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(1)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

(2)

A power series is a series of the form

S(x ) :=

X

k =0

ak(x − x0)k

(3)

An extended real number R is said to be the radius of convergence of a power series S(x ) :=

P

k =0

ak(x − x0)k if and only if S(x ) converges absolutely for |x − x0| < R and S(x ) diverges for |x − x0| > R.

(4)

An extended real number R is said to be the radius of convergence of a power series S(x ) :=

P

k =0

ak(x − x0)k if and only if S(x ) converges absolutely for |x − x0| < R and S(x ) diverges for |x − x0| > R.

(5)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(6)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i)S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(7)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(8)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i)S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii)S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(9)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(10)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii)S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii)and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(11)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(12)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii)and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(13)

Let S(x ) = P

k =0

ak(x − x0)k be a power series centered at x0, If R = 1

lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞, then R is the radius of convergence of S.

In fact,

(i) S(x) converges absolutely for each x ∈ (x0− R, x0+R), (ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x0+R),

(iii) and (when R is finite), S(x) diverges for each x /∈ [x0− R, x0+R].

(14)

Fix x ∈R, x 6= x0, and set ρ := 1 lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞. To apply the Root Test to S(x ), consider

r (x ) := lim sup

n→∞

|ak(x − x0)k|1/k = |x − x0| · lim sup

k →∞

|ak|1/k.

(15)

Fix x ∈R, x 6= x0,and set ρ := 1 lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞. To apply the Root Test to S(x ),consider

r (x ) := lim sup

n→∞

|ak(x − x0)k|1/k = |x − x0| · lim sup

k →∞

|ak|1/k.

(16)

Fix x ∈R, x 6= x0, and set ρ := 1 lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞. To apply the Root Test to S(x ), consider

r (x ) := lim sup

n→∞

|ak(x − x0)k|1/k = |x − x0| · lim sup

k →∞

|ak|1/k.

(17)

Fix x ∈R, x 6= x0, and set ρ := 1 lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞. To apply the Root Test to S(x ),consider

r (x ) := lim sup

n→∞

|ak(x − x0)k|1/k = |x − x0| · lim sup

k →∞

|ak|1/k.

(18)

Fix x ∈R, x 6= x0, and set ρ := 1 lim sup

k →∞

|ak|1/k, with the convention that 1

∞ =0 and 1

0 = ∞. To apply the Root Test to S(x ), consider

r (x ) := lim sup

n→∞

|ak(x − x0)k|1/k = |x − x0| · lim sup

k →∞

|ak|1/k.

(19)

ρ =0. By our convention,ρ =0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.

(20)

ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1,so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.

(21)

ρ =0. By our convention,ρ =0 implies that r (x ) = ∞ > 1, so by the Root Test,S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.

(22)

ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1,so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.

(23)

ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test,S(x) does not converge for any x 6= x0. Hence,the radius of convergence of S is R = 0 = ρ.

(24)

ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.

(25)

ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence,the radius of convergence of S is R = 0 = ρ.

(26)

ρ =0. By our convention, ρ = 0 implies that r (x ) = ∞ > 1, so by the Root Test, S(x) does not converge for any x 6= x0. Hence, the radius of convergence of S is R = 0 = ρ.

(27)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.

(28)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.

(29)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.

(30)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence,the radius of convergence of S is R = ∞ = ρ.

(31)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.

(32)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence,the radius of convergence of S is R = ∞ = ρ.

(33)

ρ = ∞. Then r (x ) = 0 < 1, so by the Root Test, S(x) converges absolutely for all x ∈R. Hence, the radius of convergence of S is R = ∞ = ρ.

(34)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(35)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(36)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ,it follows from the Root Test that S(x ) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(37)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(38)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ,it follows from the Root Test that S(x ) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(39)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ,we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(40)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(41)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ,we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(42)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(43)

ρ ∈ (0, ∞). Then r (x ) = |x − x0|

ρ . Since r (x ) < 1 if and only if |x − x0| < ρ, it follows from the Root Test that S(x) converges absolutely when x ∈ (x0− ρ, x0+ ρ). Similarly, since r (x ) > 1 if and only if |x − x0| > ρ, we also have that S(x) diverges when x /∈ [x0− ρ, x0+ ρ]. This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.

