Advanced Calculus (I)

Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

6.1 Introduction

Definition Let S =

∞

P

k =1

ak be an infinite series whose terms ak belong toR

(i)

The partial sums of S of order n are the numbers defined, for each n ∈N, by

Sn :=

n

Xak.

6.1 Introduction

Definition Let S =

∞

P

k =1

ak be an infinite series whose terms ak belong toR

(i)

The partial sums of S of order n are the numbers defined, for each n ∈N, by

Sn :=

n

Xak.

6.1 Introduction

Definition Let S =

∞

P

k =1

ak be an infinite series whose terms ak belong toR

(i)

The partial sums of S of order n are the numbers defined, for each n ∈N, by

Sn :=

n

Xak.

6.1 Introduction

Definition Let S =

∞

P

k =1

ak be an infinite series whose terms ak belong toR

(i)

The partial sums of S of order n are the numbers defined, for each n ∈N, by

Sn :=

n

Xak.

Definition (ii)

S is said to converge if and only if its sequence of partial sums {sn} converges to some s ∈ R as n → ∞; i.e., for every  > 0 there is an N ∈N such that n ≥ N implies that

|s_n− s| < . In this case we shall write

∞

X

k =1

ak =s

and called s the sum, or value, of the seriesP∞

a

Definition (ii)

S is said to converge if and only if its sequence of partial sums {sn} converges to some s ∈ R as n → ∞; i.e., for every  > 0 there is an N ∈N such that n ≥ N implies that

|s_n− s| < . In this case we shall write

∞

X

k =1

ak =s

and called s the sum, or value, of the seriesP∞

a

Definition (ii)

S is said to converge if and only if its sequence of partial sums {sn} converges to some s ∈ R as n → ∞; i.e., for every  > 0 there is an N ∈N such that n ≥ N implies that

|s_n− s| < . In this case we shall write

∞

X

k =1

ak =s

and called s the sum, or value, of the seriesP∞

a

Definition (iii)

S is said to diverge if and only if its sequence of partial sums {sn} does not converge as n → ∞. When s_n diverges to +∞ as n → ∞, we shall also write

∞

X

k =1

ak = ∞.

Definition (iii)

S is said to diverge if and only if its sequence of partial sums {sn} does not converge as n → ∞. When s_n diverges to +∞ as n → ∞, we shall also write

∞

X

k =1

ak = ∞.

Definition (iii)

S is said to diverge if and only if its sequence of partial sums {sn} does not converge as n → ∞. When s_n diverges to +∞ as n → ∞, we shall also write

∞

X

k =1

ak = ∞.

Example:

Prove that the sequence 1

k converges but the series

∞

P

k =1

1 k diverges to +∞.

Example:

Prove that the sequence 1

k converges but the series

∞

P

k =1

1 k diverges to +∞.

Proof:

The sequence 1/k converges to zero (by Example 2.2).

On the other hand, by the Comparison Theorem for Integrals,

n

X

k =1

1 k ≥

n

X

k =1

Z k +1 k

1 xdx =

Z n+1 1

1

xdx = log(n + 1).

We conclude that sn → ∞ as n → ∞. 2

Proof:

The sequence 1/k converges to zero (by Example 2.2).

On the other hand, by the Comparison Theorem for Integrals,

n

X

k =1

1 k ≥

n

X

k =1

Z k +1 k

1 xdx =

Z n+1 1

1

xdx = log(n + 1).

We conclude that sn → ∞ as n → ∞. 2

Proof:

The sequence 1/k converges to zero (by Example 2.2).

On the other hand, by the Comparison Theorem for Integrals,

n

X

k =1

1 k ≥

n

X

k =1

Z k +1 k

1 xdx =

Z n+1 1

1

xdx = log(n + 1).

We conclude that sn → ∞ as n → ∞. 2

Proof:

The sequence 1/k converges to zero (by Example 2.2).

On the other hand, by the Comparison Theorem for Integrals,

n

X

k =1

1 k ≥

n

X

k =1

Z k +1 k

1 xdx =

Z n+1 1

1

xdx = log(n + 1).

We conclude that sn → ∞ as n → ∞. 2

Proof:

The sequence 1/k converges to zero (by Example 2.2).

On the other hand, by the Comparison Theorem for Integrals,

n

X

k =1

1 k ≥

n

X

k =1

Z k +1 k

1 xdx =

Z n+1 1

1

xdx = log(n + 1).

We conclude that sn → ∞ as n → ∞. 2

Theorem (Divergence Test)

Let {ak}k ∈Nbe a sequence of real numbers. If ak does not converge to zero, then the series

∞

P

k =1

ak diverges.

Theorem (Divergence Test)

Let {ak}k ∈Nbe a sequence of real numbers. If ak does not converge to zero, then the series

∞

P

k =1

ak diverges.

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition,the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞, a contradiction. 2

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition, the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞, a contradiction. 2

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition,the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞, a contradiction. 2

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition, the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞,a contradiction. 2

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition, the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞, a contradiction. 2

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition, the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞,a contradiction. 2

Proof:

Suppose to the contrary that

∞

P

k =1

ak converges to some s ∈R. By definition, the sequence of partial sums sn :=

n

P

k =1

ak converges to s as n → ∞. Therefore,

ak =sk − s_{k −1}→ s − s = 0 as k → ∞, a contradiction. 2

Theorem (Telescopic Test)

If {ak} is a convergent real sequence, then

∞

X

k =1

(ak − a_{k +1}) =a1− lim

k →∞ak.

Theorem (Telescopic Test)

If {ak} is a convergent real sequence, then

∞

X

k =1

(ak − a_{k +1}) =a1− lim

k →∞ak.

