the harmonic series and the integral test

1. 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.

2. 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∞

3. 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∞

4. 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∞

5. 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.

6. 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.

7. 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.

8. 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.

9. 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.

10. The Harmonic Series and the Integral Test
However the fact that lim an 0 does not guarantee
that their sum converges.

11. 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 , ..

12. 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

13. 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

14. 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

15. 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
+ + +

16. 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
+ + + + + + +

17. 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 ..+ + + + + + + + = ∞

18. 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,{ }

19. 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:

20. 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

21. 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+ + + +

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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
= ∞

28. 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.

29. The Harmonic Series and the Integral Test
The divergent harmonic series
Σn=1
∞
1/n = 1
2
1
3 ...1+ + + = ∞
Area = ∞
y = 1/x

30. 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

31. 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

32. 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

33. 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

34. 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
∞

35. 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.

36. 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.

37. 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),

38. 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)

39. 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

40. Σn=1
Theorem (p-series)
The Harmonic Series and the Integral Test
converges if and only if p > 1.
∞
np
1

41. 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.

42. 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

43. 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

44. 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:

45. 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

46. 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

47. The Harmonic Series and the Integral Test
By the same logic we have their discrete versions.

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

49. 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.

50. 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.

51. 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.

52. 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

53. 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

54. 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

55. 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

56. The Harmonic Series and the Integral Test
The Ceiling Theorem

57. 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.

58. 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

59. 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

60. 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

61. 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

62. 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.

63. 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.

64. 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

65. 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.

66. 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
∞