Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
The Integral and Comparison Tests
The Integral Test
THE INTEGRAL TEST: Suppose f is a continuous, positive, decreasing function on [1, ∞)
∞
X
and let an = f (n). Then the series
an is convergent if and only if the improper integral
n=1
Z∞
f (x)dx is convergent. In other words:
1
(a) If
Z∞
f (x)dx is convergent, then
(b) If
1
an is convergent.
n=1
1
Z∞
∞
X
f (x)dx is divergent, then
∞
X
an is divergent.
n=1
REMARK: Don’t use the Integral Test to evaluate series, because in general
Z∞
∞
X
an 6= f (x)dx
n=1
1
EXAMPLES:
Z∞
∞
X
1
1
1
1.
is divergent, because f (x) =
is continuous, positive, decreasing and
dx is
n
x
x
n=1
divergent by the p-test for improper integrals, since p = 1 ≤ 1.
1
∞
Z
∞
X
1
1
1
2.
is divergent, because f (x) = 1/2 is continuous, positive, decreasing and
dx
1/2
1/2
n
x
x
n=1
1
is divergent by the p-test for improper integrals, since p = 1/2 ≤ 1.
Z∞
∞
X
1
1
1
3.
is convergent, because f (x) = 2 is continuous, positive, decreasing and
dx is
2
n
x
x2
n=1
convergent by the p-test for improper integrals, since p = 2 > 1.
1
REMARK 1: When we use the Integral Test it is not necessary to start the series or the integral
at n = 1. For instance, in testing the series
Z∞
∞
X
1
1
we use
dx
n+1
x+1
n=5
5
REMARK 2: It is not necessary that f be always decreasing. It has to be ultimately decreasing,
that is, decreasing for x larger than some number N.
EXAMPLE: Determine whether the series
∞
X
ln n
n
n=1
1
converges or diverges.
Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
EXAMPLE: Determine whether the series
∞
X
ln n
n
n=1
converges or diverges.
ln x
is positive and continuous for x > 1. However, looking at
Solution: The function f (x) =
x
the graph of this function we conclude that f is not decreasing.
At the same time one can show that this function is ultimately decreasing. In fact,
′
1
· x − ln x · 1
ln x
1 − ln x
(ln x)′ · x − ln x · x′
′
x
f (x) =
=
=
=
2
2
x
x
x
x2
Note that 1 − ln x < 0 for all sufficiently large x which means that f ′ (x) < 0 and therefore f
is ultimately decreasing. So, we can apply the Integral Test:
Z∞
ln x
dx = lim
t→∞
x
Zt
ln x
(ln x)2
dx = lim
t→∞
x
2
1
1
Since this integral diverges, the series
∞
X
ln n
n=1
EXAMPLE: Determine whether the series
n
∞
X
n=2
t
(ln t)2
=∞
t→∞
2
= lim
1
also diverges.
1
converges or diverges.
n ln n
1
is continuous, positive and decreasing on [2, ∞), therefore
Solution: The function f (x) =
x ln x
we can apply the Integral Test:
Z∞
2
1
dx = lim
t→∞
x ln x
Zt
1
dx = lim ln(ln x)]t2 = lim [ln(ln t) − ln(ln 2)] = ∞
t→∞
t→∞
x ln x
2
Since this integral diverges, the series
∞
X
n=2
EXAMPLE: Determine whether the series
1
also diverges.
n ln n
∞
X
n=2
1
converges or diverges.
n ln2 n
2
Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
EXAMPLE: Determine whether the series
∞
X
n=2
1
converges or diverges.
n ln2 n
1
is continuous, positive and decreasing on [2, ∞), therefore
Solution: The function f (x) =
x ln2 x
we can apply the Integral Test:
t
Z∞
Zt
1
1
1
1
1
1
lim
lim −
= lim −
=
+
2 dx = t→∞
2 dx = t→∞
t→∞
ln x 2
ln t ln 2
ln 2
x ln x
x ln x
2
2
Since this integral converges, the series
∞
X
n=2
mentioned before,
∞
X
n=2
1
also converges. We also note that, as it was
n ln2 n
1
1
6=
.
2
ln 2
n ln n
THEOREM (p-Test): The p-series
∞
X
1
is convergent if p > 1 and divergent if p ≤ 1.
p
n
n=1
Proof: We distinguish three cases:
∞
X
1
1
diverges by the Divergence Test.
Case I: If p < 0, then lim p = ∞, therefore
p
n→∞ n
n
n=1
∞
X
1
1
1
1
Case II: If p = 0, then lim p = lim 0 = lim
= 1, therefore
diverges by the
p
n→∞ n
n→∞ n
n→∞ 1
n
n=1
Divergence Test.
