Calculus (I)

Calculus (I)

W^EN-C^HING L^IEN

Department of Mathematics National Cheng Kung University

2008

10.4

Theorem (1) If p >0, then lim

n→∞

√np=1

10.4

Theorem (1) If p >0, then lim

n→∞

√np=1

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

n

q1 p

→1. 2

pf:

(i)If p =1, trivial

(ii)If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

q1 →1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

n

q1 p

→1. 2

pf:

(i) If p =1, trivial

(ii)If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

q1 →1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

n

q1 p

→1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

q1 →1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

n

q1 p

→1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii)If p <1, then√ⁿp = 1

q1 →1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1,then√ⁿp = 1

n

q1 p

→1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii)If p <1, then√ⁿp = 1

q1 →1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1,then√ⁿp = 1

n

q1 p

→1. 2

pf:

(i) If p =1, trivial

(ii) If p >1, set xn =√ⁿp−1 (>0),

then (1+xn)ⁿ=p ≥1+n·xn (the binomial theorem)

⇒ 0<xn ≤ p−1 n

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√np=1

(iii) If p <1, then√ⁿp = 1

q1 →1. 2

Theorem (2)

nlim→∞

√n

n=1

Theorem (2)

nlim→∞

√n

n=1

pf:

Let xn =√ⁿ

n−1≥0

⇒ n= (1+xn)ⁿ ≥ n(n−1) 2 xn2

⇒ 0≤xn ≤

r 2 n−1

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√n

n=1 2

pf:

Let xn =√ⁿ

n−1≥0

⇒ n= (1+xn)ⁿ ≥ n(n−1) 2 xn2

⇒ 0≤xn ≤

r 2 n−1

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√n

n=1 2

pf:

Let xn =√ⁿ

n−1≥0

⇒ n= (1+xn)ⁿ ≥ n(n−1) 2 xn2

⇒ 0≤xn ≤

r 2 n−1

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√n

n=1 2

pf:

Let xn =√ⁿ

n−1≥0

⇒ n= (1+xn)ⁿ ≥ n(n−1) 2 xn2

⇒ 0≤xn ≤

r 2 n−1

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√n

n=1 2

pf:

Let xn =√ⁿ

n−1≥0

⇒ n= (1+xn)ⁿ ≥ n(n−1) 2 xn2

⇒ 0≤xn ≤

r 2 n−1

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√n

n=1 2

pf:

Let xn =√ⁿ

n−1≥0

⇒ n= (1+xn)ⁿ ≥ n(n−1) 2 xn2

⇒ 0≤xn ≤

r 2 n−1

⇒ lim

n→∞

xn=0

⇒ lim

n→∞

√n

n=1 2

Theorem (3)

nlim→∞(1+^x_n)ⁿ=e^x

Theorem (3)

nlim→∞(1+^x_n)ⁿ=e^x

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i)For x =0, trivial.

(ii)For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii)For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

pf:

(i) For x =0, trivial.

(ii) For x 6=0, ln(1+ x

n)ⁿ =n ln(1+ x

n) =x· ln(1+^x_n)−ln 1

x n

nlim→∞

ln(1+ x

n)−ln 1

x n

= d

dt(ln t)|t=1=1

∴ lim

n→∞

ln(1+ ^x_n)ⁿ=x

∴ lim

n→∞

(1+ ^x_n)ⁿ=e^x 2

Remark:(Cauchy sequence)

Theorem

(1)Every convergent sequence is a Cauchy sequence (2)Every Cauchy sequence is convergent

Remark:(Cauchy sequence)

Theorem

(1)Every convergent sequence is a Cauchy sequence (2)Every Cauchy sequence is convergent

Remark:(Cauchy sequence)

Theorem

(1)Every convergent sequence is a Cauchy sequence (2)Every Cauchy sequence is convergent

Remark:(Cauchy sequence)

Theorem

(1)Every convergent sequence is a Cauchy sequence (2)Every Cauchy sequence is convergent

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

pf :

(1)

Let an →L

We want to show that ∀ǫ >0, ∃an index K s.t.

|an−am|< ǫ for all m,n≥K

For givenǫ >0,

∃ K s.t. |an−L|< ǫ

2 for n ≥K Thus, if m,n≥K ,

|an−am| ≤ |an−L|+|L−am|< ǫ 2 + ǫ

2 =ǫ 2

(2)

