Advanced Calculus (I)

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Advanced Calculus (I)

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.

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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.

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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

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Proof:

Set sn =

n

P

k =1

ak for n ∈N. If

∞

P

k =1

∞

P

k =1

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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

∞

P

k =1

∞

P

k =1

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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

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Proof:

Set sn =

n

P

k =1

∞

P

k =1

∞

P

k =1

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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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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

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∞

X

k =1

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

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Corollary:

[p-Series Test]

The series

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∞

X

k =1

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

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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

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

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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

x^p = lim

x →∞

x^1−p 1 − p

n

1

= lim

n→∞

n^1−p− 1 1 − p

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Proof:

Z ∞ 1

dx

x^p = lim

x →∞

x^1−p 1 − p

n

1

= lim

n→∞

n^1−p− 1 1 − p

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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

x^p = lim

x →∞

x^1−p 1 − p

n

1

= lim

n→∞

n^1−p− 1 1 − p

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Proof:

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

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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

x^p = lim

x →∞

x^1−p 1 − p

n

1

= lim

n→∞

n^1−p− 1 1 − p

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Proof:

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

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Proof:

Z ∞ 1

dx

x^p = lim

x →∞

x^1−p 1 − p

n

1

= lim

n→∞

n^1−p− 1 1 − p

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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 = ∞.

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(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

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(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

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(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

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(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

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(i)If

∞

P

k =1

bk < ∞, then

∞

P

k =1

ak < ∞.

(ii)If

∞

P

k =1

ak = ∞, then

∞

P

k =1

bk = ∞.

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Example:

Determine whether the following series converges or diverges.

∞

X

k =1

3k k²+k

rlog k k

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Example:

Determine whether the following series converges or diverges.

∞

X

k =1

3k k²+k

rlog k k

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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.

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n→∞

an

bn

(i) If 0 < L < ∞, then

∞

P

k =1

∞

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.

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n→∞

an

bn

(i)If 0 < L < ∞, then

∞

P

k =1

∞

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.

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n→∞

an

bn

(i) If 0 < L < ∞, then

∞

P

k =1

∞

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.

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n→∞

an

bn

(i) If 0 < L < ∞, then

∞

P

k =1

∞

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.

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n→∞

an

bn

(i) If 0 < L < ∞, then

∞

P

k =1

∞

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.

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n→∞

an

bn

(i) If 0 < L < ∞, then

∞

P

k =1

∞

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.

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n→∞

an

bn

(i) If 0 < L < ∞, then

∞

P

k =1

∞

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.

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Example:

(1)

∞

P

k =1

√k

k k

(2)

∞

P

k =1

√ 1 k 3^{k −1}

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Example:

(1)

∞

P

k =1

√k

k k

(2)

∞

P

k =1

√ 1 k 3^{k −1}

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Example:

(1)

∞

P

k =1

√k

k k

(2)

∞

P

k =1

√ 1 k 3^{k −1}

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Advanced Calculus (I)