Advanced Calculus (I)

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Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN Advanced Calculus (I)

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6.3 Absolute convergence

Definition Let S =

∞

P

k =1

ak be an infinite series.

(i)S is said to converge absolutely if and only if

∞

P

k =1

|ak| < ∞.

(ii)S is said to converge conditionally if and only if S converges but not absolutely.

(3)

6.3 Absolute convergence

Definition Let S =

∞

P

k =1

∞

P

k =1

|ak| < ∞.

(4)

6.3 Absolute convergence

Definition Let S =

∞

P

k =1

∞

P

k =1

|ak| < ∞.

(5)

6.3 Absolute convergence

Definition Let S =

∞

P

k =1

∞

P

k =1

|ak| < ∞.

(6)

6.3 Absolute convergence

Definition Let S =

∞

P

k =1

∞

P

k =1

|ak| < ∞.

(7)

6.3 Absolute convergence

Definition Let S =

∞

P

k =1

∞

P

k =1

|ak| < ∞.

(8)

Remark:

If

∞

P

k =1

ak converges absolutely, then

∞

P

k =1

ak converges, but not conversely. In particular, there exist conditionally convergent series.

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Remark:

If

∞

P

k =1

ak converges absolutely, then

∞

P

k =1

ak converges, but not conversely. In particular, there exist conditionally convergent series.

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Proof:

Suppose that

∞

P

k =1

ak converges absolutely. Given > 0, choose N ∈N so that (6) holds. Then

m

X

k =n

ak

≤

m

X

k =n

|ak| <

for M > n ≥ N. Hence, by the Cauchy Criterion,

∞

P

k =1

ak

converges.

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Proof:

Suppose that

∞

P

k =1

ak converges absolutely. Given > 0, choose N ∈N so that (6) holds. Then

m

X

k =n

ak

≤

m

X

k =n

|ak| <

∞

P

k =1

ak

converges.

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Proof:

Suppose that

∞

P

k =1

m

X

k =n

ak

≤

m

X

k =n

|ak| <

∞

P

k =1

ak

converges.

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Proof:

Suppose that

∞

P

k =1

ak converges absolutely. Given > 0, choose N ∈N so that (6) holds. Then

m

X

k =n

ak

≤

m

X

k =n

|ak| <

for M > n ≥ N. Hence,by the Cauchy Criterion,

∞

P

k =1

ak

converges.

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Proof:

Suppose that

∞

P

k =1

m

X

k =n

ak

≤

m

X

k =n

|ak| <

∞

P

k =1

ak

converges.

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Proof:

Suppose that

∞

P

k =1

m

X

k =n

ak

≤

m

X

k =n

|ak| <

for M > n ≥ N. Hence,by the Cauchy Criterion,

∞

P

k =1

ak

converges.

(16)

Proof:

Suppose that

∞

P

k =1

m

X

k =n

ak

≤

m

X

k =n

|ak| <

∞

P

k =1

ak

converges.

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Proof:

Suppose that

∞

P

k =1

m

X

k =n

ak

≤

m

X

k =n

|ak| <

∞

P

k =1

ak

converges.

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We shall finish the proof by showing that S :=

∞

P

k =1

(−1)^k/k converges conditionally. Since the harmonic series

diverges,S does not converge absolutely. On the other hand, the tails of S look like

∞

X

j=k

(−1)^j

j = (−1)^k 1

k − 1

k + 1 + 1

k + 2 − 1

k + 3 + · · ·

.

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∞

P

k =1

(−1)^k/k converges conditionally.Since the harmonic series

diverges, S does not converge absolutely.On the other hand, the tails of S look like

∞

X

j=k

(−1)^j

j = (−1)^k 1

k − 1

k + 1 + 1

k + 2 − 1

k + 3 + · · ·

.

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∞

P

k =1

diverges,S does not converge absolutely. On the other hand,the tails of S look like

∞

X

j=k

(−1)^j

j = (−1)^k 1

k − 1

k + 1 + 1

k + 2 − 1

k + 3 + · · ·

.