Let{an}be a Cauchy sequence Want to prove that {an}is convergent

(2)

Let{an}be a Cauchy sequence Want to prove that {an}is convergent

(2)

Let{an}be a Cauchy sequence Want to prove that {an}is convergent

Lemma:

For any n ∈N, an<bn, In ≡[an,bn] (a)∀n ∈N, In⊇In+1, then

∞

T

n=1

In6=φ (b)If lim

n→∞

(bn−an) =0, then

∞

T

n=1

Inis a set containing exactly one point

Lemma:

For any n ∈N, an<bn, In ≡[an,bn] (a)∀n ∈N, In⊇In+1, then

∞

T

n=1

In6=φ (b)If lim

n→∞

(bn−an) =0, then

∞

T

n=1

Inis a set containing exactly one point

Lemma:

For any n ∈N, an<bn, In ≡[an,bn] (a)∀n ∈N, In⊇In+1, then

∞

T

n=1

In6=φ (b)If lim

n→∞

(bn−an) =0, then

∞

T

n=1

Inis a set containing exactly one point

Lemma:

For any n ∈N, an<bn, In ≡[an,bn] (a)∀n ∈N, In⊇In+1, then

∞

T

n=1

In6=φ (b)If lim

n→∞

(bn−an) =0, then

∞

T

n=1

Inis a set containing exactly one point

Lemma:

For any n ∈N, an<bn, In ≡[an,bn] (a)∀n ∈N, In⊇In+1, then

∞

T

n=1

In6=φ (b)If lim

n→∞

(bn−an) =0, then

∞

T

n=1

Inis a set containing exactly one point

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n But lim (b −a ) =0, ⇒a=b 2

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n But lim (b −a ) =0, ⇒a=b 2

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n But lim (b −a ) =0, ⇒a=b 2

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n But lim (b −a ) =0, ⇒a=b 2

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n

pf:

(a)

a1 ≤a2≤. . .an≤ · · · ≤bn ≤. . .b2≤b1

Set a=sup{ak :k ∈N},

∵completeness ⇒a∈R

And an ≤a≤bn, ∀n ⇒a∈

∞

T

n=1

In

(b) If a,b ∈

∞

T

n=1

In,

then 0≤ |a−b| ≤bn−an, ∀n But lim (b −a ) =0, ⇒a=b 2

proof of (2):

∃n1 ∈N s.t. |ak −aj|< 1

2², ∀k,j ≥n1

∃n2 ∈N s.t. |ak −aj|< 1

2²⁺¹, ∀k,j ≥n2 (∴n2≥n1) We can construct a sequence nl ≤nl+1, ∀l ∈N

and|ak −aj|< 1

2^l+1, ∀k,j ≥nl

proof of (2):

∃n1 ∈N s.t. |ak −aj|< 1

2², ∀k,j ≥n1

∃n2 ∈N s.t. |ak −aj|< 1

2²⁺¹, ∀k,j ≥n2 (∴n2≥n1) We can construct a sequence nl ≤nl+1, ∀l ∈N

and|ak −aj|< 1

2^l+1, ∀k,j ≥nl

proof of (2):

∃n1 ∈N s.t. |ak −aj|< 1

2², ∀k,j ≥n1

∃n2 ∈N s.t. |ak −aj|< 1

2²⁺¹, ∀k,j ≥n2 (∴n2≥n1) We can construct a sequence nl ≤nl+1, ∀l ∈N

and|ak −aj|< 1

2^l+1, ∀k,j ≥nl

proof of (2):

∃n1 ∈N s.t. |ak −aj|< 1

2², ∀k,j ≥n1

∃n2 ∈N s.t. |ak −aj|< 1

2²⁺¹, ∀k,j ≥n2 (∴n2≥n1) We can construct a sequence nl ≤nl+1, ∀l ∈N

and|ak −aj|< 1

2^l+1, ∀k,j ≥nl

proof of (2):

∃n1 ∈N s.t. |ak −aj|< 1

2², ∀k,j ≥n1

∃n2 ∈N s.t. |ak −aj|< 1

2²⁺¹, ∀k,j ≥n2 (∴n2≥n1) We can construct a sequence nl ≤nl+1, ∀l ∈N

and|ak −aj|< 1

2^l+1, ∀k,j ≥nl

Set Il ≡

anl − 1

2^l,anl + 1 2^l

(i)

