WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

WEN-CHINGLIEN **Advanced Calculus (I)**

A power series is a series of the form

S(x ) :=

∞

X

k =0

ak(x − x0)^{k}

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{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, x_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{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, x_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{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, x_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{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, x_{0}+R),
(ii)S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{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, x_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{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, x_{0}+R),
(ii)S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{1/k}, with the convention that 1

∞ =0 and 1

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

In fact,

_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{1/k}, with the convention that 1

∞ =0 and 1

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

In fact,

_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Let S(x ) = P

k =0

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

lim sup

k →∞

|a_{k}|^{1/k}, with the convention that 1

∞ =0 and 1

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

In fact,

_{0}+R),
(ii) S(x) converges uniformly on any closed interval

[a, b] ⊆ (x0− R, x_{0}+R),

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

WEN-CHINGLIEN **Advanced Calculus (I)**

Fix x ∈**R, x 6= x**0, and set ρ := 1
lim sup

k →∞

|a_{k}|^{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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

Fix x ∈**R, x 6= x**0,and set ρ := 1
lim sup

k →∞

|a_{k}|^{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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

Fix x ∈**R, x 6= x**0, and set ρ := 1
lim sup

k →∞

|a_{k}|^{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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

Fix x ∈**R, x 6= x**0, and set ρ := 1
lim sup

k →∞

|a_{k}|^{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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

Fix x ∈**R, x 6= x**0, and set ρ := 1
lim sup

k →∞

|a_{k}|^{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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

**R. Hence,**the radius of
convergence of S is R = ∞ = ρ.

WEN-CHINGLIEN **Advanced Calculus (I)**

**R.** Hence, the radius of
convergence of S is R = ∞ = ρ.

WEN-CHINGLIEN **Advanced Calculus (I)**

**R. Hence,**the radius of
convergence of S is R = ∞ = ρ.

WEN-CHINGLIEN **Advanced Calculus (I)**

**R. Hence, the radius of**
convergence of S is R = ∞ = ρ.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

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 /∈ [x_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

ρ ∈ (0, ∞). Then r (x ) = |x − x_{0}|

_{0}− ρ, x_{0}+ ρ]. This proves that ρ
is the radius of convergence of S, and that parts (i) and
(iii) hold.

WEN-CHINGLIEN **Advanced Calculus (I)**

|x − x_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

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

WEN-CHINGLIEN **Advanced Calculus (I)**

|x − x_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

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

WEN-CHINGLIEN **Advanced Calculus (I)**

|x − x_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

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

WEN-CHINGLIEN **Advanced Calculus (I)**

_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

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

WEN-CHINGLIEN **Advanced Calculus (I)**

_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

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

WEN-CHINGLIEN **Advanced Calculus (I)**

_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

^{k}| ≤ M_{k} for x ∈ [a, b] and
k ∈**N,**it follows from the Wirestrass M-test that S(x)
converges uniformly on [a,b]. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

_{0}| ≤ |x_{1}− x_{0}| (see Figure 7.5). Set
Mk = |ak||x_{1}− x_{0}|^{k} and observe by part (i) that

∞

P

k =0

Mk

^{k}| ≤ M_{k} for x ∈ [a, b] and
k ∈**N, it follows from the Wirestrass M-test that S(x)**
converges uniformly on [a,b]. 2

WEN-CHINGLIEN **Advanced Calculus (I)**

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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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, x_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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, x_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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, x_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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

_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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

_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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, x_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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

_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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, x_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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

_{0}+R].

WEN-CHINGLIEN **Advanced Calculus (I)**

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

WEN-CHINGLIEN **Advanced Calculus (I)**

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

WEN-CHINGLIEN **Advanced Calculus (I)**

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).

WEN-CHINGLIEN **Advanced Calculus (I)**

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).

WEN-CHINGLIEN **Advanced Calculus (I)**

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].

WEN-CHINGLIEN **Advanced Calculus (I)**

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].

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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.

