6.4. TESTING SERIES WITH MIXED TERMS
6.4
323
Testing Series With Mixed Terms
We now focus on series with general terms which can be positive as well as
negative. In some cases, series with mixed terms can be handled with techniques
developed for series with positive terms. In other cases, we will have to develop
new techniques. Before we see how, we begin with some de…nitions.
6.4.1
Absolute Versus Conditional Convergence
P
P
Given a series an , the series of absolute values is jan j = ja1 j+ja2 j+ja3 j+:::.
It is obviously a series of positive terms.
Example 6.4.1 What is the series of absolute values corresponding to
It is
X1
X ( 1)n
=
n!
n
P ( 1)n
?
n!
P ( 1)n xn
Example 6.4.2 What is the series of absolute values corresponding to
n!
where x is a …xed real number?
It is
X ( 1)n xn
X jxjn
=
n!
n
P
P
Remark
6.4.3 Obviously, if
an is a series of positive terms, then
an =
P
jan j. We will use this obvious fact below.
In working with absolute values, it is important to remember some of its
properties we list here for convenience.
Proposition 6.4.4 Let a and b denote real numbers and n a positive integer.
Then:
1. jabj = jaj jbj
n
2. jan j = jaj
3.
jaj
a
=
b
jbj
4. If a
0 then jaj = a
5. If a < 0 then jaj =
6. jaj < c ()
a
c<a<c
It turns out that knowing something about the convergence of the series of
absolute values helps us with the convergence of the original series. We begin
with a few de…nitions, then the main theorem.
324
CHAPTER 6. INFINITE SEQUENCES AND SERIES
P
De…nition 6.4.5 (Absolute convergence) A series
an is said to be absolutely
convergent
if
the
series
of
absolute
values
is
convergent.
In other
P
P
words,
an is said to be absolutely convergent if
jan j converges.
P
De…nition 6.4.6 (Conditional convergence) A series
an is said to be
conditionally convergent
if
it
converges
but
the
series
of
absolute
values
P
P diverges. In other
words,
a
is
said
to
be
conditionally
convergent
if
an
n
P
converges but
jan j diverges.
Theorem 6.4.7 If a series converges absolutely, then it also converges.
Remark 6.4.8 All the tests studied so far only apply to series with positive
terms. One way of testing a series having negative terms is to test the series
of absolute values for convergence. If the series of absolute values converges, so
will the series with mixed terms. This allows us to use the tools we have already
developed since the series of absolute values is a series of positive terms.
Remark 6.4.9 Absolute convergence also has other important consequences
which we will not study in detail here. It can be proven that when a series
converges absolutely, then the order in which we sum its terms does not matter.
However, if it converges conditionally, changes the order in which we sum the
terms may cause the series to converge to a di¤ erent sum, or even to diverge.
Riemann was the …rst one to notice this in 1867. The important consequence of
this is that unless we know a series converges absolutely, the order of summation
cannot be changed. This is very di¤ erent from what we know about …nite sums,
where the order of summation is not important.
Remark 6.4.10 Students are sometimes confused because we seem to have two
kinds of convergence: absolute and conditional. In both cases, the in…nite sum
converges to some number. The only di¤ erence is that absolute convergence is
a little bit stronger in the sense that the order of summation can be changed.
Example 6.4.11 Is
1
n
X
( 1)
n
n=1
1
1
n
X
( 1)
2n
n=1
1
absolutely convergent, conditionally conver-
gent or divergent.
n 1
1
( 1)
Since
= , the series of absolute values is the harmonic series, which
n
n
1
n 1
X
( 1)
diverges. We don’t know yet how to study the convergence of
but
n
n=1
1
n 1
X
( 1)
we will very soon. We will prove that it converges. Therefore,
n
n=1
converges conditionally.
Example 6.4.12 Is
gent or divergent.
absolutely convergent, conditionally conver-
6.4. TESTING SERIES WITH MIXED TERMS
325
1
1
n 1
X
X
( 1)
1
1
1 1
=
,
the
series
of
absolute
values
is
=
.
n
n
n
2
2
2
2 2n 1
n=1
n=1
Thus it is a geometric series with jrj < 1 hence it converges. Since the series of
1
n 1
X
( 1)
absolute values converges, we say that
converges absolutely.
2n
n=1
Since
Remark 6.4.13 Let us repeat the statement which was made above. Whether
it converges absolutely or conditionally, a series converges. One of the points of
this section is to realize that if a series converges absolutely, then it converges.