1
Case III: If p > 0, then the function f (x) = p is continuous, positive and decreasing on
x
∞
X
1
is convergent if and only if
[1, ∞), therefore we can apply the Integral Test by which
p
n
n=1
Z∞
Z∞
1
1
the improper integral
dx is convergent. But
dx is convergent if p > 1 and divergent
p
x
xp
1
if p ≤ 1 by the p-test for improper integrals. 1
REMARK: As before, when we use the p-Test it is not necessary to start the series at n = 1.
EXAMPLE: Determine whether the following series converge or diverge:
∞
∞
∞
X
X
X
1
1
1
(b)
(c)
, ǫ>0
(a)
2
1+ǫ
n
n
n
n=1
n=1
n=1
∞
∞
X
X
1
1
diverges by the p-test for series, since p = 1 ≤ 1. The series
Solution: The series
n
n2
n=1
n=1
∞
X
1
converges by the p-test for series, since p = 2 > 1. The series
converges by the p-test
1+ǫ
n
n=1
for series, since p = 1 + ǫ > 1.
∞
X
1
converges or diverges.
EXAMPLE: Determine whether the series
n+1
n=1
3
Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
EXAMPLE: Determine whether the series
∞
X
n=1
1
converges or diverges.
n+1
REMARK: Note that we CAN’T apply the p-test directly!
1
is continuous, positive and decreasing on [1, ∞), therex+1
fore we can apply the Integral Test:
Zt
Z∞
1
1
dx = lim
dx = lim ln(x + 1)]t1 = lim [ln(t + 1) − ln 2] = ∞
t→∞
t→∞
t→∞
x+1
x+1
Solution 1: The function f (x) =
1
1
Since this integral diverges, the series
Solution 2: We have
∞
X
1
also diverges.
n+1
∞
X
X1
1
=
n + 1 n=2 n
n=1
n=1
∞
∞
∞
X
X
1
1
diverges by the p-test with p = 1, it follows that
also diverges, since
Since
n
n+1
n=1
n=2
convergence or divergence is unaffected by deleting a finite number of terms.
The Comparison Tests
X
X
THE COMPARISON TEST: Suppose that
an and
bn are series with positive terms.
X
X
(a) If
bn is convergent and an ≤ bn for all n, then
an is also convergent
X
X
(b) If
bn is divergent and an ≥ bn for all n, then
an is also divergent
EXAMPLE: Use the Comparison Test to determine whether the following series converge or
diverge.
∞
∞
X
X
1
ln n
√
(b)
(a)
3
n
n−1
n=2
n=1
∞
X
1
ln n
1
> for n > e and
diverges by the p-test with p = 1, it follows that
Solution: Since
n
n
n
n=1
∞
∞
X
X
ln n
1
1
1
√
also diverges. Similarly, since √
> √
and
diverges by the p-test with
3
3
3
n
n−1
n
n
n=1
n=2
∞
X
1
√
p = 1/3, it follows that
also diverges.
3
n−1
n=2
EXAMPLE: Use the Comparison Test to determine whether the following series converge or
diverge: ∞
∞
∞
X
X
X
1
1
1
(b)
(c)
(a)
3
2
2
n +n+1
n +1
n −1
n=2
n=2
n=1
4
Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
EXAMPLE: Use the Comparison Test to determine whether the following series converge or
diverge:
∞
∞
∞
X
X
X
1
1
1
(b)
(c)
(a)
3
2
2
n +n+1
n +1
n −1
n=2
n=2
n=1
Solution:
(a) Since
1
1
< 3
3
n +n+1
n
it follows that
∞
X
n=1
n3
∞
X
1
n3
n=1
and
converges by the p-test with p = 3
1
also converges.
+n+1
(b) Since
1
1
<
n2 + 1
n2
it follows that
∞
X
n=1
n2
and
∞
X
1
n2
n=1
converges by the p-test with p = 2
1
also converges.
+1
(c) Since
2
1
<
n2 − 1
n2
it follows that
∞
X
n=2
for n ≥ 2 and
∞
∞
X
X
2
1
=
2
n2
n2
n=2
n=2
1
also converges.
n2 − 1
Another way to show it is to rewrite
∞
X
n=2
∞
X
n=1
1
1
< 2
2
n + 2n
n
it follows that
n=1
n2
1
as
−1
∞
∞
X
X
1
1
1
=
=
2
2
2
(n + 1) − 1 n=1 n + 2n + 1 − 1 n=1 n + 2n
Since
∞
X
converges by the p-test with p = 2
and
∞
X
1
n2
n=2
converges by the p-test with p = 2
∞
X
1
1
and
therefore
also converges.
