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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

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## 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.

(3)

## 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.

(4)

### 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

(5)

### 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

(6)

### 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

(7)

### 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

(8)

### 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

(9)

### 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

(10)

### 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

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### 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

(12)

### 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

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### 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

(14)

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 < ∞.

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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 < ∞.

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### Corollary:

[p-Series Test]

The series

(4)

X

k =1

1 kp converges if and only if p > 1.

(17)

### Corollary:

[p-Series Test]

The series

(4)

X

k =1

1 kp converges if and only if p > 1.

(18)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(19)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(20)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(21)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞).Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(22)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(23)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞).Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(24)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(25)

### 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 f0(x ) = −px−p−1 <0 for all x ∈ [1, ∞). Hence, f is nonnegative and decreasing on [1, ∞). Since,

Z 1

dx

xp = lim

x →∞

x1−p 1 − p

n

1

= lim

n→∞

n1−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

(26)

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ bk 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 = ∞.

(27)

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ bk 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 = ∞.

(28)

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ bk 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 = ∞.

(29)

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ bk 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 = ∞.

(30)

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ bk 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 = ∞.

(31)

Theorem (Comparison Test)

Suppose that 0 ≤ ak ≤ bk 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 = ∞.

(32)

### Example:

Determine whether the following series converges or diverges.

X

k =1

3k k2+k

rlog k k

(33)

### Example:

Determine whether the following series converges or diverges.

X

k =1

3k k2+k

rlog k k

(34)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(35)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(36)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(37)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(38)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(39)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(40)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(41)

Theorem (Limit Comparison Test)

Suppose that ak ≥ 0 and bk >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.

(42)

(1)

P

k =1

k

k k

(2)

P

k =1

√ 1 k 3k −1

(43)

(1)

P

k =1

k

k k

(2)

P

k =1

√ 1 k 3k −1

(44)

(1)

P

k =1

k

k k

(2)

P

k =1

√ 1 k 3k −1

(45)

## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

This implies that f is an open mapping which maps interior points (of X) to interior points (of Y ) and f −1 is an open mapping from Y onto X... Is this convergence uniform

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung University... In this case we say that Q is finer

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

From all the above, φ is zero only on the nonnegative sides of the a, b-axes. Hence, φ is an NCP function.. Graph of g functions given in Example 3.9.. Graphs of generated NCP