Proof:

By telescoping, we have

sn:=

n

X

k =1

(ak− ak +1) =a1− an+1.

Hence sn → a₁− lim

k →∞ak as n → ∞.2

Proof:

By telescoping,we have

sn:=

n

X

k =1

(ak− ak +1) =a1− an+1.

Hence sn → a₁− lim

k →∞ak as n → ∞.2

Proof:

By telescoping, we have

sn:=

n

X

k =1

(ak− ak +1) =a1− an+1.

Hence sn → a₁− lim

k →∞ak as n → ∞.2

Proof:

By telescoping, we have

sn:=

n

X

k =1

(ak− ak +1) =a1− an+1.

Hence sn → a₁− lim

k →∞ak as n → ∞.2

Theorem (Geometric Series) The series

∞

P

k =1

x^k converges if and only if |x | < 1, in which

case ∞

X

k =1

x^k = x 1 − x. (see also Exercise 1.)

Theorem (Geometric Series) The series

∞

P

k =1

x^k converges if and only if |x | < 1, in which

case ∞

X

k =1

x^k = x 1 − x. (see also Exercise 1.)

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1,then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1,then set sn =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1,then set sn =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Proof:

If |x | ≥ 1, then

∞

P

k =1

x^k diverges by the Divergence Test. If

|x| < 1, then set s_n =

n

P

k =1

x^k and observe by the telescoping that

(1 − x )Sn = (1 − x )(x + x²+ · · · +xⁿ)

=x + x²+ · · · +xⁿ− x²− x³− · · · − xⁿ⁺¹

=x − xⁿ⁺¹.

Hence,

sn= x

1 − x − xⁿ⁺¹ 1 − x

for all n ∈N. Since xⁿ⁺¹ → 0 as n → ∞ for all |x| < 1 (see Example 2.20), we conclude that sn → x

(1 − x ) as n → ∞.

2

Hence,

sn= x

1 − x − xⁿ⁺¹ 1 − x

for all n ∈N. Since xⁿ⁺¹ → 0 as n → ∞ for all |x| < 1 (see Example 2.20),we conclude that sn → x

(1 − x ) as n → ∞.

2

Hence,

sn= x

1 − x − xⁿ⁺¹ 1 − x

for all n ∈N. Since xⁿ⁺¹ → 0 as n → ∞ for all |x| < 1 (see Example 2.20), we conclude that sn → x

(1 − x ) as n → ∞.

2

Hence,

sn= x

1 − x − xⁿ⁺¹ 1 − x

for all n ∈N. Since xⁿ⁺¹ → 0 as n → ∞ for all |x| < 1 (see Example 2.20),we conclude that sn → x

(1 − x ) as n → ∞.

2

Hence,

sn= x

1 − x − xⁿ⁺¹ 1 − x

for all n ∈N. Since xⁿ⁺¹ → 0 as n → ∞ for all |x| < 1 (see Example 2.20), we conclude that sn → x

(1 − x ) as n → ∞.

2

Theorem (Cauchy Criterion)

Let {ak} be a real sequence. Then the infinite series

∞

P

k =1

ak converges if and only if for every  > 0 there is an N ∈N such that

m > n ≥ N imply     

m

X

k =n

ak

< .

Theorem (Cauchy Criterion)

Let {ak} be a real sequence. Then the infinite series

∞

P

k =1

ak converges if and only if for every  > 0 there is an N ∈N such that

m > n ≥ N imply     

m

X

k =n

ak

< .

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29),sn

converges if and only if given  > 0 there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1, the proof is complete. 2

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29), sn

converges if and only if given  > 0there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1, the proof is complete. 2

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29),sn

converges if and only if given  > 0 there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1, the proof is complete. 2

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29), sn

converges if and only if given  > 0there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1,the proof is complete. 2

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29), sn

converges if and only if given  > 0 there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1, the proof is complete. 2

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29), sn

converges if and only if given  > 0 there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1,the proof is complete. 2

Proof:

Let sn represent the sequence of partial sums of

∞

P

k =1

ak

and set s0 =0. By Cauchy’s Theorem (Theorem 2.29), sn

converges if and only if given  > 0 there is an N ∈N such that m, n ≥ N imply |sm− s_n−1| < . Since

sm− sn−1 =

m

X

k =n

ak

for all integers m > n ≥ 1, the proof is complete. 2

Theorem

Let {ak} and {bk} be real sequences. If

∞

P

k =1

ak and

∞

P

k =1

bk

are convergent series, then

∞

X

k =1

(ak +bk) =

∞

X

k =1

ak +

∞

X

k =1

bk

and ∞

X

k =1

(αak) = α

∞

X

k =1

ak

Theorem

Let {ak} and {bk} be real sequences. If

∞

P

k =1

ak and

∞

P

k =1

bk

are convergent series, then

∞

X

k =1

(ak +bk) =

∞

X

k =1

ak +

∞

X

k =1

bk

and ∞

X

k =1

(αak) = α

∞

X

k =1

ak

Theorem

Let {ak} and {bk} be real sequences. If

∞

P

k =1

ak and

∞

P

k =1

bk

are convergent series, then

∞

X

k =1

(ak +bk) =

∞

X

k =1

ak +

∞

X

k =1

bk

and ∞

X

k =1

(αak) = α

∞

X

k =1

ak

Exmple:

(1)

∞

P

k =1

1 k (k + 1)

(2)

∞

P

k =1

 1 − 1

k

k

Exmple:

(1)

∞

P

k =1

1 k (k + 1)

(2)

∞

P

k =1

 1 − 1

k

k

Exmple:

(1)

∞

P

k =1

1 k (k + 1)

(2)

∞

P

k =1

 1 − 1

k

k

Thank you.