25 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) n∞ 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.n∞ n∞

If = a1 + a2 + a3 + … = L is a Σi=1

∞ai


On the other hand,




∞ai

Theorem:



n∞


Hence lim sn – sn-1 n∞

If = a1 + a2 + a3 + … = L is a Σi=1

∞ai


On the other hand,




∞ai

Theorem:



n∞


Hence lim sn – sn-1 = lim an n∞ n∞

If = a1 + a2 + a3 + … = L is a Σi=1

∞ai


On the other hand,




∞ai

Theorem:



n∞


Hence lim sn – sn-1 = lim an = 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 , ..




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..

but their sum




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 , 0, 1

5 , ..

but their sum 1+ 12 + 12




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13+ + +




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14+ + + + + + +




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞

An important sequence that goes to 0 but sums to ∞is the harmonic sequence: {1/n} = 1

2 ,13 ,

14 , ..1,{ }




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }

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




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }

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

1




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }

To see that they sum to ∞, sum in blocks as shown:12

13

110...1 + + + +




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + +

> 910




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... +

> 910




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... +

> 910 > 90

100




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... +

> 910 > 90

100= 910




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... + 1

101+ 11000... +

> 910 > 90

100= 910




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... + 1

101+ 11000... +

> 910 > 90

100= 910 > 1000

900




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... + 1

101+ 11000... +

> 910 > 90

100= 910 > 1000 = 10

900 9




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... + 1

101+ 11000... + + …

> 910 > 90

100= 910 > 1000 = 10

900 9

= ∞




12 ,

12 ,

13 ,

13 ,

13 ,

14 ,

14 ,

14 ,

14 , 0, 1

5 , ..


13

13

13

14

14

14

14

15 ..+ + + + + + + + = ∞


2 ,13 ,

14 , ..1,{ }


13

110...1 + + + + 1

11+ 1100... + 1

101+ 11000... + + …

> 910 > 90

100= 910 > 1000 = 10

900 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


∫1 f(x) dx CGs. ∞

Σi=1

∞

ai

Theorem:



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

CGs if and only if


∫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:


(p-series)


np1

Proof:

Σi=1

CGs if and only if CGs. ∞

np1

By the integral test,

∫1 xp

1∞

dx

Σi=1

Theorem:


(p-series)


np1

Proof: By the integral test,

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

Σi=1


np1

∫1 xp

1∞

dx

Σi=1

Theorem:


(p-series)


np1



So CGs if and only if p > 1. Σi=1

∞

np1

Σi=1


np1

∫1 xp

1∞

dx

Σi=1

Theorem:


(p-series)


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 xp

1∞

dx

Σi=1

Theorem:


(p-series)


np1




Example:

a. CGs since 3/2 > 1.Σi=1

∞

n3/21

b. Σi=1

∞

n1

Σi=1

∞

np1

Σi=1


np1

∫1 xp

1∞

dx

Σi=1

Theorem:


(p-series)


np1




Example:


∞

n3/21

b. DGs since 1/2 < 1.Σi=1

∞

n1

Σi=1

∞

np1

Σi=1


np1

∫1 xp

1∞

dx

Σi=1

Theorem:


(p-series)


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 xp

1∞

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)∞



a

b

∫a

b

y = f(x)

y = g(x)∞

(The Ceiling theorem)



a

b

∫a

b

y = f(x)

y = g(x)∞


y = f(x)

y = g(x)

N



a

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



a

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


The Harmonic Series and the Integral TestBy the same logic we have their discrete versions.

The Harmonic Series and the Integral TestBy the same logic we have their discrete versions.The Floor Theorem

The Harmonic Series and the Integral TestBy the same logic we have their discrete versions.The Floor TheoremLet {an} and {bn} be two sequences and an > bn > 0.


Suppose bn = ∞, then an = ∞. Σi=k

∞

Σi=k

∞

By the same logic we have their discrete versions.The Floor TheoremLet {an} and {bn} be two sequences and an > bn > 0.



∞

Σi=k

∞

Example: Does CG or DG?

By the same logic we have their discrete versions.

Σi=2

∞

Ln(n)

1

The Floor TheoremLet {an} and {bn} be two sequences and an > bn > 0.



∞

Σi=k

∞


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


Σi=2

∞

Ln(n)

1




∞

Σi=k

∞


Ln(n) 1 > n .

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


Σi=2

∞

Ln(n)

1

so




∞

Σi=k

∞


Ln(n) 1 > n .

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

Σi=2 n

1Hence Σi=2 Ln(n)

2


>

Σi=2

∞

Ln(n)

1

so




∞

Σi=k

∞


Ln(n) 1 > n .

1For 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




∞

Σi=k

∞


Ln(n) 1 > n .

1For 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.




∞

Σi=k

∞


Ln(n) 1 > n .

1For 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.


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 Harmonic Series and the Integral TestThe Ceiling Theorem

The Harmonic Series and the Integral TestThe Ceiling TheoremLet {an} and {bn} be two sequences and an > bn > 0.


Suppose that an CGs, then bn CGs. Σi=k

Σi=k

The Ceiling TheoremLet {an} and {bn} be two sequences and an > bn > 0.



Σi=k



Σi=1

∞

n2 + 4

2

2 2



Σi=k



Σi=1

∞

n2 + 4

2

Compare with n2 + 4

2n2 2



Σi=k


n2 + 4 2>n2

2. we have


Σi=1

∞

n2 + 4

2

Compare with n2 + 4

2n2 2



Σi=k


n2 + 4 2>n2

2.

Σ n2 2

we have


Σi=1

∞

n2 + 4

2

Compare with n2 + 4

2n2 2

= 2Σ n2 1



Σi=k


n2 + 4 2>n2

2.

Σ n2 2

we have

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


Σi=1

∞

n2 + 4

2

Compare with n2 + 4

2n2 2

= 2Σ n2 1



Σi=k


n2 + 4 2>n2

2.

Σ n2 2

we have


n2 + 1


Σi=1

∞

n2 + 4

2

Compare with n2 + 4

2n2 2

2we see that Σ CGs also.

= 2Σ n2 1



Σi=k


n2 + 4 2>n2

2.

Σ n2 2

we have


n2 + 1


Σi=1

∞

n2 + 4

2

Compare with n2 + 4

2n2 2

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 n2 1

25 the harmonic series and the integral test

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Transcript of 25 the harmonic series and the integral test