LECTURE 31: THE COMPARISON TESTS AND POWER SERIES
MINGFENG ZHAO
March 25, 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
Theorem 4 (Ratio Test). Let
∞
X
ak be an infinite series with positive terms and r = lim
k→∞
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.
Theorem 5 (Comparison test). Let
∞
X
ak and
k=1
I. If 0 < ak ≤ bk for all k ≥ 1, and
II. If 0 < bk ≤ ak for all k ≥ 1 and
∞
X
bk be series with positive terms, then
k=1
∞
X
bk converges, then
∞
X
k=1
∞
X
k=1
∞
X
k=1
k=1
bk diverges, then
1
ak also converges.
ak diverges as well.
ak+1
, then
ak
2
MINGFENG ZHAO
Limit comparison test
Theorem 6 (Limit comparison test). Let
∞
X
ak and
k=1
I. If 0 < L < ∞, then
II. If L = 0 and
∞
X
∞
X
ak and
k=1
∞
X
III. If L = ∞ and
bk diverges, then
k=1
bk be series with positive terms and L = lim
k→∞
k=1
bk
, then
ak
bk either both converge or both diverge.
k=1
bk converges, then
k=1
∞
X
∞
X
∞
X
k=1
∞
X
ak converges.
ak diverges.
k=1
Remark 1. The comparison test and limit comparison test are the same essentially, but the limit comparison
test is much easier to use, because it’s easy to find a comparable function to compute a limit at infinity.
Example 1. Look at the series
∞
X
k=1
k3
k3
k3
1
,
for
the
term
,
it
behaves
like
=
as k → ∞, so let’s compare
4
4
4
2k + 1
2k + 1
2k
2k
1
it with . We get
k
k3
2k4 +1
1
k
=
k4
1
→ ,
2k 4 + 1
2
as k → ∞.
∞
∞
∞
X
X
X
1
1
1
is a p-series with p = 1, then
diverges, which implies that
diverges. By the limit comparison
k
k
2k
k=1
k=1
k=1
∞
X
k3
test, then the series
diverges.
4
2k + 1
Since
k=1
Example 2. For the series
∞
X
5k 4 − 2k 2 + 3
k=1
compare it with
=
k 2 · (5k 4 − 2k 2 + 3)
5k 6 − 2k 4 + 3k 2
5
=
→ ,
6
6
2k − k + 5
2k − k + 5
2
∞
∞
X
X
1
1
is
a
p-series
with
p
=
2,
then
converges.
2
k
k2
k=1
∞
X
5k 4 − 2k 2 + 3
k=1
5k 4 − 2k 2 + 3
5
, it behaves like
as k → ∞, so let’s
2k 6 − k + 5
2k 2
1
. We get
k2
5k4 −2k2 +3
2k6 −k+5
1
k2
Since
2k 6 − k + 5
. For the term
2k 6 − k + 5
k=1
converges.
as k → ∞.
By the limit comparison test, then the series
LECTURE 31: THE COMPARISON TESTS AND POWER SERIES
Example 3. For the series
with
1
3
√
√
3
3
k
k
k
1
= 5 as k → ∞, so let’s compare it
.
For
the
term
,
it
behaves
like
k2 − k
k2 − k
k2
3
k
k=2
∞
X
√
3
. We get
5
k3
√
3
k
k2 −k
1
5
1
5
k3 · k3
k2
= 2
= 2
→ 1,
k −k
k −k
as k → ∞.
k3
Since
∞
X
k=2
√
3
k2
∞
X
1
5
k=2
k3
∞
X
5
1
is a p-series with p =
> 1, then
5 converges. By the limit comparison test, then the series
3
k3
k=2
k
converges.
−k
Summary for choosing a test
The followings are to test when a series of positive terms
P
ak for convergence:
I. Begin with the Divergence Test.
II. Is the series a special series (telescoping series, or geometric series, p-series)?
III. If the general kth term of the series looks like a positive decreasing function you can integrate, then try the
Integral Test.
IV. If the general kth term of series involves k!, k k or ak , where a is a constant, the Ratio Test is advisable.
V. If the general kth term of the series is a rational function of k, use the Comparison Test or the Limit
Comparison Test with the families of series given in Step II.
Absolute and conditional convergences of the series
Definition 1. For a series
say that the series
∞
X
∞
X
ak , we say that the series
k=1
ak conditionally coverages if the series
Example 5. For the series
k=1
∞
X
(−1)k
k2
k=1
converges.
k
∞
X
∞
X
(−1)k
k=1
k
∞
X
|ak | converges. We
k=1
ak converges, but the series
k=1
Example 4. For the series
the series
ak absolutely converges if the series
k=1
k=1
∞
X
(−1)k
∞
X
∞
X
|ak | diverges.
k=1
∞ ∞
∞
X
X
X
(−1)k 1
(−1)k
=
, since
converges,
then
the
series
absolutely
k2 k2
k2
k=1
, notice that
converges, then the series
k=1
∞ ∞
X
(−1)k X
1
=
diverges. On the other hand, one can show that
k k
k=1
∞
X
(−1)k
k=1
k=1
k
k=1
conditionally converges.
4
MINGFENG ZHAO
Theorem 7. If the series
∞
X
|ak | converges, then the series
k=1
∞
X
ak converges.
k=1
∞
X
cos(k) 1
≤ 1 and
,
since
converges, by the comparison test, then
k2 k2
k2
k2
k=1
k=1
∞ ∞
∞
X
X
X
cos(k) cos(k)
cos(k)
converges, that is,
absolutely
converges.
By
Theorem
7,
then
converges.
k2 2
k
k2
Example 6. For the series
k=1
∞
X
cos(k)
k=1
Example 7. For the series
∞
X
k=1
k=1
1
1
, let f (x) =
for all x ≥ 100, then f (x) is positive and
k(ln k)(ln ln k)
x(ln x)(ln ln x)
decreasing. Notice that
Z
∞
Z
f (x) dx
100
For
1
dx, we have
x(ln x)(ln ln x)
Z
1
dx
x(ln x)(ln ln x)
Z
∞
Hence we get
100
∞
=
100
Z
=
1
1
1
1
du Let u = ln ln x, then du =
· dx =
dx
u
ln x x
x(ln x)
=
ln |u| + C
=
ln | ln ln x| + C
∞
1
dx.
x(ln x)(ln ln x)
Since u = ln ln x.
f (x) dx = ln | ln ln x||100 = ∞. By the integral test, we know that the series
∞
X
k=1
1
is
k(ln k)(ln ln k)
divergent.
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