Section 10.3: The Integral and Comparison Tests
Theorem: (The Integral Test)
Suppose that an = f (n), where f is a positive, continuous, decreasing function on [1, ∞).
Z ∞
∞
X
f (x)dx converges.
Then the series
an converges if and only if the improper integral
1
n=1
Example: Determine whether the series
∞
X
2
ne−n converges or diverges.
n=1
2
Let f (x) = xe−x . Clearly f is positive and continuous on [1, ∞). Moreover, f is decreasing
on [1, ∞) since
2
2
2
f 0 (x) = e−x − 2x2 e−x = e−x (1 − 2x2 ) < 0 for x ≥ 1.
Then
Z
∞
xe
−x2
Z
1
N
2
xe−x dx
1
N
1 −x2 = lim − e N →∞
2
1
1
1 −N 2
= lim − e
+
N →∞
2
2e
1
.
=
2e
dx =
lim
N →∞
The series converges by the Integral Test.
Example: Determine whether the series
∞
X
ln n
n=1
n
converges or diverges.
ln x
Let f (x) =
. Clearly f is positive and continuous on [1, ∞). Moreover, f is eventually
x
decreasing since
1 − ln x
f 0 (x) =
< 0 for x > e.
x2
Then
Z ∞
Z N
ln x
ln x
dx = lim
dx
N
→∞
x
x
1
1
N
1
2
= lim ln |x| N →∞ 2
1
1
= lim ln |N |2 = ∞.
N →∞ 2
The series diverges by the Integral Test.
Example: Determine whether the series
∞
X
n=1
Let f (x) =
x2
1
converges or diverges.
n2 + 1
1
. Clearly f is positive, continuous, and decreasing on [1, ∞). Then
+1
Z ∞
Z N
1
1
dx = lim
dx
2
2
N →∞ 1
x +1
x +1
1
N
= lim tan−1 x
N →∞
1
lim tan−1 (N ) − tan−1 (1)
N →∞
π π
=
−
2
4
π
=
.
4
=
The series converges by the Integral Test.
∞
∞
X
X
1
1
Definition: A series of the form
is called a p-series. The special case
, where
p
n
n
n=1
n=1
p = 1 is called the harmonic series.
∞
X
1
converge?
Example: For which values of p does the series
np
n=1
If p < 0, then
1
= ∞.
n→∞ np
lim an = lim
n→∞
Similarly, if p = 0, then
lim an = lim 1 = 1.
n→∞
n→∞
In either case, the series diverges by the Divergence Test.
1
If p > 0, then f (x) = p is clearly continuous, positive, and decreasing on [1, ∞). Recall
x
that the improper integral
Z ∞
dx
xp
1
converges if p > 1 and diverges if p ≤ 1.
By the Integral Test, the series converges if p > 1 and diverges if p ≤ 1.
Theorem: (The p-series Test)
∞
X
1
The p-series
converges if p > 1 and diverges if p ≤ 1.
p
n
n=1
Example: Determine whether the following series converge or diverge:
(a)
∞
X
n=1
5
√
n n
The series converges by the p-series Test with p = 3/2 > 1
∞
X
n=1
∞
X 5
5
√ =
.
n n n=1 n3/2
∞
X
1
√
(b)
3
n2
n=1
The series diverges by the p-series Test with p = 2/3 < 1
∞
∞
X
X
1
1
√
.
=
3
2/3
2
n
n
n=1
n=1
(c)
∞
X
3n
√
n7
n=1
The series converges by the p-series Test with p = 5/2 < 1
∞
∞
∞
X
X
X
3n
3n
3
√ =
=
.
7/2
5/2
7
n
n
n
n=1
n=1
n=1
Theorem: (Direct Comparison Test)
X
X
Suppose
an and
bn are series with positive terms.
X
X
1. If an ≤ bn for all n and
bn converges, then
an converges.
2. If an ≥ bn for all n and
X
bn diverges, then
X
an diverges.
Note: If the larger series diverges, then no conclusion can be made about the smaller series.
