## Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

## 6.2 Series with nonnegative terms

Theorem

Suppose that ak ≥ 0 for k ≥ N. Then

∞

P

k =1

ak converges if 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

k =1

ak

**≤ M for all n ∈ N.**

## 6.2 Series with nonnegative terms

Theorem

Suppose that ak ≥ 0 for k ≥ N. Then

∞

P

k =1

ak converges if 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

k =1

ak

**≤ M for all n ∈ N.**

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N.** If

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N.**Since ak ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N,snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N.**Since ak ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

Conversely, suppose that |sn**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N,snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19),sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19),sn converges. 2

**Proof:**

Set sn =

n

P

k =1

ak for n ∈**N. If**

∞

P

k =1

ak converges, then sn

converges as n → ∞. Since every convergent sequence is bounded (Theorem 2.8),

∞

P

k =1

ak has bounded partial sums.

**| ≤ M for n ∈ N. Since a**k ≥ 0
for k ≥ N, snis an increasing sequence when n ≥ N.

Hence by the Monotone Convergence Theorem (Theorem 2.19), sn converges. 2

Theorem (Integral Test)

Suppose that f : [1, ∞) →**R is positive and decreasing on**
[1, ∞). Then

∞

P

k =1

f (k ) converges if and only if f is improperly integrable on [1, ∞), i.e., if and only if

Z ∞ 1

f (x )dx < ∞.

Theorem (Integral Test)

Suppose that f : [1, ∞) →**R is positive and decreasing on**
[1, ∞). Then

∞

P

k =1

f (k ) converges if and only if f is improperly integrable on [1, ∞), i.e., if and only if

Z ∞ 1

f (x )dx < ∞.

**Corollary:**

[p-Series Test]
The series

(4)

∞

X

k =1

1
k^{p}
converges if and only if p > 1.

**Corollary:**

[p-Series Test]
The series

(4)

∞

X

k =1

1
k^{p}
converges if and only if p > 1.

**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,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

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

**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,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

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

**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,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

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

**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,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

**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,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

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

**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,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

**Proof:**

^{−p} and observe that f^{0}(x ) = −px^{−p−1} <0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

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

**Proof:**

^{−p} and observe that f^{0}(x ) = −px^{−p−1} <0 for
all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on
[1, ∞). Since,

Z ∞ 1

dx

x^{p} = lim

x →∞

x^{1−p}
1 − p

n

1

= lim

n→∞

n^{1−p}− 1
1 − p

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ b_{k} for large k.

(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ b_{k} for large k.

(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ b_{k} for large k.

(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ b_{k} for large k.

(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ b_{k} for large k.

(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ b_{k} for large k.

(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

**Example:**

Determine whether the following series converges or diverges.

∞

X

k =1

3k
k^{2}+k

rlog k k

**Example:**

Determine whether the following series converges or diverges.

∞

X

k =1

3k
k^{2}+k

rlog k k

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i)If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii) If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii) If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i) If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii) If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii) If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i)If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii)If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii) If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i) If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii) If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii) If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i) If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii)If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii)If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i) If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii) If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii) If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i) If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii) If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii)If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and b_{k} >0 for large k and
L := lim

n→∞

an

bn

exists as an extended real number.

(i) If 0 < L < ∞, then

∞

P

k =1

ak converges if and only if

∞

P

k =1

bk

converges.

(ii) If L = 0 and

∞

P

k =1

ak converges, then

∞

P

k =1

ak converges.

(iii) If L = 0 and

∞

P

k =1

bk diverges, then

∞

P

k =1

ak diverges.

**Example:**

(1)

∞

P

k =1

√k

k k

(2)

∞

P

k =1

√ 1
k 3^{k −1}

**Example:**

(1)

∞

P

k =1

√k

k k

(2)

∞

P

k =1

√ 1
k 3^{k −1}

**Example:**

(1)

∞

P

k =1

√k

k k

(2)

∞

P

k =1

√ 1
k 3^{k −1}