LECTURE 30: THE RATIO, AND COMPARISON TESTS
MINGFENG ZHAO
March 23, 2015
Theorem 1 (Divergence test). If a series
series
∞
X
∞
X
ak converges, then lim ak = 0. That means, if lim ak 6= 0, then the
k→∞
k=1
k→∞
ak diverges.
k=1
Theorem 2 (Integral test). If f (x) is a continuous, positive, decreasing function for x ≥ 1, let ak = f (k) for k = 1, 2, · · · ,
then
∞
X
Z
ak
∞
and
f (x) dx
1
k=1
either both converge or both diverge. In the case of convergence, the value of the integral is NOT equal to the value of
the series.
Theorem 3 (p-series). Let p be a real number, then

∞
 converges, if p > 1,
X
1
=
 diverges,
kp
if p ≤ 1.
k=1
Example 1. For the series
∞
X
k=1
√
k2 + 1
, notice that
k
√
k2 + 1
k
√
=
=
=
By the divergence test, we know that the series
k2 + 1
√
k2
r
k2 + 1
k2
r
1
1 + 2 → 1,
k
∞
X
k=1
√
as k → ∞.
k2 + 1
diverges.
k
1
2
MINGFENG ZHAO
Example 2. For the series
∞
X
2
2
ke−2k , let f (x) = xe−2x , then f (x) is continuous, positive and decreasing. Notice that
k=1
Z
∞
Z
∞
f (x) dx =
1
Z
For
2
xe−2x dx = lim
b→∞
1
Z
b
2
xe−2x dx.
1
2
xe−2x dx, then
Z
xe
−2x2
Z
dx
du
e−2u ·
Let u = x2 , then du = 2xdx
2
Z
1
e−2u du
2
1
− e−2u + C
4
2
1
− e−2x + C Since u = x2 .
4
=
=
=
=
Hence we get
Z
∞
f (x) dx = lim
b→∞
1
Z
So
"
#
b
2
e−2
e−2b
e−2
1 −2x2 = lim
−
=
.
− e
b→∞
4
4
4
4
1
∞
f (x) dx converges. By integral test, we know that the series
1
∞
X
2
ke−2k converges.
k=1
Properties of convergent series
Theorem 4 (Properties of Convergent Series). a. Suppose
∞
X
cak converges and
k=1
b. Suppose
A ± B.
c. Suppose
∞
X
∞
X
k=1
cak = c
∞
X
k=1
ak converges to A and c is a real number, then the series
k=1
ak = cA.
k=1
ak converges to A and
k=1
∞
X
∞
X
∞
X
bk converges to B, then the series
k=1
ak converges and
∞
X
d. If M is a positive integer, then
k=1
(ak ±bk ) converges, and
k=1
bk diverges, then
k=1
∞
X
∞
X
∞
X
ak and
∞
X
∞
X
(ak ±bk ) =
k=1
(ak ± bk ) diverges.
k=1
ak either both converges or both divergent. In general, whether a
k=M
series converges does not depend on a finite number of terms added to or removed from the series. However, the value
of a convergent series does change if nonzero terms are added or removed.
#
" ∞
k
X
2
22k−1
Example 3. Evaluate the infinite series S =
5·
−
.
3
7k
k=1
LECTURE 30: THE RATIO, AND COMPARISON TESTS
3
k
22k−1
2
2
10
and bk =
for k ≥ 1, then {ak }∞
=
k=1 is a geometric sequence with initial term a = 5 ·
3
7k
3
3
2
2
1 4
∞
and ratio r = . For {bk }k=1 , it is a geometric sequence with initial term a = · = and
3
2 7
7
Let ak = 5 ·
ratio r
=
=
=
22(k+1)−1
7k+1
22k−1
7k
22k+1
7k
·
22k−1 7k+1
4
7
So we get
S
=
=
∞
X
"
#
k
2
22k−1
−
5·
3
7k
k=1
k
k X
∞
∞
X
2
4
2
·
5·
−
3
7
7
k=1
=
Example 4. For the series
∞ X
1
k=1
k
−
k=1
10
3
1−
2
3
=
10 −
=
28
.
