Series with Both Positive and Negative Terms
S. F. Ellermeyer
March 16, 2005
1
Absolute Convergence
Consider the series
1
X
1
1
1
ak = 2 + 2 + 2
1
2
3
k=1
1
42
1
52
1
1
1
1
+ 2+ 2+ 2
2
6
7
8
9
1
102
.
The general (kth) term in this series has absolute value 1=k 2 (that is, jak j =
1=k 2 ) and the series consists of three positive terms followed by three negative
terms followed by three positive terms and so on (continuing forever in this
fashion).
Does the above series converge? To answer this question, we need to
consider three other series:
1
X
1
1
1
1
1
1
1
1
1
1
jak j = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 +
,
1
2
3
4
5
6
7
8
9
10
k=1
1
X
k=1
2 jak j = 2
+2
1
1
1
1
1
1
+
2
+
2
+
2
+
2
+
2
12
22
32
42
52
62
1
1
1
1
+2 2 +2 2 +2
+
2
7
8
9
102
,
and
1
X
k=1
(jak j
ak ) = 0 + 0 + 0 + 2
+0+0+0+2
1
1
1
1
+2 2 +2 2
2
4
5
6
1
+
102
.
P
1
The …rst of the above three series is the p–series 1
k=1 k2 . We are already
familiar with this series and we know that it converges (having proved this
using the Integral Test).
P
1
The second of the above series is 1
k=1 2 k2 . Since each term of this
series is a constant multiple of the corresponding term of the p–series, this
series also converges.
Finally, each term in the third series is non–negative and is less than
or equal to the corresponding term in the second series. This means that
this third series also converges by the Standard Comparison Test. Hence, all
three of the above series converge. P
To prove that the original series, 1
k=1 ak , also converges, we now simply
note that
1
1
X
X
ak =
(jak j (jak j ak )) .
k=1
k=1
P1
In other words, each term ofPthe series P
k=1 ak is the di¤erence of corre1
sponding terms in the series 1
ja
j
and
ak ). However,
k
k=1
k=1 (jak j
P1 both
of the latter series are convergent. This P
means that the series
k=1 ak is
P
also convergent. (Recall that if
ck and a series
dk are both
Pa series
convergent, then the the series
(ck dk ) must also be convergent.)
The reasoning that has been used in
P the above example can be generalized: Suppose that we have a series,
ak , that might have both positive
P
and negative terms, and suppose also that we knowP
that the series
jak j
is convergent. Then wePalso know that the series
2 jak j is convergent.
Furthermore, the series
(jak j ak ) has all non–negative terms and
0
jak j
ak
2 jak j
for all P
k. (In fact jak j ak must always be equal
P to 0 or to 2 jak j.) Since the
series
2 jak j is convergent, then the series
(jak j ak ) is also convergent
by the Standard Comparison Test. Upon noting that
1
X
k=1
ak =
P1
1
X
k=1
(jak j
(jak j
ak )) ,
we seeP
that the series k=1 ak must also be convergent.
In summary, if the
P1
1
series k=1 jak j is convergent, then the series k=1 ak must also be convergent. Since this fact is very useful in the study of series, we will state it as a
theorem.
2
Theorem 1 If the series
convergent.
P
jak j is convergent, then the series
P
ak is also
Let us look at another example in which we can use Theorem 1.
Example 2 Explain why the series
1
2
1
22
1
1
1
1
+ 4+ 5+ 6
3
2
2
2
2
1
27
1
28
1
29
1
+
210
converges.
P1
Answer: If we call the above series
1
X
k=1
jak j =
k=1
1
X
1
2
k=1
ak , then
k
and we know the latter series is a convergent geometric series
P (with r =
1=2). We can then conclude from Theorem 1 that the series 1
k=1 ak is also
convergent. In fact, since we know that
1
X
1
2
k=1
we can conclude that
1
X
k
= 1,
ak < 1.
k=1
P1
(It might make a fun exercise to try to compute the exact value
P of k=1 ak .)
It is important to noteP
that Theorem 1 tells us that if
jak j converges,
then it must
be
true
that
a
also
converges.
