Alternating Series and Conditional Convergence

1. Alternating Series and Conditional Convergence
An alternating series is a series with alternating
positive and negative terms.

2. Alternating Series and Conditional Convergence
An alternating series is a series with alternating
positive and negative terms. Alternating series usually
are given as Σn=1
∞
(–1)nan = –a1 + a2 – a3 + a4 …. or as
Σn=1
∞
(–1)n+1an = a1 – a2 + a3 – a4 … where an > 0.

3. Alternating Series and Conditional Convergence
An alternating series is a series with alternating
positive and negative terms. Alternating series usually
are given as Σn=1
∞
(–1)nan = –a1 + a2 – a3 + a4 …. or as
Σn=1
∞
(–1)n+1an = a1 – a2 + a3 – a4 … where an > 0
From here on, we will write an alternating series as
Σn=1
∞
an where the sequence {an} has alternating signs.

4. Alternating Series and Conditional Convergence
An alternating series is a series with alternating
positive and negative terms. Alternating series usually
are given as Σn=1
∞
(–1)nan = –a1 + a2 – a3 + a4 …. or as
Σn=1
∞
(–1)n+1an = a1 – a2 + a3 – a4 … where an > 0
Theorem (Alternating Series):
Suppose {an} is a alternating sequence where {lanl} is
a decreasing sequence and lim an = 0,
then the alternating series converges.
From here on, we will write an alternating series as
Σn=1
∞
an where the sequence {an} has alternating signs.
Σn=1
∞
an

5. Alternating Series and Conditional Convergence
An alternating series is a series with alternating
positive and negative terms. Alternating series usually
are given as Σn=1
∞
(–1)nan = –a1 + a2 – a3 + a4 …. or as
Σn=1
∞
(–1)n+1an = a1 – a2 + a3 – a4 … where an > 0
Theorem (Alternating Series):
Suppose {an} is a alternating sequence where {lanl} is
a decreasing sequence and lim an = 0,
then the alternating series converges.
From here on, we will write an alternating series as
Σn=1
∞
an where the sequence {an} has alternating signs.
Σn=1
∞
an
Again the decreasing-condition is to prevent the
sequence from oscillating forever.

6. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.

7. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±

8. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±
We may visualize Sn geometrically as follows:

9. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±
We may visualize Sn geometrically as follows:
0 1
S1
1 1

10. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±
We may visualize Sn geometrically as follows:
0 1
S1
1
S2
–1/2
1
1–1/2
1

11. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±
We may visualize Sn geometrically as follows:
0 1
S1
1
S2
–1/2
S3
1/3
1–1/2
1–1/2+1/3
1

12. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±
We may visualize Sn geometrically as follows:
0 1
S1
1
S2
–1/2
S3
1/3
S4
–1/4
1–1/2
1–1/2+1/3
1
1–1/2+1/3–1/4

13. Alternating Series and Conditional Convergence
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –The convergence of
may be demonstrated on the real line.
Let Sn = be the partial sum.1
2
1
3 ...1 + 1
4
– – 1
n±
We may visualize Sn geometrically as follows:
0 1
S1
1
S2
–1/2
S3
1/3
S4
–1/4
Because  0 and it’s decreasing,
So the red dot must be funneled to a limit.
and the series converges.
1
n
1–1/2
1–1/2+1/3
1
1–1/2+1/3–1/4

14. Alternating Series and Conditional Convergence
Let’s note the difference between the following two
alternating convergent series.
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –
1
2
1
8
...+Σn=1
(–1)n+1 =1
2n
∞
1
4
– 1
16–

15. Alternating Series and Conditional Convergence
Let’s note the difference between the following two
alternating convergent series.
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –
1
2
1
8
...+Σn=1
(–1)n+1 =1
2n
∞
1
4
– 1
16–
Σn=1
(–1)n+1 converges, but1
n
∞
Σn=1
1
n
∞
diverges,

16. Alternating Series and Conditional Convergence
Let’s note the difference between the following two
alternating convergent series.
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –
1
2
1
8
...+Σn=1
(–1)n+1 =1
2n
∞
1
4
– 1
16–
Σn=1
(–1)n+1 converges, but1
n
∞
Σn=1
(–1)n+1 1
2n
∞
Σn=1
1
n
∞
diverges, where as
converge.both Σn=1
1
2n
∞
and

17. Alternating Series and Conditional Convergence
Let’s note the difference between the following two
alternating convergent series.
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –
1
2
1
8
...+Σn=1
(–1)n+1 =1
2n
∞
1
4
– 1
16–
Σn=1
(–1)n+1 converges, but1
n
∞
Σn=1
(–1)n+1 1
2n
∞
Σn=1
1
n
∞
diverges, where as
converge.both Σn=1
1
2n
∞
and
Two Types of Convergence of Alternating Series

18. Alternating Series and Conditional Convergence
Let’s note the difference between the following two
alternating convergent series.
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –
1
2
1
8
...+Σn=1
(–1)n+1 =1
2n
∞
1
4
– 1
16–
Σn=1
(–1)n+1 converges, but1
n
∞
Σn=1
(–1)n+1 1
2n
∞
Σn=1
1
n
∞
diverges, where as
converge.both Σn=1
1
2n
∞
and
if also converges, we say the alternating series
converges absolutely,
Σ|an|
Two Types of Convergence of Alternating Series
Given a convergent alternating seriesΣan:

19. Alternating Series and Conditional Convergence
Let’s note the difference between the following two
alternating convergent series.
1
2
1
3
...1 +Σn=1
(–1)n+1 =1
n
∞
1
4
– –
1
2
1
8
...+Σn=1
(–1)n+1 =1
2n
∞
1
4
– 1
16–
Σn=1
(–1)n+1 converges, but1
n
∞
Σn=1
(–1)n+1 1
2n
∞
Σn=1
1
n
∞
diverges, where as
converge.both Σn=1
1
2n
∞
and
if also converges, we say the alternating series
converges absolutely,
Σ|an|
if diverges, we say the alternating series
converges conditionally.
Σ|an|
Two Types of Convergence of Alternating Series
Given a convergent alternating seriesΣan:

20. Alternating Series and Conditional Convergence
The alternating harmonic series converges
1
n(–1)n
Σn=1
∞
conditionally (because diverges).
1
nΣn=1
∞

21. Alternating Series and Conditional Convergence
The alternating harmonic series converges
1
n(–1)n
Σn=1
∞
conditionally (because diverges).
1
nΣn=1
∞
The alternating p–series , p < 1 converges
1
np(–1)n
Σn=1
∞
conditionally (because diverges).
1
npΣn=1
∞

22. Alternating Series and Conditional Convergence
The alternating harmonic series converges
1
n(–1)n
Σn=1
∞
conditionally (because diverges).
1
nΣn=1
∞
The alternating p–series , p < 1 converges
1
np(–1)n
Σn=1
∞
conditionally (because diverges).
1
npΣn=1
∞
Hence when we answer questions about the
convergence of alternating series, we always identify
whether it converges absolutely or conditionally

23. Alternating Series and Conditional Convergence
The alternating harmonic series converges
1
n(–1)n
Σn=1
∞
conditionally (because diverges).
1
nΣn=1
∞
The alternating p–series , p < 1 converges
1
np(–1)n
Σn=1
∞
conditionally (because diverges).
1
npΣn=1
∞
Hence when we answer questions about the
convergence of alternating series, we always identify
whether it converges absolutely or conditionally.

24. Alternating Series and Conditional Convergence
The alternating harmonic series converges
1
n(–1)n
Σn=1
∞
conditionally (because diverges).
1
nΣn=1
∞
The alternating p–series , p < 1 converges
1
np(–1)n
Σn=1
∞
conditionally (because diverges).
1
npΣn=1
∞
Hence when we answer questions about the
convergence of alternating series, we always identify
whether it converges absolutely or conditionally.
Example A. Discuss the convergence of the
following series.
a. (–1)nΣn=1
∞
= –1 + 1 – 1 + 1 … diverges since {an}
does not converge to 0.

25. Example A.
b. an = 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
(–1)n

26. 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

27. 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

28. 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominating terms are 3n
n2 .
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

29. 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

30. 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
n∞
Lim 3n
n2 + n – 1 3n
n2
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

31. 3n
n2 + n – 1.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
n∞
Lim 3n
n2 + n – 1 3n
n2
= 1
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

32. 3n
n2 + n – 1.
So {an} and are almost–multiple of each other.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
n∞
Lim 3n
n2 + n – 1 3n
n2
= 1
3n
n2
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

33. 3n
n2 + n – 1.
So {an} and are almost–multiple of each other.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
n∞
Lim 3n
n2 + n – 1 3n
n2
= 1
3n
n2
But Σn=1
∞ 3n
n2 = Σn=1
∞
3
n diverges.
Since an  0, converges.Σn=1
∞
an
Example A.
b. an = (–1)n

34. 3n
n2 + n – 1.
So {an} and are almost–multiple of each other.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
n∞
Lim 3n
n2 + n – 1 3n
n2
= 1
3n
n2
But Σn=1
∞ 3n
n2 = Σn=1
∞
3
n diverges.
Therefore Σn=1
3n
n2 + n – 1
diverges.
Since an  0, converges.Σn=1
∞
an
∞
Example A.
b. an = (–1)n

35. 3n
n2 + n – 1.
So {an} and are almost–multiple of each other.
Alternating Series and Conditional Convergence
We need to identify what type of convergence it is.
The dominate terms are 3n
n2 .
Compare this to an.
n∞
Lim 3n
n2 + n – 1 3n
n2
= 1
3n
n2
But Σn=1
∞ 3n
n2 = Σn=1
∞
3
n diverges.
Therefore Σn=1
3n
n2 + n – 1
diverges.
So
Since an  0, converges.Σn=1
∞
an
Σn=1
3n
n2 + n – 1
(–1)n
∞
∞
converges conditionally.
Example A.
b. an = (–1)n

36. Alternating Series and Conditional Convergence
Summary: Given an alternating sequence {an}

37. Alternating Series and Conditional Convergence
Summary: Given an alternating sequence {an}
1. check if an  0, if no, the series diverges

38. Alternating Series and Conditional Convergence
Summary: Given an alternating sequence {an}
1. check if an  0, if no, the series diverges
2. if yes, check what type of convergence by
investigating the convergence of Σn=1
∞
|an|.

39. Alternating Series and Conditional Convergence
Summary: Given an alternating sequence {an}
1. check if an  0, if no, the series diverges
2. if yes, check what type of convergence by
investigating the convergence of
If it converges, the alternating series converges
absolutely.
Σn=1
∞
|an|.

40. Alternating Series and Conditional Convergence
Summary: Given an alternating sequence {an}
1. check if an  0, if no, the series diverges
2. if yes, check what type of convergence by
investigating the convergence of
If it converges, the alternating series converges
absolutely. If it diverges, the alternating series
converges conditionally.
Σn=1
∞
|an|.