(44)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(45)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(46)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges.Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(47)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N,it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(48)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges.Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(49)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N,it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(50)

|x − x0| ≤ |x1− x0| (see Figure 7.5). Set Mk = |ak||x1− x0|k and observe by part (i) that

P

k =0

Mk

converges. Since |ak(x − x0)k| ≤ Mk for x ∈ [a, b] and k ∈N, it follows from the Wirestrass M-test that S(x) converges uniformly on [a,b]. 2

(51)

If the limit

R = lim

k →∞

ak

ak +1

exists as an extended real number, then R is the radius of convergence of the power series S(x ) =

P

k =0

ak(x − x0)k.

(52)

If the limit

R = lim

k →∞

ak

ak +1

exists as an extended real number, then R is the radius of convergence of the power series S(x ) =

P

k =0

ak(x − x0)k.

(53)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(54)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i)R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(55)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(56)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i)R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii)R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(57)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(58)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii)R = 0, in which case the interval of convergence of S is {x0}, and

(iii)0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(59)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(60)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii)0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(61)

The interval of convergence of a power series S(x) is the largest interval on which S(x) converges.

By Theorem 7.21, for a given power series S =

P

k =0

ak(x − x0)k, there are only three possibilities:

(i) R = ∞, in which case the interval of convergence of S is (−∞, ∞),

(ii) R = 0, in which case the interval of convergence of S is {x0}, and

(iii) 0 < R < ∞, in which case the interval of convergence of S is (x0− R, x0+R), [x0− R, x0+R), (x0− R, x0+R], or [x0− R, x0+R].

(62)

The interval of convergence may contain none, one, or both its endpoints.

(63)

The interval of convergence may contain none, one, or both its endpoints.

(64)

If f (x ) = P

k =0

ak(x − x0)k is a power series with positive radius of convergence R, then f is continuous on (x0− R, x0+R).

(65)

If f (x ) = P

k =0

ak(x − x0)k is a power series with positive radius of convergence R, then f is continuous on (x0− R, x0+R).

(66)

Suppose that [a, b] is nondegenerate. If f (x ) :=

P

k =0

ak(x − x0)k converges on [a,b], then f(x) is continuous and converges uniformly on [a,b].

(67)

Suppose that [a, b] is nondegenerate. If f (x ) :=

P

k =0

ak(x − x0)k converges on [a,b], then f(x) is continuous and converges uniformly on [a,b].

(68)

If a power series S(x ) =

P

k =0

ak(x − x0)k converges at some x1 >x0, then S(x) converges uniformly on [x0,x1] and absolutely on [x0,x1). It might not converge

absolutely at x = x1.

(69)

If a power series S(x ) =

P

k =0

ak(x − x0)k converges at some x1 >x0, then S(x) converges uniformly on [x0,x1] and absolutely on [x0,x1). It might not converge

absolutely at x = x1.

(70)

If an ∈ R for n ∈ N, then x := lim sup

n→∞

(n|an|1/(n−1)) = y := lim sup

n→∞

|an|1/n

(71)

If an ∈ R for n ∈ N, then x := lim sup

n→∞

(n|an|1/(n−1)) = y := lim sup

n→∞

|an|1/n

(72)

Let  > 0. Since n1/(n−1) → 1 as n → ∞,choose N ∈N so that n ≥ N implies 1 −  < n1/(n−1) <1 + , i.e.,

(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).

(73)

Let  > 0.Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 −  < n1/(n−1) <1 + , i.e.,

(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).

(74)

Let  > 0. Since n1/(n−1) → 1 as n → ∞,choose N ∈N so that n ≥ N implies 1 −  < n1/(n−1) <1 + ,i.e.,

(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).

(75)

Let  > 0. Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 −  < n1/(n−1) <1 + , i.e.,

(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).

(76)

Let  > 0. Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 −  < n1/(n−1) <1 + ,i.e.,

(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).

(77)

Let  > 0. Since n1/(n−1) → 1 as n → ∞, choose N ∈ N so that n ≥ N implies 1 −  < n1/(n−1) <1 + , i.e.,

(1 − )|an|1/(n−1) < (n|an)1/(n−1) < (1 + )|an|1/(n−1).