However,
if
it
turns out that
k
P1
the series k=1 P
jak j diverges, then Theorem 1 does not
P1tell us anything
about the seriesP 1
a
.
It
is
possible
that
the
series
k=1 k
k=1 jak j is divergent
1
but the series k=1 ak is convergent! For example, in the next section we
will prove that the series
1
X
k=1
ak = 1
1 1
+
2 3
3
1 1
+
4 5
(which is known as the alternating harmonic series) is convergent. This is
true even though we know that the harmonic series,
1
X
k=1
jak j = 1 +
1 1 1 1
+ + + +
2 3 4 5
is divergent.
We P
conclude this section by introducing
some vocabulary: If we have a
P
series, P ak , for which the series
jak j is convergent, then we say that the
series
ak is absolutely convergent or convergent in absolute value.
Theorem 1 tells us that any absolutely
convergent series must be convergent.
P
If, however,
we
have
a
series,
a
,
which
is convergent
k
P
P but for which the
series
jak j is divergent, then we say that the series
ak is conditionally
convergent. Thus, for example, the series
1
1
1
+
+
12 22 32
1
42
1
52
1
1
1
1
+
+
+
62 72 82 92
and
1
1
1
1
1
1
1
1
+ 4+ 5+ 6
2
3
7
2 2
2
2
2
2
2
28
are both absolutely convergent, but the series
1
1 1
+
2 3
1
29
1
102
1
+
210
1 1
+
4 5
is not absolutely convergent. (In the upcoming section, we will prove that
the latter series is convergent, meaning that this series is conditionally convergent.)
2
The Alternating Series Test
We begin this section by proving that the alternating harmonic series,
1
1 1
+
2 3
1 1
+
4 5
,
converges. We then generalize our argument to obtain a theorem that is
known as the Alternating Series Test.
4
First let us consider the alternating harmonic series (given above). This
series is not absolutely convergent, so we cannot apply Theorem 1 to it.
Instead, we will examine its sequence of partial sums: Since
s1 = 1
s2 = 1
1
,
2
we easily observe that
s2 < s 1 .
Also, since
1 1
+
2 3
s3 = 1
we see that
s3 =
and
1
2
s3 = 1
1
1
3
1
2
1
1
= s2 + > s 2
3
3
+
(something positive) < s1
= s1
from which we conclude that
s2 < s 3 < s 1 .
Now note that
1 1
+
2 3
s4 = 1
so
s4 =
1
2
1
and
s4 =
1
+
1 1
+
2 3
1
3
1
4
1
4
= s2 + (something positive) > s2
1
= s3
4
(something positive) < s3 .
We now have that
s2 < s 4 < s 3 < s 1 .
Just to make sure we understand how this sequence of partial sums is ordered,
let’s consider s5 :
1 1 1 1
s5 = 1
+
+
2 3 4 5
5
gives us
s5 =
1 1
+
2 3
1
1
4
+
1
= s4 + (something positive) > s4
5
and
s5 =
1
1 1
+
2 3
1
4
1
5
(something positive) < s3 .
= s3
We now have
s2 < s 4 < s 5 < s 3 < s 1 .
By continuing with this sort of reasoning, we observe that
< s7 < s5 < s3 < s1 .
s2 < s 4 < s 6 < s 8 <
Note that the sequence of partial sums is bounded. (A lower bound for the sequence is s2 and an upper bound is s1 .) Also, the sequence of even–numbered
partial sums is monotone increasing and the sequence of odd–numbered partial sums is monotone decreasing. Thus, both of these sequences have limits.
In addition, if we look at any two consecutive partial sums,
sn = 1
and
1 1
+
2 3
sn+1 = 1
1 1
+
2 3
1
+
4
1
+
4
+ ( 1)n+1
we see that
jsn+1
+ ( 1)n+1
sn j = ( 1)n+2
1
n
1
1
+ ( 1)n+2
,
n
n+1
1
1
=
n+1
n+1
which shows us that
lim jsn+1
n!1
sn j = lim
n!1
1
= 0.