3
−
2
7
1−
4
7
2
3
∞
∞ ∞
X
X
X
1
1
1
1
1
diverges
and
converges,
then
the
series
−
,
since
k2
k
k2
k k2
k=1
k=1
k=1
diverges.
∞ X
1
1
Example 5. For the series
−
, this is a telescoping series, we know that it’s convergent and
k k+1
k=1
∞
∞
∞ X
X
X
1
1
1
1
−
= 1. On the other hand, both
and
diverge.
k k+1
k
k+1
k=1
k=1
k=1
Ration test
Theorem 5 (Ratio Test). Let
∞
X
ak be an infinite series with positive terms and r = lim
k=1
I. If 0 ≤ r < 1, the series converges.
II. If r > 1, the series diverges.
III. If r = 1, the ration test is inconclusive.
k→∞
ak+1
, then
ak
4
MINGFENG ZHAO
Example 6. For the series
∞
X
e−k (k 2 + 4), let ak = e−k (k 2 + 4), then
k=1
ak+1
ak
e−(k+1) ((k + 1)2 + 4)
e−k (k 2 + 4)
=
= e−1 ·
Since e−1 < 1, by the ratio test, then the series
k 2 + 2k + 5
→ e−1 ,
k2 + 4
∞
X
as k → ∞.
e−k (k 2 + 4) converges.
k=1
Definition 1. For any positive integer n, we define the factorial of n, n!, by
n! := n · (n − 1) · (n − 2) · · · 2 · 1.
And the factorial of 0 is defined as:
0! := 1.
Example 7. For 4!, then
4! = 4 · 3 · 2 · 1 = 24.
Example 8. For the series
∞
X
10k
k=1
k!
, let ak =
ak+1
ak
10k
, then
k!
=
10k+1
(k+1)!
10k
k!
=
10k+1
k!
·
10k
(k + 1)!
=
10 ·
=
By the ratio test, then the series
∞
X
10k
k=1
Example 9. For the series
∞
X
kk
k=1
k!
k!
, let ak =
ak+1
ak
k · (k − 1) · · · 2 · 1
(k + 1) · k · (k − 1) · · · 2 · 1
1
10 ·
→ 0, as k → ∞.
k+1
converges.
kk
, then
k!
=
(k+1)k+1
(k+1)!
kk
k!
=
(k + 1)k+1
k!
·
kk
(k + 1)!
LECTURE 30: THE RATIO, AND COMPARISON TESTS
=
k · (k − 1) · · · 2 · 1
(k + 1)k+1
·
k
k
(k + 1) · k · (k − 1) · · · 2 · 1
=
(k + 1)k+1
1
·
kk
k+1
(k + 1)k
kk
k
1
=
1+
→ e,
k
5
=
Since e > 1, by the ratio test, then the series
∞
X
kk
k=1
k!
as k → ∞.
diverges.
Comparison test
Theorem 6 (Comparison test). Let
∞
X
ak and
II. If 0 < bk ≤ ak for all k ≥ 1 and
Example 10. Look at the series
∞
X
k=1
bk be series with positive terms, then
k=1
k=1
I. If 0 < ak ≤ bk for all k ≥ 1, and
∞
X
∞
X
bk converges, then
∞
X
k=1
∞
X
k=1
∞
X
k=1
k=1
bk diverges, then
ak also converges.
ak diverges as well.
1
, notice that k 2 (k + 1) = k 3 + k 2 ≥ k 3 , then
k 2 (k + 1)
0<
1
1
≤ 3,
+ 1)
k
k 2 (k
for all k ≥ 1.
∞
∞
∞
X
X
X
1
1
1
is
a
p-series
with
p
=
3,
then
converges.
By
the
comparison
test,
then
the
series
k3
k3
k 2 (k + 1)
k=1
k=1
k=1
converges.
Since
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca