X27 the harmonic series and the integral test

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 goes to zero.

Theorem:





If = a1 + a2 + a3 + … = L is a Σi=1

∞

ai

convergent series, then lim an = 0.n∞

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

∞

ai

Theorem:





If = a1 + a2 + a3 + … = L is a Σi=1

∞

ai



∞

ai

Theorem:

This means the for the sequence of partial sums,

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





n∞

If = a1 + a2 + a3 + … = L is a Σi=1

∞

ai



∞

ai

Theorem:





n∞

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

If = a1 + a2 + a3 + … = L is a Σi=1

∞

ai


On the other hand,




∞

ai

Theorem:





n∞

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

Since an = sn – sn-1,n∞

If = a1 + a2 + a3 + … = L is a Σi=1

∞

ai


On the other hand,




∞

ai

Theorem:





n∞

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

Since an = sn – sn-1, so lim an = lim sn – sn-1 = L – L = 0.n∞

n∞

If = a1 + a2 + a3 + … = L is a Σi=1

∞

ai


On the other hand,




However the fact that lim an 0 does not guarantee

that their sum CGs to a finite number.

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

+ + +

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

+ + + + + + +

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞

is the harmonic sequence: {1/n} = 12 ,

13 ,

14 , ..1,{ }

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }

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

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


1

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + +

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + +

> 910

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... +

> 910

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... +

> 910 > 90

100

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... +

> 910 > 90

100= 9

10

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... + 1101

+ 11000

... +

> 910 > 90

100= 9

10

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... + 1101

+ 11000

... +

> 910 > 90

100= 9

10>

1000900

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... + 1101

+ 11000

... +

> 910 > 90

100= 9

10>

1000=

10900 9

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... + 1101

+ 11000

... + + …

> 910 > 90

100= 9

10>

1000=

10900 9

= ∞

Example:

The sequence 1,




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum 1+ 12

+ 12

13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞



13 ,

14 , ..1,{ }


12

13

110

...1 + + + + 111

+ 1100

... + 1101

+ 11000

... + + …

> 910 > 90

100= 9

10>

1000=

10900 9

= ∞

Hence the harmonic series DGs.


The following theorem and theorems in the next

section give various methods of determining if a

series is convergent or divergent.





We shall assume all series are positive series, i.e.

all terms in the series are positive unless stated

otherwise.

Σi=1

∞

ai

Theorem:





(Integral Test) If an = f(n) > 0, then

CGs if and only if



otherwise.

∫1 f(x) dx CGs. ∞

Σi=1

∞

ai

Theorem:





(Integral Test) If an = f(n) > 0, then

CGs if and only if



otherwise.

∫1 f(x) dx CGs. ∞

Combine this with the p-theorem from before, we

have the following theorem about the convergence of

the p-series:

Σi=1

Theorem:


(p-series) CGs if and only if p > 1. ∞

np1

Σi=1

Theorem:



np1

Proof:

Σi=1

CGs if and only if CGs. ∞

np1

By the integral test,

∫1 x

p1

∞

dx

Σi=1

Theorem:



np1

Proof: By the integral test,

By the p-theorem, this integral CGs if and only if p >1.

Σi=1


np1

∫1 x

p1

∞

dx

Σi=1

Theorem:



np1



So CGs if and only if p > 1. Σi=1

∞

np1

Σi=1


np1

∫1 x

p1

∞

dx

Σi=1

Theorem:



np1



So CGs if and only if p > 1.

Example:

a. Σi=1

∞

n3/21

b. Σi=1

∞

n1

Σi=1

∞

np1

Σi=1


np1

∫1 x

p1

∞

dx

Σi=1

Theorem:



np1




Example:

a. CGs since 3/2 > 1.Σi=1

∞

n3/21

b. Σi=1

∞

n1

Σi=1

∞

np1

Σi=1


np1

∫1 x

p1

∞

dx

Σi=1

Theorem:



np1




Example:


∞

n3/21

b. DGs since 1/2 < 1.Σi=1

∞

n1

Σi=1

∞

np1

Σi=1


np1

∫1 x

p1

∞

dx

Σi=1

Theorem:



np1




Example:


∞

n3/21

b. DGs since 1/2 < 1.Σi=1

∞

n1

This theorem applies to series that are p-series

except for finitely many terms (eventual p-series).

Σi=1

∞

np1

Σi=1


np1

∫1 x

p1

∞

dx

Recall the following theorems of improper integrals.