n+1
This tells us that the distance between the even–numbered and odd–numbered
partial sums approaches zero as n ! 1. This, in turn, tells us that the limit
of each of these sequences (the even–numbered and odd–numbered partial
sums) must be the same! In other words the entire sequence of partial sums
converges to some de…nite limit. Therefore, the alternating harmonic series
6
converges. Perhaps some numerical computations are in order here to help
see what is happening in the above reasoning:
s1 = 1
s2 = 1
s3 = 1
s4 = 1
s5 = 1
s6 = 1
s7 = 1
1
2
1
2
1
2
1
2
1
2
1
2
= 0:5
+
+
+
+
+
1
= 0:83
3
1 1
= 0:583
3 4
1 1 1
+ = 0:783
3 4 5
1 1 1 1
+
= 0:616
3 4 5 6
1 1 1 1 1
+
+ = 0:759523809.
3 4 5 6 7
The reasoning that was used in showing that the alternating harmonic series
converges can be generalized. We will state this result as Theorem 3. First,
let us de…ne exactly what we mean by an alternating series. An alternating
series is a series in which every other term is positive and every other term
is negative. Thus an alternating series is a series,
a1 + a2 + a3 +
,
for which either all of the odd–numbered terms are positive and all of the
even–numbered terms are negative or vice–versa.
P
Theorem 3 (Alternating Series Test) Suppose that
ak is an alternating series and suppose that the sequence
jan j is monotone decreasing and that
P
limn!1 jan j = 0. Then the series
ak converges.
Proof.
Let us assume that all of the odd–numbered terms of the series
P1
a
are
positive numbers and that all of the even–numbered terms are
k=1 k
negative numbers. (The proof would be entirely similar if it P
were the other
way around.) Also, let sn be the sequence of partial sums of 1
k=1 ak .
Observe that
s 1 = a1
s2 = a1 + a2 = s1 + (something negative) < s1
7
so
s2 < s 1 .
Next, observe that
s 3 = a1 + a2 + a3
so
s3 = (a1 + a2 ) + ja3 j = s2 + (something positive) > s2
and
s3 = a1 + ( ja2 j + ja3 j) = s1 + (something negative) < s1 .
(Note the “something negative” in the above calculation is ja3 j ja2 j. We
know that this number is negative because we are assuming that the sequence
jan j is monotone decreasing, meaning that ja3 j < ja2 j.) We now have
s2 < s 3 < s 1 .
If we continue with this type of reasoning, we see that
< s7 < s5 < s3 < s1 .
s2 < s 4 < s 6 < s 8 <
This shows that the sequence sn is bounded and it also shows that the sequence of even–numbered partial sums is monotone increasing (and hence
has a limit) and the sequence of odd–numbered partial sums is monotone
decreasing (and hence has a limit). The fact that the limits of these two
subsequences must be the same follows from the fact that
lim jsn+1
n!1
sn j = lim jan+1 j = 0.
n!1
This completes the proof of the theorem.
We now provide another example to illustrate the Alternating Series Test.
Example 4 Let us use show that the series
1
X
k=3
( 1)k+1
ln (k)
ln (3)
=
k
3
ln (4) ln (5)
+
4
5
ln (6)
+
6
converges.
First, we observe that this series is indeed an alternating series: The
numbers ln (k) =k are all positive numbers and the ( 1)k+1 makes the series
alternate.
8
Now consider the sequence
( 1)n+1
ln (n)
ln (n)
=
.
n
n
If we let f (x) = ln (x) =x, then
f 0 (x) =
1
ln (x)
< 0 for all x
x2
e.
This shows that the function f is monotone decreasing on the interval [e; 1)
and hence that the sequence ln (n) =n is monotone decreasing for all n 3.
Finally, note that
ln (x)
ln (n)
= lim
= 0 (by L’Hopital’s Rule).
x!1
n!1
n
x
lim
Since all of the hypothesis of Theorem 3 are satis…ed, we conclude that the
series
1
X
ln (k)
( 1)k+1
k
k=3
converges.
Remark 5 The Alternating Series Test can never be used to conclude that
a series diverges!
3
The Ratio Test
Let us recall the basic facts about a geometric series
X
rk
where r is a constant and the summation index is k. We know that the
geometric series converges if jrj < 1 and diverges if jrj 1. Furthermore, in
the case of convergence, we can actually compute the sum of the series. This
sum depends on where (which value of k) we begin adding. For example,
1
X
rk =
k=0
9
1
1
r
and
1
X
r
rk =
1
k=1
r
.