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∞

P

k =1

diverges, S does not converge absolutely.On the other hand, the tails of S look like

∞

X

j=k

(−1)^j

j = (−1)^k 1

k − 1

k + 1 + 1

k + 2 − 1

k + 3 + · · ·

.

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∞

P

k =1

diverges, S does not converge absolutely. On the other hand,the tails of S look like

∞

X

j=k

(−1)^j

j = (−1)^k 1

k − 1

k + 1 + 1

k + 2 − 1

k + 3 + · · ·

.

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∞

P

k =1

diverges, S does not converge absolutely. On the other hand, the tails of S look like

∞

X

j=k

(−1)^j

j = (−1)^k 1

k − 1

k + 1 + 1

k + 2 − 1

k + 3 + · · ·

.

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By grouping pairs of terms together,

it is easy to see that the sum inside the parentheses is greater than 0 but less than 1/k ,

i.e.,

∞

X

j=k

(−1)^j j

< 1 k.

Hence

∞

P

k =1

(−1)^k

k converges by Corlllary 6.9. 2

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i.e.,

∞

X

j=k

(−1)^j j

< 1 k.

Hence

∞

P

k =1

(−1)^k

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i.e.,

∞

X

j=k

(−1)^j j

< 1 k.

Hence

∞

P

k =1

(−1)^k

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i.e.,

∞

X

j=k

(−1)^j j

< 1 k.

Hence

∞

P

k =1

(−1)^k

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i.e.,

∞

X

j=k

(−1)^j j

< 1 k.

Hence

∞

P

k =1

(−1)^k

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Definition

The supremum s of the set of adherent points of a sequence {xk} is called the limit supremum of {x_k}.

(Notation: s := lim sup

k →∞

xk.)

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Definition

The supremum s of the set of adherent points of a sequence {xk} is called the limit supremum of {x_k}.

(Notation: s := lim sup

k →∞

xk.)

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Remark:

Let x ∈R and {xk} be a real sequence.

(i)If lim sup

k →∞

xk <x , then xk <x for large k.

(ii)If lim sup

k →∞

xk >x , then xk >x for infinitely many k.

(iii)If xk → x as k → ∞, then lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Remark:

(i)If lim sup

k →∞

(ii)If lim sup

k →∞

xk =x .

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Proof:

(i)

Let s := lim sup

k →∞

xk <x but suppose to the contrary that there exist natural numbers k1<k2< · · · such that xkj ≥ x for j ∈ N. If {x_k} is unbounded above, then ∞ is an adherent point of {xk} so s = ∞, a contradiction.

If {xkj} is bounded above by (by C), then it is bounded (since x ≤ xk ≤ C for all j ∈ N). Hence, by the

Bolzano-Weierstrass Theorem and the fact that xkj ≥ x, {x_k} has an adherent point ≥ x, i.e., s ≥ x, another contradiction.

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Proof:

(i)

Let s := lim sup

k →∞

xk <x but suppose to the contrary that there exist natural numbers k1<k2< · · · such that xkj ≥ x for j ∈ N. If {xk} is unbounded above, then ∞ is an adherent point of {xk} so s = ∞, a contradiction.

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Proof:

(i)

Let s := lim sup

k →∞

xk <x but suppose to the contrary that there exist natural numbers k1<k2< · · · such that xkj ≥ x for j ∈ N. If {x_k} is unbounded above,then ∞ is an adherent point of {xk} so s = ∞, a contradiction.

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Proof:

(i)

Let s := lim sup

k →∞

xk <x but suppose to the contrary that there exist natural numbers k1<k2< · · · such that xkj ≥ x for j ∈ N. If {xk} is unbounded above, then ∞ is an adherent point of {xk} so s = ∞, a contradiction.

(43)

Proof:

(i)

Let s := lim sup

k →∞

xk <x but suppose to the contrary that there exist natural numbers k1<k2< · · · such that xkj ≥ x for j ∈ N. If {x_k} is unbounded above,then ∞ is an adherent point of {xk} so s = ∞, a contradiction.