Similarly, if the smaller series converges, then no conclusion can be made about the larger
series.
Example: Determine whether the following series converge or diverge:
(a)
∞
X
n=1
3n
1
+n
The series converges by the Direct Comparison Test,
∞
X
n=1
∞
X 1
1
≤
< ∞.
3n + n n=1 3n
The last series converges by the Geometric Series Test with r = 1/3 < 1.
√
∞
X
n
(b)
2
n +4
n=1
The series converges by the Direct Comparison Test,
√
∞
∞ √
∞
X
X
n
n X 1
≤
=
< ∞.
2+4
2
3/2
n
n
n
n=1
n=1
n=1
The last series converges by the p-series Test with p = 3/2 > 1.
(c)
∞
X
n=1
2n
√
3n5 − n
The series diverges by the Direct Comparison Test,
∞
X
n=1
∞
∞
∞
X 2n
X 2n
X
2n
2
√
√
√
√
>
=
=
< ∞.
5
5/2
5
3n − n n=1 3n
3n
3n3/2
n=1
n=1
The last series converges by the p-series Test with p = 3/2 > 1.
Note: It is not always possible to make a direct comparison between two similar series. For
∞
∞
X
X
1
1
instance, consider the series
. This series is similar to
which is a convergent
n
4 −3
4n
n=1
n=1
geometric series. However,
∞
∞
X
X
1
1
<
.
n
n
4
4 −3
n=1
n=1
Thus, the Direct Comparison Test cannot be used.
Theorem: (Limit Comparison Test)
X
X
Suppose
an and
bn are series with positive terms. If
lim
n→∞
an
= L,
bn
where 0 < L < ∞, then the series either both converge or both diverge.
Example: Determine whether the following series converge or diverge:
(a)
∞
X
n=1
1
4n − 3
Compare this series to
∞
∞ n
X
X
1
1
=
n
4
4
n=1
n=1
which converges by the Geometric Series Test with r =
1
4n
4n
·
=
lim
= lim
n→∞ 4n − 3
n→∞ 4n − 3
n→∞
1
lim
1
< 1. Now
4
1
1−
3
4n
= 1 ∈ (0, ∞).
The given series converges by the Limit Comparison Test.
(b)
∞
X
n5 − n4
n=1
n6 + 1
Compare this series to
∞
X
1
which diverges by the p-series Test with p = 1. Now
n
n=1
n5 − n4 n
n6 − n5
·
=
lim
= 1 ∈ (0, ∞).
n→∞ n6 + 1
1 n→∞ n6 + 1
lim
The given series diverges by the Limit Comparison Test.
X
Suppose that we have shown
an is a convergent series and we want to approximate
the sum S of the series. Since S = limn→∞ Sn , the nth partial sum is a good approximation
for n sufficiently large. The question is, how large is sufficiently large?
Definition: The remainder of this approximation is defined as
Rn = S − Sn .
Theorem: (Remainder Estimate for the Integral Test)
X
If
an converges by the Integral Test and Rn = S − Sn , then
Z ∞
Z ∞
f (x)dx.
f (x)dx ≤ Rn ≤
n
n+1
∞
X
1
Example: Consier the series
.
n2
n=1
(a) Approximate the sum of the series by using the sum of the first 10 terms. Estimate
the error involved in this approximation.
The approximate sum is
S ≈ S10 = 1 +
1
1
1 1
+ +
+ ··· +
≈ 1.5498.
4 9 16
100
An estimate for the remainder is
∞
Z
Rn ≤
10
dx
1
= .
2
x
10
The size of the error is at most 0.1.
(b) How many terms are required to ensure that the sum is accurate to within 0.001?
By the Remainder Estimate Theorem,
Z
∞
Rn ≤
n
dx
1
= .
2
x
n
To ensure that the error is less than 0.001, set
1
≤ 0.001
n
1
1
≤
n
1000
1000 ≤ n.
Therefore, we need at least n = 1000 terms.