WEN-CHINGLIEN **Advanced Calculus (I)**

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

n→∞

(n|an|^{1/(n−1)}) = y := lim sup

n→∞

|a_{n}|^{1/n}

WEN-CHINGLIEN **Advanced Calculus (I)**

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

n→∞

(n|an|^{1/(n−1)}) = y := lim sup

n→∞

|a_{n}|^{1/n}

WEN-CHINGLIEN **Advanced Calculus (I)**

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

(1 − )|an|^{1/(n−1)} < (n|an)^{1/(n−1)} < (1 + )|an|^{1/(n−1)}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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

(1 − )|an|^{1/(n−1)} < (n|an)^{1/(n−1)} < (1 + )|an|^{1/(n−1)}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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

(1 − )|an|^{1/(n−1)} < (n|an)^{1/(n−1)} < (1 + )|an|^{1/(n−1)}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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

(1 − )|an|^{1/(n−1)} < (n|an)^{1/(n−1)} < (1 + )|an|^{1/(n−1)}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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

(1 − )|an|^{1/(n−1)} < (n|an)^{1/(n−1)} < (1 + )|an|^{1/(n−1)}.

WEN-CHINGLIEN **Advanced Calculus (I)**

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

(1 − )|an|^{1/(n−1)} < (n|an)^{1/(n−1)} < (1 + )|an|^{1/(n−1)}.

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} If nk n ≥ N, then

|a_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(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 + )y^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

|a_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(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 + )y^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} If nk n ≥ N, then

|a_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(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 + )y^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

|a_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(n−1)} as k → ∞. Since y is the
supremum of the set of adherent points of |an|^{1/n},it

^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(n−1)} as k → ∞. Since y is the
supremum of the set of adherent points of |an|^{1/n}, it

^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

|a_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(n−1)} as k → ∞. Since y is the
supremum of the set of adherent points of |an|^{1/n},it

^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(n−1)} as k → ∞. Since y is the
supremum of the set of adherent points of |an|^{1/n}, it

^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(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 + )y^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly,the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(n−1)} as k → ∞. Since y is the
supremum of the set of adherent points of |an|^{1/n}, it

^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(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 + )y^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly,the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

zk := |an_{k} k n ≥ N, then

_{n}_{k}|^{1/(n}^{k}^{−1)} <z_{k}^{n/(n−1)} → z^{n/(n−1)} as k → ∞. Since y is the
supremum of the set of adherent points of |an|^{1/n}, it

^{n/(n−1)}. Let n → ∞ and then let → 0. We
obtain x ≤ y . Similarly, the left-most inequality above
implies y ≤ x . 2

WEN-CHINGLIEN **Advanced Calculus (I)**

If f (x ) = P

k =0

ak(x − x0)^{k} is a power series with positive
radius of convergence R, then f^{0}(x ) =

∞

P

k =1

kak(x − x0)^{k −1}
for x ∈ (x0− R, x_{0}+R).

WEN-CHINGLIEN **Advanced Calculus (I)**

If f (x ) = P

k =0

ak(x − x0)^{k} is a power series with positive
radius of convergence R, then f^{0}(x ) =

∞

P

k =1

kak(x − x0)^{k −1}
for x ∈ (x0− R, x_{0}+R).

WEN-CHINGLIEN **Advanced Calculus (I)**

If f (x ) =

∞

P

k =0

ak(x − x0)^{k} has a positive radius of
convergence R, then f ∈**C**^{∞}(x0− R, x_{0}+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.**

WEN-CHINGLIEN **Advanced Calculus (I)**

If f (x ) =

∞

P

k =0

ak(x − x0)^{k} has a positive radius of
convergence R, then f ∈**C**^{∞}(x0− R, x_{0}+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.**

WEN-CHINGLIEN **Advanced Calculus (I)**

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)^{k}dx .

(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)^{k}dx .

WEN-CHINGLIEN **Advanced Calculus (I)**

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)^{k}dx .

(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)^{k}dx .

WEN-CHINGLIEN **Advanced Calculus (I)**

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)^{k}dx .

(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)^{k}dx .

WEN-CHINGLIEN **Advanced Calculus (I)**

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)^{k}dx .

(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)^{k}dx .

WEN-CHINGLIEN **Advanced Calculus (I)**

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)^{k}dx .

(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)^{k}dx .

WEN-CHINGLIEN **Advanced Calculus (I)**

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)^{k}dx .

(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)^{k}dx .

WEN-CHINGLIEN **Advanced Calculus (I)**

(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].

WEN-CHINGLIEN **Advanced Calculus (I)**

(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].

WEN-CHINGLIEN **Advanced Calculus (I)**

(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].

WEN-CHINGLIEN **Advanced Calculus (I)**

(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].

WEN-CHINGLIEN **Advanced Calculus (I)**