Il+1⊆Il, ∀l ∈N If b ∈Il+1,

b−anl+1

< 1

2^l+1 Since

anl+1 −anl

< 1

2^l+1, thus|b−anl|< 1

2^l ⇒b∈Il

Set Il ≡

anl − 1

2^l,anl + 1 2^l

(i)

Il+1⊆Il, ∀l ∈N If b ∈Il+1,

b−anl+1

< 1

2^l+1 Since

anl+1 −anl

< 1

2^l+1, thus|b−anl|< 1

2^l ⇒b∈Il

Set Il ≡

anl − 1

2^l,anl + 1 2^l

(i)

Il+1⊆Il, ∀l ∈N If b ∈Il+1,

b−anl+1

< 1

2^l+1 Since

anl+1 −anl

< 1

2^l+1, thus|b−anl|< 1

2^l ⇒b∈Il

Set Il ≡

anl − 1

2^l,anl + 1 2^l

(i)

Il+1⊆Il, ∀l ∈N If b ∈Il+1,

b−anl+1

< 1

2^l+1 Since

anl+1 −anl

< 1

2^l+1, thus|b−anl|< 1

2^l ⇒b∈Il

Set Il ≡

anl − 1

2^l,anl + 1 2^l

(i)

Il+1⊆Il, ∀l ∈N If b ∈Il+1,

b−anl+1

< 1

2^l+1 Since

anl+1 −anl

< 1

2^l+1, thus|b−anl|< 1

2^l ⇒b∈Il

Set Il ≡

anl − 1

2^l,anl + 1 2^l

(i)

Il+1⊆Il, ∀l ∈N If b ∈Il+1,

b−anl+1

< 1

2^l+1 Since

anl+1 −anl

< 1

2^l+1, thus|b−anl|< 1

2^l ⇒b∈Il

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p Givenǫ >0,∃l ∈N s.t. 1

2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒

p−an_l+1

< 1

2^l+1

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p Givenǫ >0,∃l ∈N s.t. 1

2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒ p−an

< 1

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p Givenǫ >0,∃l ∈N s.t. 1

2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒

p−an_l+1

< 1

2^l+1

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p

Givenǫ >0,∃l ∈N s.t. 1 2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒ p−an

< 1

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p

Givenǫ >0,∃l ∈N s.t. 1 2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒

p−an_l+1

< 1

2^l+1

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p

Givenǫ >0,∃l ∈N s.t. 1 2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒ p−an

< 1

(ii)

nlim→∞|Il|=0, so by lemma

∃a point p in R s.t.

∞

T

l=1

Il ={p} Want to prove that lim

n→∞

an=p

Givenǫ >0,∃l ∈N s.t. 1 2^l < ǫ We have

an−anl+1

< 1

2^l+2 if n>nl+1

p ∈Il+1 ⇒

p−an_l+1

< 1

2^l+1

Thus, |p−an| ≤

an−an_l+1

+

an_l+1 −p

≤ 1

2^l+2 + 1 2^l+1

< 1 2^l

< ǫ, for n≥nl+1

∴ lim

n→∞

an=p 2

Thus, |p−an| ≤

an−an_l+1

+

an_l+1 −p

≤ 1

2^l+2 + 1 2^l+1

< 1 2^l

< ǫ, for n≥nl+1

∴ lim

n→∞

an=p 2

Thus, |p−an| ≤

an−an_l+1

+

an_l+1 −p

≤ 1

2^l+2 + 1 2^l+1

< 1 2^l

< ǫ, for n≥nl+1

∴ lim

n→∞

an=p 2

Thus, |p−an| ≤

an−an_l+1

+

an_l+1 −p

≤ 1

2^l+2 + 1 2^l+1

< 1 2^l

< ǫ, for n≥nl+1

∴ lim

n→∞

an=p 2

Thus, |p−an| ≤

an−an_l+1

+

an_l+1 −p

≤ 1

2^l+2 + 1 2^l+1

< 1 2^l

< ǫ, for n≥nl+1

∴ lim

n→∞

an=p 2

Example:

∀n ∈N, Sn=

n

P

k=1

1

k, is{Sn}convergent ?

Thank you.