An important observation about a geometric series is that the ratios of successive terms of the series are constant. This is easy to see. For the geometric
series
1
X
rk = 1 + r + r2 + r3 +
,
k=0
the nth term in the series is an = rn and the (n + 1)st term in the series is
an+1 = rn+1 . Therefore the ratio of the (n + 1)st term to the nth term is
an+1
rn+1
= n =r
an
r
which is constant.
The Ratio Test is a tool for studying series that are “like” geometric
series. By “like” geometric series, we mean that the ratio of the absolute
value of the (n + 1)st term to the nth term of the series is “almost constant”
for all large values of n. To be more speci…c, suppose that we have a series
1
X
ak
k=1
(none of whose terms is zero) and suppose that we know that
lim
n!1
an+1
=r
an
where r is a real number. This means that
an+1
an
r
for all large values of n. Now consider the sequence of partial sums of the
10
series
P1
k=1
jak j:
s1 = ja1 j = ja1 j (1)
a2
a2
= ja1 j 1 +
a1
a1
a2
a2
s3 = ja1 j + ja2 j + ja3 j = ja1 j + ja1 j
+ ja1 j
a1
a1
a2
a2 a3
a2 a3 a4
s4 = ja1 j 1 +
+
+
a1
a1 a2
a1 a2 a3
..
.
etc.
s2 = ja1 j + ja2 j = ja1 j + ja1 j
a3
a2
a2
= ja1 j 1 +
+
a2
a1
a1
If every one of the ratios jan+1 =an j is approximately equal to the constant r,
then, from the above calculation of partial sums, we see that
s1 = ja1 j (1)
s2 ja1 j (1 + r)
s3
s4
ja1 j 1 + r + r2
ja1 j 1 + r + r2 + r3
..
.
etc.
P
P1
k
Thus, the series 1
k=1 jak j “looks like” the geometric series
k=1 ja1 j r . Of
course, we are only assuming that
an+1
an
r
for all large values ofP
n, but that does not a¤ect our argument about convergence/divergence of
jak j, since convergence versus divergence of a series
does not depend on the …rst …nitely many terms of the series.
The precise statement of the Ratio Test is given in the following theorem.
P
Theorem 6 (Ratio Test) Let
ak be a series, none of whose terms is 0,
and suppose that
an+1
lim
= r.
n!1
an
11
a3
a2
1. If r < 1, then the series
verges).
P
jak j converges (and hence
2. If r > 1 (including the possibility that r = 1), then
Remark 7 If
lim
n!1
P
P
ak also con-
ak diverges.
an+1
= 1,
an
then the Ratio Test tells us nothing about convergence/divergence of
We conclude with three examples that illustrate the Ratio Test.
P
ak .
Example 8 In this example, we will show that the series
1
X
k=1
k
k+2
2k
=
3
+1+
2
3
5
6
+
6
8
4
+
converges.
The key here is to consider the ratios of successive terms in the series.
We will begin by doing some speci…c computations. (Note that since all terms
of this series are positive, then jan+1 =an j = an+1 =an for all n.)
1
a2
2
= 3 = = 0:6
a1
3
2
a3
=
a2
a4
=
a3
a5
=
a4
a6
=
a5
a7
=
a6
a8
=
a7
5 3
6
=
1
6
8
5
6
7
10
6
8
8
12
7
10
9
14
8
12
10
16
9
14
5
6
4
6
=
8
3
5
4
7
=
10
6
5
7
6
8
7
=
8
12
3
0:578
6 3
6
8
=
5 3
8
6
7 4
10
=
6 4
8
5
80
84
36
40
7
10
4
0:492
56
60
0:522
9
=
14
108
112
6
10
=
16
140
144
7
12
0:517
0:513
4
0:531
By looking at the above computations, it appears that perhaps
an+1
1
=
n!1 an
2
lim
(and we will show that this is in fact true).
