NOTES
When we say lim
k
→∞
a
k
b
k
=
L
>
0,
we mean that the limit exists and
is positive. In particular, we mean
that lim
k
→∞
a
k
b
k
= ∞
.
THEOREM 3.4
(Limit Comparison Test)
Suppose that
a
k
,
b
k
>
0 and that for some (finite) value,
L
,
lim
k
→∞
a
k
b
k
=
L
>
0. Then,
either
∞
∑
k
=
1
a
k
and
∞
∑
k
=
1
b
k
both
converge or they
both
diverge.
PROOF
If lim
k
→∞
a
k
b
k
=
L
>
0, this says that we can make
a
k
b
k
as close to
L
as desired. So, in particular,
we can make
a
k
b
k
within distance
L
2
of
L
. That is, for some number
N
>
0,
L
−
L
2
<
a
k
b
k
<
L
+
L
2
,
for all
k
>
N
or
L
2
<
a
k
b
k
<
3
L
2
.
(3.6)

##### We have textbook solutions for you!

**The document you are viewing contains questions related to this textbook.**

**The document you are viewing contains questions related to this textbook.**

Expert Verified

644
CHAPTER 8
.
.
Infinite Series
8-34

EXAMPLE 3.8
Using the Limit Comparison Test
Investigate the convergence or divergence of the series
∞
∑
k
=
3
1
k
3
−
5
k
.
EXAMPLE 3.9
Using the Limit Comparison Test
Investigate the convergence or divergence of the series
∞
k
2
−
2
k
+
7
k
5
+
5
k
4
−
3
k
3
+
2
k
−
1
.
5
10
15
20
1.50

k
=
1
FIGURE 8.29

8-35
SECTION 8.3
.
.
The Integral Test and Comparison Tests
645
Notice that for
k
large, the general term looks like
k
2
k
5
=
1
k
3
(since the terms with the
largest exponents tend to dominate the expression, for large values of
k
). From the Limit
Comparison Test, for
b
k
=
1
k
3
, we have
lim
k
→∞
a
k
b
k
=
lim
k
→∞
k
2
−
2
k
+
7
k
5
+
5
k
4
−
3
k
3
+
2
k
−
1
1
1
k
3
=
lim
k
→∞
(
k
2
−
2
k
+
7)
(
k
5
+
5
k
4
−
3
k
3
+
2
k
−
1)
k
3
1
=
lim
k
→∞
(
k
5
−
2
k
4
+
7
k
3
)
(
k
5
+
5
k
4
−
3
k
3
+
2
k
−
1)
1
k
5
1
k
5
=
lim
k
→∞
1