(The Floor Theorem)


(The Floor Theorem)

y = f(x)

y = g(x)∞


(The Floor Theorem)

If f(x) > g(x) > 0 and g(x) dx = ∞, ∫a

b

y = f(x)

y = g(x)∞


(The Floor Theorem)

If f(x) > g(x) > 0 and g(x) dx = ∞, then f(x) = ∞. ∫a

b

∫a

b

y = f(x)

y = g(x)∞


(The Floor Theorem)


b

∫a

b

y = f(x)

y = g(x)∞

(The Ceiling theorem)


(The Floor Theorem)


b

∫a

b

y = f(x)

y = g(x)∞


y = f(x)

y = g(x)

N


(The Floor Theorem)


b

∫a

b

y = f(x)

y = g(x)∞


If f(x) > g(x) > 0 and f(x) dx = N converges∫a

b

y = f(x)

y = g(x)

N


(The Floor Theorem)


b

∫a

b

y = f(x)

y = g(x)∞


If f(x) > g(x) > 0 and f(x) dx = N converges then

g(x) dx converges also.

∫a

b

∫a

b

y = f(x)

y = g(x)

N



By the same logic we have their discrete versions.



The Floor Theorem



The Floor Theorem

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


Suppose bn = ∞, then an = ∞. Σi=k

∞

Σi=k

∞


The Floor Theorem




∞

Σi=k

∞

Example: Does CG or DG?


Σi=2

∞

Ln(n) 1

The Floor Theorem




∞

Σi=k

∞


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


Σi=2

∞

Ln(n) 1

The Floor Theorem




∞

Σi=k

∞


Ln(n) 1 >

n . 1

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


Σi=2

∞

Ln(n) 1

so

The Floor Theorem




∞

Σi=k

∞


Ln(n) 1 >

n . 1

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

Σi=2 n

1Hence Σi=2 Ln(n)

2


>

Σi=2

∞

Ln(n) 1

so

The Floor Theorem




∞

Σi=k

∞


Ln(n) 1 >

n . 1

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

Σi=2 n

1Hence Σi=2 Ln(n)

2


> = ∞ because it’s harmonic.

Σi=2

∞

Ln(n) 1

so

The Floor Theorem




∞

Σi=k

∞


Ln(n) 1 >

n . 1

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

Σi=2 n

1

Therefore

Hence Σi=2 Ln(n)

2



Σi=2

∞

Ln(n) 1

Σi=2 Ln(n)

2

so

= ∞ or that it DGs.

The Floor Theorem




∞

Σi=k

∞


Ln(n) 1 >

n . 1

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

Σi=2 n

1

Therefore

Hence Σi=2 Ln(n)

2



Σi=2

∞

Ln(n) 1

Σi=2 Ln(n)

2

so

= ∞ or that it DGs.

The Floor Theorem


Note that no conclusion can be drawn about Σan if that

Σ bn < ∞ i.e. Σ an may CG or it may DG. (Why so?)


The Ceiling Theorem


The Ceiling Theorem



Suppose that an CGs, then bn CGs.Σi=k

Σi=k

The Ceiling Theorem




Σi=k


The Ceiling Theorem


Σi=1

∞

n2 + 4 2



Σi=k


The Ceiling Theorem


Σi=1

∞

n2 + 4 2

Compare with n2 + 4

2n22



Σi=k


n2 + 4 2>

n22

. we have

The Ceiling Theorem


Σi=1

∞

n2 + 4 2

Compare with n2 + 4

2n22



Σi=k


n2 + 4 2>

n22

.

Σ n22

we have

The Ceiling Theorem


Σi=1

∞

n2 + 4 2

Compare with n2 + 4

2n22

= 2Σ n21



Σi=k


n2 + 4 2>

n22

.

Σ n22

we have

CGs since it’s the p–series with p = 2 > 1,

The Ceiling Theorem


Σi=1

∞

n2 + 4 2

Compare with n2 + 4

2n22

= 2Σ n21



Σi=k


n2 + 4 2>

n22

.

Σ n22

we have


n2 + 4

The Ceiling Theorem


Σi=1

∞

n2 + 4 2

Compare with n2 + 4

2n22

2we see that Σ CGs also.

= 2Σ n21



Σi=k


n2 + 4 2>

n22

.

Σ n22

we have


n2 + 4

The Ceiling Theorem


Σi=1

∞

n2 + 4 2

Compare with n2 + 4

2n22

2

Note that no conclusion can be drawn about Σbn if that

Σan = ∞ i.e. Σ bn may CG or it may DG. (Why so?)

we see that Σ CGs also.

= 2Σ n21

X27 the harmonic series and the integral test

Education