Since the nth term in the series is
n+2
2n
an =
n
and the (n + 1)st term in the series is
an+1 =
n+1
(n + 1) + 2
2 (n + 1)
=
n+3
2n + 2
n+1
,
we see that
an+1
=
an
n+3 n+1
2n+2
n+2 n
2n
n+3
=
2n + 2
n+3 n
2n+2
n+2 n
2n
n+3
2n + 2
n+3
=
2n + 2
2n (n + 3)
(n + 2) (2n + 2)
n
2n2 + 6n
2n2 + 6n + 4
=
n
Our goal is to show that limn!1 jan+1 =an j = 1=2. First, note that
1+
n+3
= lim
n!1 2n + 2
n!1 2 +
lim
3
n
2
n
1
= .
2
Thus, what we need to do is study the limit problem
lim
n!1
2n2 + 6n
2n2 + 6n + 4
n
(which is a 11 indeterminate form problem) and show that this limit is equal
to 1. This will then give us that
an+1
n+3
= lim
n!1 an
n!1 2n + 2
lim
2n2 + 6n
2n2 + 6n + 4
13
n
=
1
1
1= .
2
2
In order to use L’Hopital’s Rule, let
x
2x2 + 6x
2x2 + 6x + 4
y=
Then
.
2x2 + 6x
2x2 + 6x + 4
ln (y) = x ln
so
2x2 +6x
2x2 +6x+4
ln
ln (y) =
.
1
x
Since
ln
lim
2x2 +6x
2x2 +6x+4
1
x
x!1
is a 0=0 indeterminate form problem, we may try to apply L’Hopital’s Rule.
Consider
lim
2x2 +6x+4
2x2 +6x
(2x2 +6x+4)(4x+6) (2x2 +6x)(4x+6)
(2x2 +6x+4)2
1
x2
x!1
1
2x2 +6x
= lim
x!1
4(4x+6)
2x2 +6x+4
1
x2
3
(16x + 24x2 )
= lim
x!1 4x4 + 24x3 + 44x2 + 24x
= 0.
This tells us that
ln
lim
2x2 +6x
2x2 +6x+4
1
x
x!1
=0
(by L’Hopital’s Rule) and hence that
lim ln (y) = 0.
x!1
Therefore
lim y = lim
x!1
x!1
2x2 + 6x
2x2 + 6x + 4
x
= 1.
and it then follows that
lim
n!1
2n2 + 6n
2n2 + 6n + 4
14
n
= 1.
We have now proved that
1
an+1
= .
n!1 an
2
lim
Since 1=2 < 1, the series
1
X
k=1
converges by the Ratio Test.
k+2
2k
k
Example 9 We will use the Ratio Test to show that the series
1
X
kk
k=1
22 33 44
=1+
+
+
+
k!
2!
3!
4!
diverges. (Actually, the easiest way to show that this series diverges is just
to use the Basic Divergence Test, but we will use the Ratio Test.)
Here we have
nn
an =
n!
and
(n + 1)n+1
an+1 =
(n + 1)!
so
an+1
(n + 1)n+1 n!
=
an
(n + 1)!nn
(n + 1) (n + 1)n n!
=
(n + 1) n!nn
(n + 1)n
=
nn
n
n+1
=
n
n
1
= 1+
.
n
The limit problem
an+1
1
lim
= lim 1 +
n!1
n!1 an
n
15
n
is of the 11 indeterminate form. We can however use L’Hopital’s Rule to
do this problem and we …nd that the answer is e, which is greater than 1.
Therefore the series
1
X
kk
k!
k=1
diverges by the Ratio Test.
Example 10 (an example where the Ratio Test does not apply) If we
consider either the harmonic series
1
X
1
k=1
k
or the alternating harmonic series
1
X
1
( 1)k+1 ,
k
k=1
then
an+1
=
an
and we see that
lim
n!1
1
n+1
1
n
=
n
n+1
n
an+1
= lim
= 1.
n!1 n + 1
an
Since this limit is equal to 1, we cannot use the Ratio Test to draw any conclusions about either the harmonic series or the alternating harmonic series.
As we have shown by other means, the harmonic series diverges and the alternating harmonic series converges. This example serves to illustrate the
point that the Ratio Test gives no information about convergence/divergence
if
an+1
lim
= 1.
n!1
an
16