So, given a series with mixed terms, one can look at the series of absolute values
(which will be a series of positive terms) and use all the tools we have to study
series of positive terms. If the series of positive terms is found to converge, then
the original series will also converge. In this case, we see that this allows us
to study the convergence of a series with mixed terms using the tools developed
for series of positive terms. This will not help however, if the series of positive
terms is found to diverge. In that case, we must use tools which are designed
for series having mixed terms. This is what we do next.
Remark 6.4.14 Another important fact about series which converge absolutely,
which we will not prove, is related to whether we can rearrange the terms in an
in…nite series. It is a well known fact that in a …nite sum, the terms can be
rearranged, in other words, in a …nite sum we can add the terms in any order
we want. This is not true in general for in…nite sums. However, it is true for
in…nite series which converge absolutely.
First, we look at series whose terms are alternatively positive and negative.
6.4.2
Alternating Series Test
De…nition 6.4.15 (Alternating series) An alternating series is a series whose
terms are alternatively positive and negative. We usually write an alternating
1
1
X
X
n
n 1
( 1) bn , where bn > 0.
( 1)
bn or
series as
n=1
n=1
Example 6.4.16 1
Example 6.4.17
1 1
+
2 3
1+2
1
+ :::: is an alternating series.
4
3+4
5 + ::: is an alternating series.
Theorem 6.4.18 (Alternating series test) If the series
1
X
n=1
bn > 0 satis…es
1. bn+1
bn for all n from some point on, and
2. lim bn = 0
n!1
then the series is convergent.
( 1)
n 1
bn , where
326
CHAPTER 6. INFINITE SEQUENCES AND SERIES
Remark 6.4.19 If conditions 1 or 2 are not satis…ed, we cannot conclude anything. We must use another test.
1
n 1
X
( 1)
Example 6.4.20 Determine if the series
, known as the alternatn
n=1
ing harmonic series, converges.
1
In this series, bn =
which decreases to 0. By the alternating series test, this
n
series converges.
1
n
X
( 1) 3n
converges.
4n 1
n=1
3n
3
This is an alternating series with bn =
: However. lim bn = =
6 0. So,
4n 1
4
n
( 1) 3n
we cannot conclude. Here, the test for divergence can help us. lim
n!1 4n
1
does not exist, therefore the series diverges.
Example 6.4.21 Determine if
Like for series tested using the integral test, we can approximate alternating
series with Sn within any required accuracy. Using the same notation as we did
for the integral test, we have:
Theorem 6.4.22 If S =
1
X
( 1)
n 1
bn and bn satis…es the conditions of the
n=1
alternating series test, then
jRn j = jS
Sn j
bn+1
In other words, the error by approximating the sum of a convergent alternating
series by the sum of the …rst n terms is no greater than the n + 1 term.
1
n 1
X
( 1)
with an error less than :001.
n!
n=1
First, we need to …nd n such that jRn j < :001. It is enough to …nd n such
1
< :001. We do not know how to solve equations with a
that bn+1 =
(n + 1)!
1
factorial. So, we try values of n. We note that when n = 5, bn+1 =
= 1:
6!
1
388 888 889 10 3 and when n = 6, bn+1 =
= 1: 984 126 984 10 4 < :001.
7!
6
n 1
X
( 1)
91
So, we use n = 6. We get that
=
= 0:631 94
n!
144
n=1
Example 6.4.23 Find the sum of
6.4.3
Ratio and Root Tests
The next two tests are extremely important. They are used very often.
6.4. TESTING SERIES WITH MIXED TERMS
327
P
Theorem 6.4.24 (Ratio test) Let
an be a series of non-zero terms and
an+1
exists or is in…nite then
suppose that L = lim
n!1
an
1. If L < 1,
2. If L > 1,
P
P
an converges absolutely
an diverges
3. If L = 1, the test provides no conclusion.
Remark 6.4.25 The ratio test works best with series which involve factorials
and other products.
Theorem
6.4.26 (Root test) Let
p
lim n jan j exists or is in…nite then
n!1
1. If L < 1,
2. If L > 1,
P
P
P
an be a series and suppose that L =
an converges absolutely
an diverges
3. If L = 1, the test provides no conclusion.
Remark 6.4.27 If L = 1 in the ratio test, do not try the root test, L will also
more than likely be 1. Indeed, in most cases when L = 1 in one of the tests, it
is also 1 in the other.
Remark 6.4.28 The root test works with series whose general term is a power
of n.
Example 6.4.29 Test
We form
1
n
X
( 1) n3
.
3n
n=1
3
an+1
an
=
=
=
!
(n + 1)
3n+1
n3
3n
3
(n + 1) 3n
n3
3n+1
3
1 n+1
3
n
1
as n ! 1
3
Therefore, the series converges absolutely (hence converges).
