The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

This means that for the sequence of partial sums,

lim sn = lim (a1 + a2 + … + an) = L converges.

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

n∞

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

n∞

lim sn-1 = lim (a1 + a2 +..+ an-1)n∞ n∞

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

On the other hand,

This means that for the sequence of partial sums,

lim sn = lim (a1 + a2 + … + an) = L converges.

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

n∞

lim sn-1 = lim (a1 + a2 +..+ an-1) = L.n∞ n∞

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

On the other hand,

This means that for the sequence of partial sums,

lim sn = lim (a1 + a2 + … + an) = L converges.

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

n∞

lim sn-1 = lim (a1 + a2 +..+ an-1) = L.n∞ n∞

Hence lim sn – sn-1n∞

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

On the other hand,

This means that for the sequence of partial sums,

lim sn = lim (a1 + a2 + … + an) = L converges.

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

n∞

lim sn-1 = lim (a1 + a2 +..+ an-1) = L.n∞ n∞

Hence lim sn – sn-1 = lim ann∞ n∞

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

On the other hand,

This means that for the sequence of partial sums,

lim sn = lim (a1 + a2 + … + an) = L converges.

Proof: Let = a1 + a2 .. = L be a convergent series.Σi=1

∞

ai

Theorem:

The Harmonic Series and the Integral Test

If we add infinitely many terms and obtain a finite sum,

it must be the case that the terms get smaller and

smaller and go to zero.

n∞

lim sn-1 = lim (a1 + a2 +..+ an-1) = L.n∞ n∞

Hence lim sn – sn-1 = lim an = L – L = 0.n∞ n∞

If = a1 + a2 + a3 + … = L is aΣi=1

∞

ai

convergent series, then lim an = 0.n∞

On the other hand,

This means that for the sequence of partial sums,

lim sn = lim (a1 + a2 + … + an) = L converges.

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges.

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

+ + +

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

+ + + + + + +

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + +

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + +

> 9

10

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + + 1

11+ 1

100… +

> 9

10

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + + 1

11+ 1

100… +

> 9

10 > 90

100

= 9

10

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + + 1

11+ 1

100… + 1

101+ 1

1000… +

> 9

10 > 90

100

= 9

10

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + + 1

11+ 1

100… + 1

101+ 1

1000… +

> 9

10 > 90

100

= 9

10

>

1000

=

10

900 9

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + + 1

11+ 1

100… + 1

101+ 1

1000… + + …

> 9

10 > 90

100

= 9

10

>

1000

=

10

900 9

= ∞

Example A.

The sequence 1,

The Harmonic Series and the Integral Test

However the fact that lim an 0 does not guarantee

that their sum converges. Here is an example.

1

2 ,

1

2 ,

1

3 ,

1

3 ,

1

3 ,

1

4 ,

1

4 ,

1

4 ,

1

4 , 0,1

5 , ..

but their sum 1+1

2

+ 1

2

1

3

1

3

1

3

1

4

1

4

1

4

1

4

1

5 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 1

2 ,

1

3 ,

1

4 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:

1

2

1

3

1

10…1+ + + + 1

11+ 1

100… + 1

101+ 1

1000… + + …

> 9

10 > 90

100

= 9

10

>

1000

=

10

900 9

= ∞

Hence the harmonic series diverges.

The Harmonic Series and the Integral Test

The divergent harmonic series

Σn=1

∞

1/n = 1

2

1

3 …1+ + + = ∞

Area = ∞

y = 1/x

The Harmonic Series and the Integral Test

The divergent harmonic series

Σn=1

∞

1/n = 1

2

1

3 …1+ + + = ∞

∫1

x

1 dx = ∞

∞

is the discrete version of the improper integral

Area = ∞

Area = ∞

y = 1/x

y = 1/x

The Harmonic Series and the Integral Test

The divergent harmonic series

Σn=1

∞

1/n = 1

2

1

3 …1+ + + = ∞

∫1

x

1 dx = ∞

The series 1/np are called p-series and they

correspond to the p-functions 1/xp where 1 ≤ x < ∞.

