EXAMPLE 3.4
Finding the Number of Terms Needed
for a Given Accuracy
Determine the number of terms needed to obtain an approximation to the sum of the
series
∞
∑
k
=
1
1
k
3
correct to within 10
−
5
.
Solution
Again, we already used the Integral Test to show that the series in question
converges. Then, by Theorem 3.2, we have that the remainder satisfies
0
≤
R
n
≤
∞
n
1
x
3
dx
=
lim
R
→∞
R
n
1
x
3
dx
=
lim
R
→∞
−
1
2
x
2
R
n
=
lim
R
→∞
−
1
2
R
2
+
1
2
n
2
=
1
2
n
2
.
So, to ensure that the remainder is less than 10
−
5
, we require that
0
≤
R
n
≤
1
2
n
2
≤
10
−
5
.
Solving this last inequality for
n
yields
n
2
≥
10
5
2
or
n
≥
10
5
2
=
100
√
5
≈
223
.
6
.

8-31
SECTION 8.3
.
.
The Integral Test and Comparison Tests
641
So, taking
n
≥
224 will guarantee the required accuracy and consequently, we have
∞
∑
k
=
1
1
k
3
≈
224
∑
k
=
1
1
k
3
=
1
.
202047, which is correct to within 10
−
5
, as desired.
Comparison Tests
We next present two results that allow us to compare a given series with one that is already
known to be convergent or divergent, much as we did with improper integrals in section 6.6.
THEOREM 3.3
(Comparison Test)
Suppose that 0
≤
a
k
≤
b
k
, for all
k
.
(i) If
∞
∑
k
=
1
b
k
converges, then
∞
∑
k
=
1
a
k
converges, too.
(ii) If
∞
∑
k
=
1
a
k
diverges, then
∞
∑
k
=
1
b
k
diverges, too.
Intuitively, this theorem should make abundant sense: if the “larger” series converges,
then the “smaller” one must also converge. Likewise, if the “smaller” series diverges, then
the “larger” one must diverge, too.
PROOF
Given that 0
≤
a
k
≤
b
k
for all
k
, observe that the
n
th partial sums of the two series satisfy
0
≤
S
n
=
a
1
+
a
2
+ ··· +
a
n
≤
b
1
+
b
2
+ ··· +
b
n
.
(i) If
∞
∑
k
=
1
b
k
converges (say to
B
), this says that
0
≤
S
n
≤
a
1
+
a
2
+ ··· +
a
n
≤
b
1
+
b
2
+ ··· +
b
n
≤
∞
k
=
1
b
k
=
B
,
(3.5)
for all
n
≥
1. From (3.5), the sequence
{
S
n
}
∞
n
=
1
of partial sums of
∞
∑
k
=
1
a
k
is bounded. No-
tice that
{
S
n
}
∞
n
=
1
is also increasing. (Why?) Since every bounded, monotonic sequence is
convergent (see Theorem 1.4), we get that
∞
∑
k
=
1
a
k
is convergent, too.
(ii) If
∞
∑
k
=
1
a
k
is divergent, we have (since all of the terms of the series are nonnegative) that
lim
n
→∞
(
b
1
+
b
2
+ ··· +
b
n
)
≥
lim
n
→∞
(
a
1
+
a
2
+ ··· +
a
n
)
= ∞
.
Thus,
∞
∑
k
=
1
b
k
must be divergent, also.
You can use the Comparison Test to test the convergence of series that look similar
to series that you already know are convergent or divergent (notably, geometric series or
p
-series).