328
CHAPTER 6. INFINITE SEQUENCES AND SERIES
1
X
2n
Example 6.4.30 Test
n!
n=1
an+1
an
2n+1
(n + 1)!
=
2n
n!
n!
2n+1
=
2n (n + 1)!
2
=
n+1
! 0 as n ! 1
Therefore, the series converges absolutely (hence converges).
1
X
1
n.
(ln
n)
n=2
Since the general term is a power of n, we use the root test. First, we …nd
s
1
1
n
=
n
jln nj
(ln n)
Example 6.4.31 Test
=
1
if n
ln n
2
1
X
1
1
Since lim
= 0,
n converges by the root test.
n!1 ln n
(ln
n)
n=2
Example 6.4.32 Find x so that
1
X
xn
converges.
n!
n=0
n+1
an+1
an
=
=
!
jxj
(n + 1)!
n
jxj
n!
jxj
n+1
0 as n ! 1
Therefore, the series converges absolutely (hence converges) for every x.
Example 6.4.33 Find x so that
1
X
2n xn converges.
n=0
Here, we use the root test.
p
n
jan j =
=
q
n
n
2n jxj
2 jxj
6.4. TESTING SERIES WITH MIXED TERMS
Therefore
lim
n!1
329
p
n
jan j = 2 jxj
The series converges absolutely (hence converges) if
2 jxj < 1
1
jxj <
2
1
1
< x<
2
2
When x =
p
1
, lim n jan j = 1, hence we must use another test.
2 n!1
When x =
When x =
verges.
1
1
X
X
1
1
, the series becomes
2n n =
1 which diverges.
2
2
n=0
n=0
1
1
n
X
X
( 1)
1
n
, the series becomes
2n n =
( 1) which di2
2
n=0
n=0
In conclusion,
1
X
2n xn converges when
n=0
Example 6.4.34 Find x so that
an+1
an
1
n
X
( 1)
(x
n
n=1
=
!
n
1) converges.
n+1
jx
=
1
1
<x< .
2
2
1j
n+1
n
jx 1j
n
n
jx 1j
n+1
jx 1j as n ! 1
Therefore, the series converges absolutely (hence converges) if
jx
1j <
1
1
<
x
0
<
x<2
1<1
When x = 0 or x = 2, the limit is 1. Hence we must use another test.
1
1
n
n
X
X
1
( 1) ( 1)
=
which diverges.
When x = 0, the series becomes
n
n
n=0
n=0
330
CHAPTER 6. INFINITE SEQUENCES AND SERIES
1
n
X
( 1)
When x = 2, the series becomes
which converges.
n
n=0
In conclusion, the series converges in (0; 2] and converges absolutely in
(0; 2).
6.4.4
Things to know
Students should know and be able to use the following tests
Absolute convergence
Alternating series test
Ratio and root tests
Students should also be able to approximate the sum of a series, and
determine the error of the approximation.
6.4.5
Problems
1. Alternating series.
(a) What is an alternating series?
(b) Give at least two examples of alternating series.
(c) When does an alternating series converge?
1
X
n 1
(d) If an alternating series
( 1)
bn converges and we approximate
n=1
it by the sum of the …rst n terms, what can be said about the error
due to the approximation?
X
X
2. If
jan j converges, what can be said about the convergence of
an ?
Explain.
X
X
3. If
an converges, what can be said about the convergence of
jan j?
Explain.
X
4. What can be said about the convergence of the series
an in each case
below? Explain.
(a) lim
an+1
=3
an
(b) lim
an+1
= :5
an
(c) lim
an+1
=1
an
6.4. TESTING SERIES WITH MIXED TERMS
(d) lim
(e) lim
(f) lim
p
n
p
n
p
n
331
jan j = 6
jan j = 1
jan j = :2
5. Test the series below for convergence or divergence.
3 3 3
+
4 5 6
1
n
X
( 1)
p
(b)
n
n=1
3 3
+
7 8
(a)
(c)
1
X
( 1)
n
n=1
:::
1
5n 3
2n + 3
6. For what value of p does the series
1
n
X
( 1)
np
n=1
1
?
7. Show that each series below is convergent. Then, …nd how many terms of
the series we need to add to …nd the sum to the indicated accuracy.
1
n+1
X
( 1)
(a)
, error less than 0:00005.
n6
n=1
(b)
(c)
1
X
( 1)
n=1
1
X
n 1
ne
n
, error less than 0:001.
n 1
( 1)
n2
, error less than 0:00005.
10n
n=1
8. Determine whether the series below are absolutely convergent, convergent
or divergent.