2−1
n2 + 2n
n
n=2
5
Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
X
X
THE LIMIT COMPARISON TEST: Suppose that
an and
bn are series with positive
terms. If
an
=c
lim
n→∞ bn
where c is a finite number and c > 0, then either both series converge or both diverge.
EXAMPLE: Use the Limit Comparison Test to determine whether
∞
X
n=2
diverges.
Solution: Put an =
n2
1
converges or
−1
1
1
,
b
=
. Then
n
n2 − 1
n2
1
2
n2
an
= lim n 1− 1 = lim 2
= lim
c = lim
n→∞
n→∞
n→∞ n − 1
n→∞ bn
2
n
n2
n2
n2 −1
n2
1
1
=1
=
1
n→∞ 1 − 2
1
−
0
n
= lim
∞
∞
X
X
1
1
converges by the p-test with p = 2, it follows that
also
Since c = 1 and
2
2
n
n −1
n=2
n=2
converges.
EXAMPLE: Use the Limit Comparison Test to determine whether the following series converge
or diverge:
(a)
∞
X
n=1
1
2 n+1
√
(b)
∞
X
n=2
1
2
4n − n − 1
(c)
∞
X
n=1
6
4n2 − n + 5
n5 + n4 + 2n − 2
(d)
∞
X
n=1
1
n1+1/n
Section 8.3 The Integral and Comparison Tests
2010 Kiryl Tsishchanka
EXAMPLE: Use the Limit Comparison Test to determine whether the following series converge
or diverge:
∞
∞
∞
∞
X
X
X
X
1
1
4n2 − n + 5
1
√
(a)
(b)
(c)
(d)
2
5
4
1+1/n
2 n+1
4n − n − 1
n + n + 2n − 2
n
n=1
n=2
n=1
n=1
Solution:
1
1
(a) Put an = √
, bn = √ . Then
2 n+1
n
c = lim
n→∞
Since c =
diverges.
an
=
bn
√
n
= lim √
= lim
n→∞ 2 n + 1
n→∞
√1
2 n+1
lim
n→∞ √1
n
√
√n
n
√
2 √n+1
n
= lim
n→∞
1
1
1 =
√
2
2+ n
∞
∞
X
X
1
1
1
√
√ diverges by the p-test with p = 1/2, it follows that
and
also
2
n
2
n
+
1
n=1
n=1
(b) Put an =
1
1
,
b
=
. Then
n
4n2 − n − 1
n2
an
= lim
c = lim
n→∞
n→∞ bn
1
4n2 −n−1
1
n2
n2
= lim
= lim
n→∞
n→∞ 4n2 − n − 1
∞
n2
n2
4n2 −n−1
n2
= lim
n→∞
1
4−
∞
1
n
−
1
n2
=
1
4
X 1
X
1
1
Since c = and
converges
by
the
p-test
with
p
=
2,
it
follows
that
also
2
2
4
n
4n
−
n
−
1
n=2
n=2
converges.
1
4n2 − n + 5
, bn = 3 . Then
(c) Put an = 5
4
n + n + 2n − 2
n
an
= lim
c = lim
n→∞
n→∞ bn
4n2 −n+5
n5 +n4 +2n−2
1
n3
4 − n1 + n52
4n5 − n4 + 5n3
=
lim
n→∞ 1 + 1 + 24 −
n→∞ n5 + n4 + 2n − 2
n
n
= lim
2
n5
=4
∞
∞
X
X
1
4n2 − n + 5
Since c = 4 and
converges
by
the
p-test
with
p
=
3,
it
follows
that
n3
n5 + n4 + 2n − 2
n=1
n=1
also converges.
1
1
(d) Put an = 1+1/n , bn = . Then
n
n
1
n−1 · n−1/n
n−1−1/n
an
n1+1/n
= lim
=
lim
= lim n−1/n
=
lim
c = lim
1
−1
−1
n→∞
n→∞
n→∞
n→∞
n→∞ bn
n
n
n
To evaluate lim n−1/n we apply L’Hospital’s Rule. To this end we first put y = n−1/n . Then
n→∞
ln y = ln(n−1/n ) = −
We have
(ln n)′
1/n
1
ln n
= lim
= lim
= lim = 0
′
n→∞
n→∞ 1
n→∞ n
n→∞ n
n
∞
X
1
= e0 = 1. Since c = 1 and
diverges by the p-test with p = 1, it
n
n=1
lim
Therefore lim n−1/n
n→∞
follows that
∞
X
n=1
ln n
n
1
n1+1/n
also diverges.
7