Σn=1

∞

∞

is the discrete version of the improper integral

Area = ∞

Area = ∞

y = 1/x

y = 1/x

The Harmonic Series and the Integral Test

The divergent harmonic series

the convergent and divergent p-series

Σn=1

∞

1/n = 1

2

1

3 …1+ + + = ∞

∫1

x

1 dx = ∞

The series 1/np are called p-series and they

correspond to the p-functions 1/xp where 1 ≤ x < ∞.

Just as the function 1/x serves as the boundary

between the convergent and divergent 1/xp,

the harmonic series 1/n serves as the boundary of

Σn=1

∞

Σn=1

∞

Σ 1/np.n=1

∞

∞

is the discrete version of the improper integral

Area = ∞

Area = ∞

y = 1/x

y = 1/x

The Harmonic Series and the Integral Test

The divergent harmonic series

the convergent and divergent p-series

Σn=1

∞

1/n = 1

2

1

3 …1+ + + = ∞

∫1

x

1 dx = ∞

The series 1/np are called p-series and they

correspond to the p-functions 1/xp where 1 ≤ x < ∞.

Just as the function 1/x serves as the boundary

between the convergent and divergent 1/xp,

the harmonic series 1/n serves as the boundary of

Σn=1

∞

Σn=1

∞

The following theorem establishes the connection

between the discrete and the continuous cases.

Σ 1/np.n=1

∞

∞

is the discrete version of the improper integral

Area = ∞

Area = ∞

y = 1/x

y = 1/x

Σn =1

Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

The Integral Test

∫f(x) dx converge or both diverge.

1

an andboth

∞

Σn =1

Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

The Integral Test

∫f(x) dx converge or both diverge.

1

an andboth

∞

In short, the series and the integrals behave the same.

We leave the proof to the end of the section.

Σn =1

Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

The Integral Test

∫f(x) dx converge or both diverge.

1

an andboth

∞

In short, the series and the integrals behave the same.

We leave the proof to the end of the section.

Note that f(x) has to be a decreasing function from

some point onward, i.e. f(x) can’t be a function that

oscillates forever.

Σn =1

Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

The Integral Test

∫f(x) dx converge or both diverge.

1

an andboth

∞

In short, the series and the integrals behave the same.

We leave the proof to the end of the section.

Note that f(x) has to be a decreasing function from

some point onward, i.e. f(x) can’t be a function that

oscillates forever.

For example, for all n,

let an = 0 = f(n)

where f(x) = sin2(πx),

Σn =1

Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

The Integral Test

∫f(x) dx converge or both diverge.

1

an andboth

∞

In short, the series and the integrals behave the same.

We leave the proof to the end of the section.

Note that f(x) has to be a decreasing function from

some point onward, i.e. f(x) can’t be a function that

oscillates forever.

For example, for all n,

let an = 0 = f(n)

where f(x) = sin2(πx),

f(x) = (sin(πx))2

(1, 0) (2, 0) (3, 0)

Σn =1

Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

The Integral Test

∫f(x) dx converge or both diverge.

1

an andboth

∞

In short, the series and the integrals behave the same.

We leave the proof to the end of the section.

Note that f(x) has to be a decreasing function from

some point onward, i.e. f(x) can’t be a function that

oscillates forever.

For example, for all n,

let an = 0 = f(n)

where f(x) = sin2(πx),

f(x) = (sin(πx))2

(1, 0) (2, 0) (3, 0)

Σn=1

∞

∫f(x) dx diverges.

1

an = 0 =

∞

Σn = 1

∞

sin2(πx) converges, but

Σn=1

Theorem (p-series)

The Harmonic Series and the Integral Test

converges if and only if p > 1.

∞

np

1

Proof: The functions y = 1/xp are positive decreasing

functions hence the integral test applies.

Σn=1

Theorem (p-series)

The Harmonic Series and the Integral Test

converges if and only if p > 1.

∞

np

1

Proof: The functions y = 1/xp are positive decreasing

functions hence the integral test applies.

∫1

xp

1

∞

dx converges if and only if p > 1,Since

Σn=1

converges if and only if p > 1.