1
n
X
( 3)
(a)
n3
n=1
(b)
1
n
X
( 10)
n!
n=1
1
n
X
( 1)
p
(c)
n
n=1
(d)
1
X
1
10n
(n + 1) 42n+1
n=1
1
X
2n
(e)
n2
n=1
332
CHAPTER 6. INFINITE SEQUENCES AND SERIES
1
X
2n
(f)
nn
n=1
(g)
(h)
1
X
sin 4n
4n
n=1
1
X
n=1
n
1
1+n
9. Find the values of x for which the series below converge.
(a)
(b)
1
X
(x
n=1
1
X
n
4)
n
xn
2n
n=1
1
X
xn
where x is a …xed real number. This is a very
n!
n=1
important series, we will study it in more detail very soon.
10. Consider the series
(a) Show that it converges for every x.
xn
must be for every x. (hint:
n!1 n!
remember the test for divergence)
(b) Using part a, conclude what lim
6.4.6
Answers
1. Alternating series.
(a) What is an alternating series?
1
1
X
X
n 1
n
Series of the form
( 1)
bn or
( 1) bn
n=1
n=1
(b) Give at least two examples of alternating series.
1
n 1
X
( 1)
p
n
n=1
1
n
X
( 1)
np
n=1
1
(c) When does an alternating series converge?
fbn g has to be a decreasing sequence which decreases to 0.
1
X
n 1
(d) If an alternating series
( 1)
bn converges and we approximate
n=1
it by the sum of the …rst n terms, what can be said about the error
due to the approximation?
jRn j bn+1 .
6.4. TESTING SERIES WITH MIXED TERMS
333
X
X
2. If
jan j converges, what can be said about the convergence of
an ?
Explain.
X
X
From a theorem studied, we know that
jan j converges=)
an converges.
X
X
3. If
an converges, what can be said about the convergence of
jan j?
Explain.
We cannot say anything.
X
4. What can be said about the convergence of the series
an in each case
below? Explain.
(a) lim
X
(b) lim
X
an+1
=3
an
an diverges.
an+1
= :5
an
an converges.
an+1
=1
an
Cannot say anything.
p
(d) lim n an = 6
X
an diverges.
p
(e) lim n an = 1
Cannot say anything.
p
(f) lim n an = :2
X
an converges.
(c) lim
5. Test the series below for convergence or divergence.
3 3 3 3 3
+
+
4 5 6 7 8
Converges.
1
n 1
X
( 1)
p
(b)
n
n=1
Converges.
1
X
3
n 5n
(c)
( 1)
2n + 3
n=1
Diverges
(a)
:::
6. For what value of p does the series
Converges if and only if p > 0.
1
n
X
( 1)
np
n=1
1
?
334
CHAPTER 6. INFINITE SEQUENCES AND SERIES
7. Show that each series below is convergent. Then, …nd how many terms of
the series we need to add to …nd the sum to the indicated accuracy.
1
n+1
X
( 1)
, error less than 0:00005.
n6
n=1
5 terms.
1
X
n 1
(b)
( 1)
ne n , error less than 0:001.
(a)
n=1
(c)
9 terms.
1
n
X
( 1)
n=1
10n
1
n2
, error less than 0:00005.
5 terms and the approximation is
5
n 1 2
X
( 1)
n
= 0:067 6.
n
10
n=1
8. Determine whether the series below are absolutely convergent, convergent
or divergent.
(a)
(b)
1
n
X
( 3)
n3
n=1
Diverges.
1
n
X
( 10)
n!
Converges absolutely.
1
n 1
X
( 1)
p
(c)
n
n=1
Converges conditionally.
1
X
10n
(d)
(n + 1) 42n+1
n=1
Converges absolutely.
1
X
2n
(e)
n2
n=1
Diverges.
1
X
2n
(f)
nn
n=1
Converges absolutely.
1
X
sin 4n
(g)
4n
n=1
Converges absolutely.
1
n
X
1
(h)
1+n
n=1
Converges absolutely.
n=1
6.4. TESTING SERIES WITH MIXED TERMS
335
9. Find the values of x for which the series below converge.
(a)
1
X
(x
4)
n
n
Converges when 3
1
X
xn
n=1
(b)
2n
Converges when
x < 5.
n=1
2 < x < 2.
1
X
xn
where x is a …xed real number. This is a very
n!
n=1
important series, we will study it in more detail very soon.
10. Consider the series
(a) Show that it converges for every x.
Use ratio test.
xn
must be for every x. (hint:
n!1 n!
remember the test for divergence)
Follow the hint.
(b) Using part a, conclude what lim