If {xkj} is bounded above by (by C),then it is bounded (since x ≤ xk ≤ C for all j ∈ N). Hence, by the

(44)

Proof:

(i)

Let s := lim sup

k →∞

If {xkj} is bounded above by (by C), then it is bounded (since x ≤ xk ≤ C for all j ∈ N). Hence, by the

(45)

Proof:

(i)

Let s := lim sup

k →∞

If {xkj} is bounded above by (by C),then it is bounded (since x ≤ xk ≤ C for all j ∈ N). Hence, by the

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Proof:

(i)

Let s := lim sup

k →∞

If {xkj} is bounded above by (by C), then it is bounded (since x ≤ xk ≤ C for all j ∈ N). Hence, by the

Bolzano-Weierstrass Theorem and the fact that xkj ≥ x, {x_k} has an adherent point ≥ x,i.e., s ≥ x , another contradiction.

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Proof:

(i)

Let s := lim sup

k →∞

Bolzano-Weierstrass Theorem and the fact that xkj ≥ x, {x_k} has an adherent point ≥ x, i.e., s ≥ x,another contradiction.

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Proof:

(i)

Let s := lim sup

k →∞

Bolzano-Weierstrass Theorem and the fact that xkj ≥ x, {x_k} has an adherent point ≥ x,i.e., s ≥ x , another contradiction.

(49)

Proof:

(i)

Let s := lim sup

k →∞

Bolzano-Weierstrass Theorem and the fact that xkj ≥ x, {x_k} has an adherent point ≥ x, i.e., s ≥ x,another contradiction.

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Proof:

(i)

Let s := lim sup

k →∞

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(ii)

If s > x ,then choose an adherent point a ∈ (x , s). By the Approximation Propert there is a subsequence {xk_j} which converges to a, i.e., xkj >x for large j.

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(ii)

If s > x , then choose an adherent point a ∈ (x , s).By the Approximation Propert there is a subsequence {xk_j} which converges to a, i.e., xkj >x for large j.

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(ii)

If s > x ,then choose an adherent point a ∈ (x , s). By the Approximation Propertthere is a subsequence {xk_j} which converges to a, i.e., xkj >x for large j.

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(ii)

If s > x , then choose an adherent point a ∈ (x , s).By the Approximation Propert there is a subsequence {xk_j} which converges to a,i.e., xkj >x for large j.

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(ii)

If s > x , then choose an adherent point a ∈ (x , s). By the Approximation Propertthere is a subsequence {xk_j} which converges to a, i.e., xkj >x for large j.

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(ii)

If s > x , then choose an adherent point a ∈ (x , s). By the Approximation Propert there is a subsequence {xk_j} which converges to a,i.e., xkj >x for large j.

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(ii)

If s > x , then choose an adherent point a ∈ (x , s). By the Approximation Propert there is a subsequence {xk_j} which converges to a, i.e., xkj >x for large j.

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(iii)

If xk converges to x,then any subsequence xk_j also converges to x (see Theorem 2.6). 2

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(iii)

If xk converges to x, then any subsequence xk_j also converges to x (see Theorem 2.6). 2

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(iii)

If xk converges to x,then any subsequence xk_j also converges to x (see Theorem 2.6). 2

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(iii)

If xk converges to x, then any subsequence xk_j also converges to x (see Theorem 2.6). 2

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Theorem (Root Test) Let ak ∈ R and r := lim sup

k →∞

|a_k|^1/k. (i)If r < 1, then

∞

P

k =1

ak converges absolutely.

(ii) If r > 1, then

∞

P

k =1

ak diverges.

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k →∞

|a_k|^1/k. (i) If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

(64)

k →∞

|a_k|^1/k. (i)If r < 1, then

∞

P

k =1

(ii)If r > 1, then

∞

P

k =1

ak diverges.

(65)

k →∞

|a_k|^1/k. (i) If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

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k →∞

|a_k|^1/k. (i) If r < 1, then

∞

P

k =1

(ii)If r > 1, then

∞

P

k =1

ak diverges.

