24 the harmonic series and the integral test x

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:
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:
If = a1 + a2 + a3 + … = L is aΣi=1
∞
ai

∞
ai
Theorem:
This means that for the sequence of partial sums,
lim sn = lim (a1 + a2 + … + an) = L converges.
n∞
If = a1 + a2 + a3 + … = L is aΣi=1
∞
ai

∞
ai
Theorem:
n∞
lim sn-1 = lim (a1 + a2 +..+ an-1)n∞ n∞
If = a1 + a2 + a3 + … = L is aΣi=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 aΣi=1
∞
ai
On the other hand,

∞
ai
Theorem:
n∞
Hence lim sn – sn-1n∞
If = a1 + a2 + a3 + … = L is aΣi=1
∞
ai
On the other hand,

∞
ai
Theorem:
n∞
Hence lim sn – sn-1 = lim ann∞ n∞
If = a1 + a2 + a3 + … = L is aΣi=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 aΣi=1
∞
ai
On the other hand,

However the fact that lim an 0 does not guarantee
that their sum converges.

Example A.
The sequence 1,
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,
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,
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,
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,
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,
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,
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,
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,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
To see that they sum to ∞, sum in blocks as shown:

Example A.
The sequence 1,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
1

Example A.
The sequence 1,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
1
2
1
3
1
10...1+ + + +

Example A.
The sequence 1,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
1
2
1
3
1
10...1+ + + +
> 9
10

Example A.
The sequence 1,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
1
2
1
3
1
10...1+ + + + 1
11+ 1
100... +
> 9
10

Example A.
The sequence 1,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
1
2
1
3
1
10...1+ + + + 1
11+ 1
100... +
> 9
10 > 90
100
= 9
10

Example A.
The sequence 1,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
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,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
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,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
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,
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 ..+ + + + + + + + = ∞
2 ,
1
3 ,
1
4 , ..1,{ }
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 divergent harmonic series
Σn=1
∞
1/n = 1
2
1
3 ...1+ + + = ∞
Area = ∞
y = 1/x

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

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
∞
∞
Area = ∞
Area = ∞
y = 1/x
y = 1/x

the convergent and divergent p-series
Σn=1
∞
1/n = 1
2
1
3 ...1+ + + = ∞
∫1
x
1 dx = ∞
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
∞
∞
Area = ∞
Area = ∞
y = 1/x
y = 1/x

the convergent and divergent p-series
Σn=1
∞
1/n = 1
2
1
3 ...1+ + + = ∞
∫1
x
1 dx = ∞
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
∞
∞
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 Integral Test
∫f(x) dx converge or both diverge.
1
an andboth
∞

Σn =1
∞
The Integral Test
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
∞
The Integral Test
1
an andboth
∞
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
∞
The Integral Test
1
an andboth
∞
oscillates forever.
For example, for all n,
let an = 0 = f(n)
where f(x) = sin2(πx),

Σn =1
∞
The Integral Test
1
an andboth
∞
oscillates forever.
let an = 0 = f(n)
f(x) = (sin(πx))2
(1, 0) (2, 0) (3, 0)

Σn =1
∞
The Integral Test
1
an andboth
∞
oscillates forever.
let an = 0 = f(n)
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)
converges if and only if p > 1.
∞
np
1

Σn=1
Theorem (p-series)
∞
np
1
Proof: The functions y = 1/xp are positive decreasing
functions hence the integral test applies.

Σn=1
Theorem (p-series)
∞
np
1
∫1
xp
1
∞
dx converges if and only if p > 1,Since
Σn=1
∞
np
1
so

Σn=1
Theorem (p-series)
∞
np
1
n1.01
1
converges and thatΣn=1
∞
So n0.99
1
Σn=1
∞
diverges.
∫1
xp
1
∞
Σn=1
∞
np
1
so

Σn=1
Theorem (p-series)
∞
np
1
n1.01
1
∞
So n0.99
1
Σn=1
∞
diverges.
∫1
xp
1
∞
Σn=1
∞
np
1
so
Recall the floor and ceiling theorems for integrals:

Σn=1
Theorem (p-series)
∞
np
1
n1.01
1
∞
So n0.99
1
Σn=1
∞
diverges.
∫1
xp
1
∞
Σn=1
∞
np
1
so
I. if f(x) ≥ g(x) ≥ 0 and g(x) dx = ∞, then f(x) = ∞.∫a
b
∫a
b

Σn=1
Theorem (p-series)
∞
np
1
n1.01
1
∞
So n0.99
1
Σn=1
∞
diverges.
∫1
xp
1
∞
Σn=1
∞
np
1
so
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

By the same logic we have their discrete versions.

The Floor Theorem

Suppose bn = ∞, then an = ∞.Σn=k
∞
Σn=k
∞
The Floor Theorem
Let {an} and {bn} be two sequences and an ≥ bn ≥ 0.

∞
Σn=k
∞
Example B. Does converge or diverge?
Σn=2
∞
Ln(n)
1
The Floor Theorem

∞
Σn=k
∞
Ln(n)
1 > n .
1For n > 1, n > Ln(n) (why?)
Σn=2
∞
Ln(n)
1
so
The Floor Theorem

∞
Σn=k
∞
Ln(n)
1 > n .
1For n > 1, n > Ln(n) (why?)
Σ
n=2 n
1therefore Σn=2 Ln(n)
1
> = ∞ diverges.
Σn=2
∞
Ln(n)
1
so
The Floor Theorem
Σn=2 n
1 = ∞Since

∞
Σn=k
∞
Ln(n)
1 > n .
1For n > 1, n > Ln(n) (why?)
Σ
n=2 n
1
> = ∞ diverges.
Σn=2
∞
Ln(n)
1
so
The Floor Theorem
No conclusion can be drawn about Σan if an ≥ bn and
Σbn < ∞.
Σn=2 n
1 = ∞Since

∞
Σn=k
∞
Ln(n)
1 > n .
1For n > 1, n > Ln(n) (why?)
Σ
n=2 n
1
> = ∞ diverges.
Σn=2
∞
Ln(n)
1
so
The Floor Theorem
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

∞
Σn=k
∞
Ln(n)
1 > n .
1For n > 1, n > Ln(n) (why?)
Σ
n=2 n
1
> = ∞ diverges.
Σn=2
∞
Ln(n)
1
so
The Floor Theorem
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 Ceiling Theorem

If an converges, then bn converges.Σn=k
Σn=k
The Ceiling Theorem

Σn=k
Example C. Does converge or diverge?
The Ceiling Theorem
Σn=1 n2 + 4
2

Σn=k
The Ceiling Theorem
Σn=1 n2 + 4
2
Compare with
n2 + 4
2
n2
2

Σn=k
n2 + 4
2>n2
2 ,we have
n2 + 4
The Ceiling Theorem
Σn=1 n2 + 4
2
Compare with
n2 + 4
2
n2
2
2ΣΣ n2
2 >so that

Σn=k
n2 + 4
2>n2
2 ,
Σn2
2
we have
= N converges since p = 2 > 1,
n2 + 4
The Ceiling Theorem
Σn=1 n2 + 4
2
Compare with
n2 + 4
2
n2
2
2
= 2Σ n2
1
ΣΣ n2
2 >so that

Σn=k
n2 + 4
2>n2
2 ,
Σn2
2
we have
n2 + 4
The Ceiling Theorem
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.

Σn=k
n2 + 4
2>n2
2 ,
Σn2
2
we have
n2 + 4
The Ceiling Theorem
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
∞
∫f(x) dx converge or both diverge.”
1
an andboth
∞
Here is an intuitive proof of the Integral Test

Σn =1
∞
1
an andboth
∞
∫ f(x) dx converges,
1
∞
I.
so if
Since ∫f(x) dx >
1
∞
a2 + a3 + …
then Σ an converges.

Σn =1
∞
1
an andboth
∞
∫ 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
∞

24 the harmonic series and the integral test x

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24 the harmonic series and the integral test x