∞

np

1

so

Σn=1

Theorem (p-series)

The Harmonic Series and the Integral Test

converges if and only if p > 1.

∞

np

1

Proof: The functions y = 1/xp are positive decreasing

functions hence the integral test applies.

n1.01

1

converges and thatΣn=1

∞

So n0.99

1

Σn=1

∞

diverges.

∫1

xp

1

∞

dx converges if and only if p > 1,Since

Σn=1

converges if and only if p > 1.

∞

np

1

so

Σn=1

Theorem (p-series)

The Harmonic Series and the Integral Test

converges if and only if p > 1.

∞

np

1

Proof: The functions y = 1/xp are positive decreasing

functions hence the integral test applies.

n1.01

1

converges and thatΣn=1

∞

So n0.99

1

Σn=1

∞

diverges.

∫1

xp

1

∞

dx converges if and only if p > 1,Since

Σn=1

converges if and only if p > 1.

∞

np

1

so

Recall the floor and ceiling theorems for integrals:

Σn=1

Theorem (p-series)

The Harmonic Series and the Integral Test

converges if and only if p > 1.

∞

np

1

Proof: The functions y = 1/xp are positive decreasing

functions hence the integral test applies.

n1.01

1

converges and thatΣn=1

∞

So n0.99

1

Σn=1

∞

diverges.

∫1

xp

1

∞

dx converges if and only if p > 1,Since

Σn=1

converges if and only if p > 1.

∞

np

1

so

Recall the floor and ceiling theorems for integrals:

I. if f(x) ≥ g(x) ≥ 0 and g(x) dx = ∞, then f(x) = ∞.∫a

b

∫a

b

Σn=1

Theorem (p-series)

The Harmonic Series and the Integral Test

converges if and only if p > 1.

∞

np

1

Proof: The functions y = 1/xp are positive decreasing

functions hence the integral test applies.

n1.01

1

converges and thatΣn=1

∞

So n0.99

1

Σn=1

∞

diverges.

∫1

xp

1

∞

dx converges if and only if p > 1,Since

Σn=1

converges if and only if p > 1.

∞

np

1

so

Recall the floor and ceiling theorems for integrals:

I. if f(x) ≥ g(x) ≥ 0 and g(x) dx = ∞, then f(x) = ∞.∫a

b

∫a

b

II. if f(x) ≥ g(x) ≥ 0 and f(x) dx < ∞, then g(x) < ∞.∫a

b

∫a

b

The Harmonic Series and the Integral Test

By the same logic we have their discrete versions.

The Floor Theorem

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

By the same logic we have their discrete versions.

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

Example B. Does converge or diverge?

By the same logic we have their discrete versions.

Σn=2

∞

Ln(n)

1

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

Example B. Does converge or diverge?

Ln(n)

1 > n .

1For n > 1, n > Ln(n) (why?)

By the same logic we have their discrete versions.

Σn=2

∞

Ln(n)

1

so

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

Example B. Does converge or diverge?

Ln(n)

1 > n .

1For n > 1, n > Ln(n) (why?)

Σ

n=2 n

1therefore Σn=2 Ln(n)

1

By the same logic we have their discrete versions.

> = ∞ diverges.

Σn=2

∞

Ln(n)

1

so

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=2 n

1 = ∞Since

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

Example B. Does converge or diverge?

Ln(n)

1 > n .

1For n > 1, n > Ln(n) (why?)

Σ

n=2 n

1therefore Σn=2 Ln(n)

1

By the same logic we have their discrete versions.

> = ∞ diverges.

Σn=2

∞

Ln(n)

1

so

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

No conclusion can be drawn about Σan if an ≥ bn and

Σbn < ∞.

Σn=2 n

1 = ∞Since

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

Example B. Does converge or diverge?

Ln(n)

1 > n .

1For n > 1, n > Ln(n) (why?)

Σ

n=2 n

1therefore Σn=2 Ln(n)

1

By the same logic we have their discrete versions.

> = ∞ diverges.