(67)

k →∞

|a_k|^1/k. (i) If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

(68)

Theorem (Ratio Test)

Let ak ∈ R with a_k 6= 0 for large k and suppose that r = lim

k →∞

|ak +1| ak

exists as an extended real number.

(i)If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

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k →∞

|ak +1| ak

(i) If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

(70)

k →∞

|ak +1| ak

(i)If r < 1, then

∞

P

k =1

(ii)If r > 1, then

∞

P

k =1

ak diverges.

(71)

k →∞

|ak +1| ak

(i) If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

(72)

k →∞

|ak +1| ak

(i) If r < 1, then

∞

P

k =1

(ii)If r > 1, then

∞

P

k =1

ak diverges.

(73)

k →∞

|ak +1| ak

(i) If r < 1, then

∞

P

k =1

(ii) If r > 1, then

∞

P

k =1

ak diverges.

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Definition A series

∞

P

j=1

bj is called a rearrangment of a series

∞

P

k =1

ak if and only if there is a 1-1 function f from N onto N such that

bf (k ) =ak, k ∈N

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Definition A series

∞

P

j=1

bj is called a rearrangment of a series

∞

P

k =1

ak if and only if there is a 1-1 function f from N onto N such that

bf (k ) =ak, k ∈N

(76)

Theorem If

∞

P

k =1

ak converges absolutely and

∞

P

j=1

bj is any rearrangment of

∞

P

k =1

ak, then

∞

P

j=1

bj converges and

∞

X

k =1

ak =

∞

X

j=1

bj.

(77)

Theorem If

∞

P

k =1

ak converges absolutely and

∞

P

j=1

bj is any rearrangment of

∞

P

k =1

ak, then

∞

P

j=1

bj converges and

∞

X

k =1

ak =

∞

X

j=1

bj.

(78)

Proof:

Let > 0. Set Sn =

n

P

k =1

ak, s =

∞

P

k =1

ak, and tm =

m

P

j=1

bj, n, m ∈N. Since

∞

P

k =1

ak converges absolutely, we can choose N ∈N (see Corollary 6.9) such that

(7)

∞

X

k =N+1

|ak| ≤ 2. Thus

(8) |sN − s| =

∞

X

k =N+1

ak

≤

∞

X

k =N+1

|ak| < 2

(79)

Proof:

Let > 0.Set Sn =

n

P

k =1

ak, s =

∞

P

k =1

ak, and tm =

m

P

j=1

bj, n, m ∈N. Since

∞

P

k =1

ak converges absolutely,we can choose N ∈N (see Corollary 6.9) such that

(7)

∞

X

k =N+1

|ak| ≤ 2. Thus

(8) |sN − s| =

∞

X

k =N+1

ak

≤

∞

X

k =N+1

|ak| < 2

(80)

Proof:

Let > 0. Set Sn =

n

P

k =1

ak, s =

∞

P

k =1

ak, and tm =

m

P

j=1

bj, n, m ∈N. Since

∞

P

k =1

(7)

∞

X

k =N+1

|ak| ≤ 2. Thus

(8) |sN − s| =

∞

X

k =N+1

ak

≤

∞

X

k =N+1

|ak| < 2

(81)

Proof:

Let > 0. Set Sn =

n

P

k =1

ak, s =

∞

P

k =1

ak, and tm =

m

P

j=1

∞

P

k =1

ak converges absolutely,we can choose N ∈N (see Corollary 6.9) such that

(7)

∞

X

k =N+1

|ak| ≤ 2. Thus

(8) |sN − s| =

∞

X

k =N+1

ak

≤

∞

X

k =N+1

|ak| < 2

(82)

Proof:

Let > 0. Set Sn =

n

P

k =1

ak, s =

∞

P

k =1

ak, and tm =

m

P

j=1

∞

P

k =1

(7)

∞

X

k =N+1

|ak| ≤ 2. Thus

(8) |sN − s| =

∞

X

k =N+1

ak

≤

∞

X

k =N+1

|ak| < 2

(83)

Proof:

Let > 0. Set Sn =

n

P

k =1

ak, s =

∞

P

k =1

ak, and tm =

m

P

j=1

∞

P

k =1

(7)

∞

X

k =N+1

|ak| ≤ 2. Thus

(8) |sN − s| =

∞

X

k =N+1

ak

≤

∞

X

k =N+1

|ak| < 2

(84)

Let f be a 1-1 function fromN onto N that satisfies bf (k ) =ak, k ∈N

and set M = max{f (1), · · · , f (N)}.Notice that {a₁, · · · ,aN} ⊆ {b₁, · · · ,bM}.

