
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
ak has bounded partial
is bounded (Theorem 2.8),
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
Set sn =
n
P
∞
P
ak for n ∈ N. If
k=1
ak converges, then sn
k=1
converges as n → ∞. Since every convergent sequence
∞
P
is bounded (Theorem 2.8),
ak has bounded partial
sums.
k=1
Conversely, suppose that |sn | ≤ M for n ∈ N. Since ak ≥ 0
for k ≥ N, sn is an increasing sequence when n ≥ N.
Hence by the Monotone Convergence Theorem (Theorem
2.19), sn converges. 2
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Integral Test)
Suppose that f : [1, ∞) → R is positive and decreasing on
∞
P
f (k) converges if and only if f is
[1, ∞). Then
k=1
improperly integrable on [1, ∞), i.e., if and only if
Z ∞
f (x)dx < ∞.
1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Integral Test)
Suppose that f : [1, ∞) → R is positive and decreasing on
∞
P
f (k) converges if and only if f is
[1, ∞). Then
k=1
improperly integrable on [1, ∞), i.e., if and only if
Z ∞
f (x)dx < ∞.
1
W EN -C HING L IEN
Advanced Calculus (I)
Corollary: [p-Series Test]
The series
(4)
∞
X
1
kp
k=1
converges if and only if p > 1.
W EN -C HING L IEN
Advanced Calculus (I)
Corollary: [p-Series Test]
The series
(4)
∞
X
1
kp
k=1
converges if and only if p > 1.
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
dx
n1−p − 1
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
n1−p − 1
dx
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
n1−p − 1
dx
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
n1−p − 1
dx
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
dx
n1−p − 1
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
n1−p − 1
dx
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
n1−p − 1
dx
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Proof:
If p = 1 or p ≤ 0, the series diverges. If p > 0 and p 6= 1,
set f (x) = x −p and observe that f 0 (x) = −px −p−1 < 0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,
n
Z ∞
n1−p − 1
dx
x 1−p = lim
= lim
x→∞ 1 − p n→∞ 1 − p
xp
1
1
has a finite limit if and only if 1 − p < 0, it follows from the
Integral Test that (4) converges if and only if p > 1. 2
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
∞
∞
P
P
ak < ∞.
bk < ∞, then
(i)If
k=1
k=1
(ii)If
∞
P
k=1
ak = ∞, then
∞
P
bk = ∞.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
∞
∞
P
P
ak < ∞.
bk < ∞, then
(i)If
k=1
k=1
(ii)If
∞
P
k=1
ak = ∞, then
∞
P
bk = ∞.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
∞
∞
P
P
ak < ∞.
bk < ∞, then
(i)If
k=1
k=1
(ii)If
∞
P
k=1
ak = ∞, then
∞
P
bk = ∞.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
∞
∞
P
P
ak < ∞.
bk < ∞, then
(i)If
k=1
k=1
(ii)If
∞
P
k=1
ak = ∞, then
∞
P
bk = ∞.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
∞
∞
P
P
ak < ∞.
bk < ∞, then
(i)If
k=1
k=1
(ii)If
∞
P
k=1
ak = ∞, then
∞
P
bk = ∞.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Comparison Test)
Suppose that 0 ≤ ak ≤ bk for large k.
∞
∞
P
P
ak < ∞.
bk < ∞, then
(i)If
k=1
k=1
(ii)If
∞
P
k=1
ak = ∞, then
∞
P
bk = ∞.
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Example:
Determine whether the following series converges or
diverges.
r
∞
X
3k
log k
k2 + k
k
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Example:
Determine whether the following series converges or
diverges.
r
∞
X
3k
log k
k2 + k
k
k=1
W EN -C HING L IEN
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k=1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k=1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k=1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k =1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k =1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k =1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k =1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Theorem (Limit Comparison Test)
Suppose that ak ≥ 0 and bk > 0 for large k and
an
L := lim
exists as an extended real number.
n→∞ bn
∞
∞
P
P
bk
ak converges if and only if
(i) If 0 < L < ∞, then
converges.
(ii) If L = 0 and
(iii) If L = 0 and
k=1
k=1
∞
P
ak converges, then
k=1
∞
P
bk diverges, then
k=1
W EN -C HING L IEN
∞
P
ak converges.
k =1
∞
P
ak diverges.
k=1
Advanced Calculus (I)
Example:
√
k
k
(1)
k=1 k
∞
P
(2)
∞
P
k=1
√
1
k3k−1
W EN -C HING L IEN
Advanced Calculus (I)
Example:
√
k
k
(1)
k=1 k
∞
P
(2)
∞
P
k=1
√
1
k3k−1
W EN -C HING L IEN
Advanced Calculus (I)
Example:
√
k
k
(1)
k=1 k
∞
P
(2)
∞
P
k=1
√
1
k3k−1
W EN -C HING L IEN
Advanced Calculus (I)
Thank you.
W EN -C HING L IEN
Advanced Calculus (I)