But for
p
6
= 1 we can easily check that
S
k
=
Z
k
1
1
x
p
d
x
=
1
p

1
1

1
k
p

1
,
42
while for
p
= 1 we have
S
k
=
Z
k
1
1
x
d
x
= log(
k
)
.
(Here log(
·
) denotes the natural logarithm.) We then see that the sequence
{
S
k
}
converges for
p >
1 and diverges for
p
≤
1. The same is thereby true for the
p
series.
3.6.
Alternating Series.
Until now we have only studied convergence tests for nonnegative
series.
The underlying tool has been the Monotonic Sequence Theorem (Proposition 2.10),
which was used to prove Proposition 3.3. Here we use the Monotonic Sequence Theorem to
obtain the following characterization of convergence for a special class of series with alternating
sign.
Proposition 3.10. (Alternating Series Test)
Let
{
a
k
}
be a positive, nonincreasing sequence
in
R
. Then
∞
X
k
=0
(

1)
k
a
k
converges
⇐⇒
lim
k
→∞
a
k
= 0
.
Proof.
The direction
=
⇒
is just Proposition 3.2 (Divergence Test).
To prove the other
direction, let
s
n
=
n
X
k
=0
(

1)
k
a
k
.
First, the picture is that
{
s
2
k
}
k
∈
N
is nonincreasing
,
while
{
s
2
k
+1
}
k
∈
N
is nondecreasing
,
and that
s
2
k
> s
2
j
+1
for every
j, k
∈
N
.
Indeed, the first two assertions follow because
s
2
k
+2

s
2
k
=
a
2
k
+2

a
2
k
+1
≤
0
,
s
2
k
+3

s
2
k
+1
=

a
2
k
+3
+
a
2
k
+2
≥
0
.
Next, because
s
2
k
> s
2
k
+1
for every
k
∈
N
, for any
j
≤
k
we have
s
2
k
> s
2
k
+1
≥
s
2
j
+1
,
s
2
j
≥
s
2
k
> s
2
k
+1
.
The result follows by exchanging
j
and
k
in the last inequality. The monotonic subsequences
{
s
2
k
}
and
{
s
2
k
+1
}
are thereby bounded below and above respectively. By the Monotonic Se
quence Theorem they therefore converge. Let
s
= lim
k
→∞
s
2
k
,
s
= lim
k
→∞
s
2
k
+1
.
Then
s

s
= lim
k
→∞
(
s
2
k

s
2
k
+1
)
= lim
k
→∞
a
2
k
+1
= 0
,
whereby
s
=
s
. The last step is to show that this fact implies that
{
s
k
}
converges. This is left
as an exercise.
Examples.
∞
X
k
=1
(

1)
k
k
p
converges for
p >
0
,
∞
X
k
=2
(

1)
k
log(
k
)
converges
.
43
3.7.
Absolute Convergence.
The Monotonic Sequence Theorem has been the tool underlying
all the convergence tests we have studied so far. We now use the Cauchy criterion to establish
a test that does not require the series to be nonnegative.
Proposition 3.11. (Absolute Convergence Test)
Let
{
a
k
}
be a real sequence. Then
∞
X
k
=0

a
k

converges
=
⇒
∞
X
k
=0
a
k
converges
.
Proof.
Let
{
p
n
}
and
{
q
n
}
be the sequences of partial sums given by
p
n
=
n
X
k
=0

a
k

,
q
n
=
n
X
k
=0
a
k
.
By hypotheses
{
p
n
}
is convergent, and thereby Cauchy. The idea of the proof is to show that
{
q
n
}
is Cauchy, and thereby convergent.
The key to doing so is the fact that for every
m, n
∈
N
we have the inequality

q
n

q
m
 ≤ 
p
n

p
m

.
This is trivially true when
m
=
n
. When
n > m
the triangle inequality yields

q
n

q
m

=
n
X
k
=
m
+1
a
k
≤
n
X
k
=
m
+1

a
k

=

p
n

p
m

.
The case
n < m
goes similarly.
Let
>
0. Because
{
p
n
}
is Cauchy there exists an
N
∈
N
such that
m, n
≥
N
=
⇒ 
p
n

p
m

<
.
Because

q
n

q
m
 ≤ 
p
n

p
m

, we immediately see that
m, n
≥
N
=
⇒ 
q
n

q
m

<
.