(85)

and set M = max{f (1), · · · , f (N)}. Notice that {a₁, · · · ,aN} ⊆ {b₁, · · · ,bM}.

(86)

and set M = max{f (1), · · · , f (N)}.Notice that {a₁, · · · ,aN} ⊆ {b₁, · · · ,bM}.

(87)

and set M = max{f (1), · · · , f (N)}. Notice that {a₁, · · · ,aN} ⊆ {b₁, · · · ,bM}.

(88)

Let m ≥ M. Then tm− s_N contains only ak0s whose indices satisfy k > N. Thus, it follows from (7) that

|tm− sN| ≤

∞

X

k =N+1

|ak| < 2. Hence by (8),

|tm− s| ≤ |tm− sN| + |sN− s| < 2 +

2 = for m ≥ M. Therefore,

s =

∞

X

j=1

bj. 2

(89)

|tm− sN| ≤

∞

X

k =N+1

|ak| < 2. Hence by (8),

|tm− s| ≤ |tm− sN| + |sN− s| < 2 +

s =

∞

X

j=1

bj. 2

(90)

|tm− sN| ≤

∞

X

k =N+1

|ak| < 2. Hence by (8),

|tm− s| ≤ |tm− sN| + |sN− s| < 2 +

s =

∞

X

j=1

bj. 2

(91)

|tm− sN| ≤

∞

X

k =N+1

|ak| < 2. Hence by (8),

|tm− s| ≤ |tm− sN| + |sN− s| < 2 +

s =

∞

X

j=1

bj. 2

(92)

|tm− sN| ≤

∞

X

k =N+1

|ak| < 2. Hence by (8),

|tm− s| ≤ |tm− sN| + |sN− s| < 2 +

s =

∞

X

j=1

bj. 2

(93)

|tm− sN| ≤

∞

X

k =N+1

|ak| < 2. Hence by (8),

|tm− s| ≤ |tm− sN| + |sN− s| < 2 +

s =

∞

X

j=1

bj. 2

(94)

Theorem

Suppose that ak ∈ R for k ∈ N.

(i)If

∞

P

k =1

ak converges absolutely, then so do

∞

P

k =1

ak+ and

∞

P

k =1

ak−. In fact,

∞

X

k =1

|a_k| =

∞

X

k =1

ak++

∞

X

k =1

ak−

and

∞

X

k =1

ak =

∞

X

k =1

ak+−

∞

X

k =1

ak−

(ii)If

∞

P

k =1

ak converges conditionally, then

∞

X

k =1

ak+ =

∞

X

k =1

ak−

= ∞.

(95)

Theorem

(i) If

∞

P

k =1

∞

P

k =1

ak+ and

∞

P

k =1

ak−. In fact,

∞

X

k =1

|a_k| =

∞

X

k =1

ak++

∞

X

k =1

ak−

and

∞

X

k =1

ak =

∞

X

k =1

ak+−

∞

X

k =1

ak−

(ii)If

∞

P

k =1

∞

X

k =1

ak+ =

∞

X

k =1

ak−

= ∞.

(96)

Theorem

(i)If

∞

P

k =1

∞

P

k =1

ak+ and

∞

P

k =1

ak−. In fact,

∞

X

k =1

|a_k| =

∞

X

k =1

ak++

∞

X

k =1

ak−

and

∞

X

k =1

ak =

∞

X

k =1

ak+−

∞

X

k =1

ak−

(ii)If

∞

P

k =1

∞

X

k =1

ak+ =

∞

X

k =1

ak−

= ∞.