(78)

zk := |ank If nk n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(79)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(80)

zk := |ank If nk n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(81)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n,it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(82)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(83)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n,it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(84)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(85)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly,the left-most inequality above implies y ≤ x . 2

(86)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(87)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly,the left-most inequality above implies y ≤ x . 2

(88)

zk := |ank k n ≥ N, then

|ank|1/(nk−1) <zkn/(n−1) → zn/(n−1) as k → ∞. Since y is the supremum of the set of adherent points of |an|1/n, it

follows from the right-most inequality above that x ≤ (1 + )yn/(n−1). Let n → ∞ and then let  → 0. We obtain x ≤ y . Similarly, the left-most inequality above implies y ≤ x . 2

(89)

If f (x ) = P

k =0

ak(x − x0)k is a power series with positive radius of convergence R, then f0(x ) =

P

k =1

kak(x − x0)k −1 for x ∈ (x0− R, x0+R).

(90)

If f (x ) = P

k =0

ak(x − x0)k is a power series with positive radius of convergence R, then f0(x ) =

P

k =1

kak(x − x0)k −1 for x ∈ (x0− R, x0+R).

(91)

If f (x ) =

P

k =0

ak(x − x0)k has a positive radius of convergence R, then f ∈C(x0− R, x0+R) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k for x ∈ (x0− R, x0+R) and k ∈ N.

(92)

If f (x ) =

P

k =0

ak(x − x0)k has a positive radius of convergence R, then f ∈C(x0− R, x0+R) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k for x ∈ (x0− R, x0+R) and k ∈ N.

(93)

Let f (x ) = P

k =0

ak(x − x0)k be a power series and let a, b ∈R with a < b.

(i)If f(x) converges on [a,b], then f is integrable on [a,b]

and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(ii) If f(x) converges on [a,b) and if

P

k =0

ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(94)

Let f (x ) = P

k =0

ak(x − x0)k be a power series and let a, b ∈R with a < b.

(i) If f(x) converges on [a,b], then f is integrable on [a,b]

and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(ii) If f(x) converges on [a,b) and if

P

k =0

ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(95)

Let f (x ) = P

k =0

ak(x − x0)k be a power series and let a, b ∈R with a < b.

(i)If f(x) converges on [a,b], then f is integrable on [a,b]

and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(ii)If f(x) converges on [a,b) and if

P

k =0

ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(96)

Let f (x ) = P

k =0

ak(x − x0)k be a power series and let a, b ∈R with a < b.

(i) If f(x) converges on [a,b], then f is integrable on [a,b]

and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(ii) If f(x) converges on [a,b) and if

P

k =0

ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(97)

Let f (x ) = P

k =0

ak(x − x0)k be a power series and let a, b ∈R with a < b.

(i) If f(x) converges on [a,b], then f is integrable on [a,b]

and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(ii)If f(x) converges on [a,b) and if

P

k =0

ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(98)

Let f (x ) = P

k =0

ak(x − x0)k be a power series and let a, b ∈R with a < b.

(i) If f(x) converges on [a,b], then f is integrable on [a,b]

and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(ii) If f(x) converges on [a,b) and if

P

k =0

ak(b − x0)k +1 (k + 1) converges, then f is improperly integrable on [a,b) and

Z b a

f (x )dx =

X

k =0

ak

Z b a

(x − x0)kdx .

(99)

(i)

By Abel’s Theorem,f(x) converges uniformly on [a,b].

Hence, by Theorem 7.14ii, f(x) is term-by-term integrable on [a,b].

(100)

(i)

By Abel’s Theorem, f(x) converges uniformly on [a,b].

Hence, by Theorem 7.14ii, f(x) is term-by-term integrable on [a,b].

(101)

(i)

By Abel’s Theorem,f(x) converges uniformly on [a,b].

Hence, by Theorem 7.14ii,f(x) is term-by-term integrable on [a,b].

(102)

(i)

By Abel’s Theorem, f(x) converges uniformly on [a,b].

Hence, by Theorem 7.14ii, f(x) is term-by-term integrable on [a,b].

Mean Value Theorem to F and G, we need to be sure the hypotheses of that result hold.... In order to apply

Suppose that E is bounded, open interval and that each f k.. is differentiable

(In this case we shall say that E has an infimum t and shall write t=inf E.).. (iv) E is said to be bounded if and only if it is bounded above

[r]

W EN -C HING L IEN Department of Mathematics National Cheng Kung

We shall actually prove that an increasing sequence converges to its supremum, and a decreasing sequence converges to its

Department of Mathematics National Cheng Kung University.. Theorem (Change

Every convergent sequence is bounded.. W EN -C HING L IEN Advanced