Σn=2

∞

Ln(n)

1

so

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

No conclusion can be drawn about Σan if an ≥ bn and

Σbn < ∞. For example, both 1/n and 1/n3/2 > bn = 1/n2

and Σ1/n2 = Σbn < ∞ since p = 2 > 1,

Σn=2 n

1 = ∞Since

The Harmonic Series and the Integral Test

Suppose bn = ∞, then an = ∞.Σn=k

∞

Σn=k

∞

Example B. Does converge or diverge?

Ln(n)

1 > n .

1For n > 1, n > Ln(n) (why?)

Σ

n=2 n

1therefore Σn=2 Ln(n)

1

By the same logic we have their discrete versions.

> = ∞ diverges.

Σn=2

∞

Ln(n)

1

so

The Floor Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

No conclusion can be drawn about Σan if an ≥ bn and

Σbn < ∞. For example, both 1/n and 1/n3/2 > bn = 1/n2

and Σ1/n2 = Σbn < ∞ since p = 2 > 1, however

Σ1/n = ∞ diverges but Σ1/n3/2 < ∞ converges,

so in general no conclusion can be made about Σan.

Σn=2 n

1 = ∞Since

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

Example C. Does converge or diverge?

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=1 n2 + 4

2

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

Example C. Does converge or diverge?

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=1 n2 + 4

2

Compare with

n2 + 4

2

n2

2

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

Example C. Does converge or diverge?

n2 + 4

2>n2

2 ,we have

n2 + 4

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=1 n2 + 4

2

Compare with

n2 + 4

2

n2

2

2ΣΣ n2

2 >so that

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

Example C. Does converge or diverge?

n2 + 4

2>n2

2 ,

Σn2

2

we have

= N converges since p = 2 > 1,

n2 + 4

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=1 n2 + 4

2

Compare with

n2 + 4

2

n2

2

2

= 2Σ n2

1

ΣΣ n2

2 >so that

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

Example C. Does converge or diverge?

n2 + 4

2>n2

2 ,

Σn2

2

we have

= N converges since p = 2 > 1,

n2 + 4

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=1 n2 + 4

2

Compare with

n2 + 4

2

n2

2

2

= 2Σ n2

1

ΣΣ n2

2 >so that

therefore N > n2 + 4

2Σ so the series converges.

The Harmonic Series and the Integral Test

If an converges, then bn converges.Σn=k

Σn=k

Example C. Does converge or diverge?

n2 + 4

2>n2

2 ,

Σn2

2

we have

= N converges since p = 2 > 1,

n2 + 4

The Ceiling Theorem

Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

Σn=1 n2 + 4

2

Compare with

n2 + 4

2

n2

2

2

No conclusion can be drawn about Σbn if an ≥ bn and

Σan = ∞ for Σbn may converge or it may diverge.

= 2Σ n2

1

ΣΣ n2

2 >so that

therefore N > n2 + 4

2Σ so the series converges.

Σn =1

“Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

∫f(x) dx converge or both diverge.”

1

an andboth

∞

Here is an intuitive proof of the Integral Test

Σn =1

“Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

∫f(x) dx converge or both diverge.”

1

an andboth

∞

Here is an intuitive proof of the Integral Test

∫ f(x) dx converges,

1

∞

I.

so if

Since ∫f(x) dx >

1

∞

a2 + a3 + …

then Σ an converges.

Σn =1

“Let an = f(n) where f(x) is a positive continuous

eventually decreasing function defined for 1 ≤ x, then

The Harmonic Series and the Integral Test

∞

∫f(x) dx converge or both diverge.”

1

an andboth

∞

Here is an intuitive proof of the Integral Test

∫ f(x) dx converges,

1

∞

I.

so if

Since ∫f(x) dx >

1

∞

a2 + a3 + …

then Σ an converges.

Let g(x) = f(x + 1) so g(x) is f(x) shifted left by 1,lI.

and that g(0) = a1, g(1) = a2, g(2) = a3,. . etc. g(x) is

decreasing hence it’s below the

rectangles. So if Σ an converges

f(x) dx converges.

1

∫then

∞