BC 3
Alternating Series
Name:
Defn: If an  0 for all n 
, an alternating series is a series of the form
a1  a2  a3  a4 
OR a1  a2  a3  a4 

In sigma notation this becomes:
  1
k 1

(1)
Consider

n 1
Find S1 , S2 , S3 ,
you.)
 1
n
n 1
k 1

ak OR
1 1 1
 1   
2 3 4
  1 a
k 1
k
k
, the alternating harmonic series.
S8 to two decimal places, plotting each on the line below. (This has been begun for
0.5
1.0
S2
S1
Describe the pattern of the Sn.
Do you think the series converges?

If it does converge and S  
n 1
 1
n
n 1
, complete the inequality involving S.
_______  S  ________
IMSA
Alt Ser 5.1
Fall 14
BC 3
Alternating Series
Name:
Keeping in mind that all of the an must be positive, what other conditions on the terms of the
series are necessary to make the alternating series converge? In other words, what are the other
conditions on the an which will make the Sn follow a pattern like the above?
Consider the alternating series a1  a2  a3  a4 
with an  0 and an  an 1  0 for all n  1. Fill in
the (positive) distances, in terms of an ai, given by each arrow.
distance =
distance =
distance =
distance =
distance =
S2
S4
S6
S
S5
S3
S1
Since it is often impossible for us to calculate the value of S directly, we use the value of one of the S n to
approximate S. For example, suppose we choose to approximate S using S4. The next real question, or
problem, is determining just how good S4 is as an approximation for S. In other words, we need to know
the distance from S4 to S (or how close S4 is to S). We can express this distance as S  S4 .
Usually, we cannot calculate this distance exactly; if we could, we would know the value of S and we
would have little reason to approximate this known value,
From the figure above, we can see that S  S4  S4  S5  a5 . Thus, we know that the error made by
using S4 as an approximation for S is less than the value of a5. Now, suppose instead, we choose to use
S5 to approximate S. Again from the figure above, we can see that S  S5 

,
which means that the error made by using S5 as an approximation for S is less than ________ ..
In general, if we choose to approximate S using Sn. we need to look at the size of
. Since the
Sn oscillate in smaller and smaller steps around S, then S is between Sn and Sn+1 for all n.
IMSA
Alt Ser 5.2
Fall 14
BC 3
Alternating Series
Name:
Consider the arbitrary stage of the diagram above, shown at
the right. Here, we can see from the diagram that
S  Sn 

.
This general result should correspond to the specific results
when we approximated S using S4 and S5. Using this result,
we have S  S6 
an+1
and S  S20 
Sn
.
S
S  Sn
In the above discussion, is it important that an  0 ? ________ Explain.
In the above discussion, is it important that an1  an ?_________ Explain.
Alternating Series Theorem (or Test): If an  0 and an  an 1  0 for all n  1 then

i.
The alternating series
  1
n1
n1
ii.
IMSA
an  converges, and

S  Sn  an1 for all n  1.
Alt Ser 5.3
Fall 14
Sn+1
BC 3
Alternating Series
Name:

(1)
Consider the alternating series

 1n1 .
n
n 1
(a)
If we approximate S by using the first 15 terms of this series, what will be the
magnitude of the error? In other words, how large is S  S15 ?
(b)
Find S15 and write an inequality (without absolute values) that bounds S.
(c)
How many terms of the series need to be used to be sure that the error made when
using the partial sum formed by these terms to approximate S is less than 0.01?

(2)
Consider the alternating series

n1
1n .
n3
(a)
If we approximate S by using the first 19 terms of this series, what will be the
magnitude of the error? In other words, how large is S  S19 ?
(b)
How many terms of the series need to be used to be sure that the error made when
using the partial sum formed by these terms to approximate S is less than 0.01?
IMSA BC 3
Alt Ser 5.4
Fall 14
BC 3
Alternating Series
Name:

Definition: Look at the series
a
n 1


i.
If both
a
n
n 1
and

If

a
converge, then we say
n
n1

ii.
.
n
a
n
n 1

an converges, but
n 1
is absolutely convergent.


an diverges, then we say
n1
a
n
is conditionally
n 1
convergent.
Notes:

1. If
a
n
n1

converges, then
a
n

must also converge. Hence if
n 1
a
n
converges,
n1

a
n
is absolutely convergent.
n 1
2.
We may now ask of any given series, whether it converges absolutely, converges
conditionally, or diverges.
Determine whether each series converges absolutely, converges conditionally, or diverges.

(1)

 1n1
n 1
n

(2)
 1n1
 n ln(n)
n2
Here are some problems from the text that you will most certainly enjoy.
Pages 488-90: For problems 4-27, look at each series and think about why the test the authors
suggest is a good one. Do 28,30,33,34,37, 54-72 (some), 73,75.
IMSA BC 3
Alt Ser 5.5
Fall 14