Series Convergence
April 2016
Math 105, 207
Elyse Yeager (with some changes done by Reza Sadoughian)
Overview
A sequence is a list of numbers. A series is the ”sum” of an infinite sequence, defined as the limit of the partial sums:
∞
N
X
X
an := lim
an . A convergent series is one in which that limit exists, and a divergent series is one in which that limit
N →∞
n=a
n=1
does not exist. If a series has all positive terms, and is divergent, then the sum is going to infinity. However, in a series with some
∞
∞
P
P
sink; we can use the divergence test
(−1)k or
terms negative, it is possible to be divergent without going to infinity. (like
k=0
k=0
to prove that these series diverge)
Geometric
A geometric series is a series of the form {an } = {c · rn } for constants c and r. Equivalently, a geometric series has terms that
differ by a constant ratio (hence the use of “r”): to get from one term to the next, you simply multiply by the constant r.
N
X
1 − rN +1
For any finite N ,
rn =
. In the infinite case, convergence depends on the absolute value of r. A large r will
1−r
n=0
cause divergence, a small r will cause convergence.
1
∞
X
if |r| < 1
1−r
rn =
DIV if |r| ≥ 1
n=0
Divergence Test
If {an } is a series and lim an 6= 0, then
n→∞
∞
X
an is divergent.
n=a
If lim an = 0, the divergence test says nothing, and we need another test.
n→∞
Integral Test
If a function f (x) is positive and decreasing, and we define a sequence {an } = {f (n)}, then
same thing: they both converge, or both diverge.
P∞
n=a
an and
R∞
a
f (x)dx do the
p-test
Using the integral test, we can determine that, for a constant p,
P∞
1
n=1 np
is convergent when p > 1 and divergent when p ≤ 1.
Comparison and Limit Comparison Tests
Suppose {an } and {bn } have positive terms. We know the convergence/divergence of one (say, by the integral test, or some other
test) and we suspect the two behave similarly, so we want to conclude something about the
Pconvergence/divergence of the other.
First,
we
may
know
that
a
≤
b
for
all
n
suitably
large.
In
this
case,
if
we
know
bn converges, then we can conclude
n
n
P
P
P
P
an converges as well. If we know
a
diverges,
then
we
can
conclude
b
diverges
as
well.
However, if we know that
an
n
n
P
converges, or if we know that
bn diverges, we can’t conclude anything. This is the Comparison Test, sometimes called the
Direct Comparison Test.
Second, we may suspect that the differences between an and bn are not great when n is large. Then we use the Limit
P
P
an
Comparison Test, which says that if lim
is a positive number (greater than zero, not infinite), then
an and
bn do the
n→∞ bn
same thing: they both converge, or both diverge.
Ratio Test
an+1
. If the limit is in [0, 1), then the series converges.
an
If the limit is greater than 1 (including infinity), the series diverges. If the limit is equal to 1, the test is inconclusive, so we have
to try another test.
Let {an } be a sequence with positive terms. We consider the limit lim
n→∞
Absolute Convergence
P
P
If
|an | converges, then it must be true that also
an converges. This is helpful to know when you have a series with some
negative
terms,
and
you
want
to
use
a
test
that
requires
positivePterms.
P
P
P
P
If
|an | converges, we say
an converges absolutely. If
an converges BUT
|an | diverges, we say
an converges
conditionally.
P
P
By theorem 8.21 of the textbook, if
|an | converges, then
an converges. (in other words, ”absolutely convergence”⇒
∞ (−1)k−1
∞ (−1)k−1
P
P
”convergence”) However, the converse is not true. For example, the series
|
converges, but the series
|=
k
k
k=1
k=1
∞ 1
P
does not converge.
k=1 k
Example: The following series are absolutely convergent (and so convergent)
1
1
1
1
1
1
1
1
+ 2 − 2 + 2 − 2 + 2 − 2 + 2 − ...
22
3
4
5
6
7
8
9
1
1
1
1
1
1
1
1
1 − 2 − 2 + 2 − 2 − 2 + 2 − 2 − 2 + ...
2
3
4
5
6
7
8
9
1
1
1
1
1
1
1
1
1 + 2 − 2 + 2 + 2 − 2 + 2 + 2 − 2 + ...
2
3
4
5
6
7
8
9
1−
Indeed, they converge, because the following series converge (theorem 8.21):
∞
1+
X 1
1
1
1
1
1
1
1
1
+ 2 + 2 + 2 + 2 + 2 + 2 + 2 + ... =
2
2
3
4
5
6
7
8
9
k2
k=1
But the series
∞
X
(−1)k−1
k=1
k
=1−
1 1 1 1 1 1
+ − + − + + ...
2 3 4 5 6 7
does not converge absolutely, which means that the series itself converges but the series
does not converge. (it is called Harmonic series). So we say that
∞
P
k=1
∞
P
k=1
(−1)k−1
k
k−1
| (−1)k
|=
∞
P
k=1
1
k
= 1 + 21 + 31 + 14 + ...
converges conditionally. (this series converges but
this is not part of the syllabus for Math105)
Remark 1
We can use Integral, Comparison, Limit Comparison and Ratio test only when the general terms of the series are all positive
(or if, after some finitely many terms, they are all positive, like 1 − 212 − 312 + 412 − 512 + 612 + 712 + 812 + 912 + ... where all terms
after the fifth one are positive). If the general terms of a series are all negative, we can first factorize a negative sign to get a
∞
∞
P
P
series like −
ak where ak ≥ 0 and then we can use any of the above tests. But if the series
ak involves infinitely many
k=1
k=1
positive terms and infinitely many negative terms, we cannot use any of the above tests. (like
first check, if the series
∞
P
∞
P
k=1
ak =
∞
P
k=1
(−1)k−1
)
k
We should
|ak | converges. Since the terms of this last series are positive, we can here use any test like Integral,
k=1
Comparison, Limit Comparison or Ratio test to obtain its convergence or divergence.
Remark 2
If the general term of a series involve factorial, like
∞
P
k=1
1
n! ,
∞
P
k=1
(n!)2
(2n)! ,
then we cannot use the integral test. (since we cannot define
1
proper functions like f (x) = x!
. These are not defined for values of x that are not integer) In this case, we can use ratio test,
Comparison, Limit Comparison or divergence test. Probably, the ratio test is the best option to try first.