CHAPTER 9
Infinite Series
9.1. Absolute Convergence
P
Definition
xn is absolutely
P (9.1.1). Let X = (xn) be a sequence in R.
convergent if |xn| converges in R. A series is conditionally convergent if it is
convergent, but not absolulely convergent.
Example.
1
1
X
X
1
1
(1)
( 1)n+1 2 is absolutely convergent since
(p = 2) converges.
2
n
n
n=1
n=1
1
X
1
(2)
( 1)n+1 is conditionally convergent since it converges by the AST but
n
n=1
1
X
1
(the harmonic series, p = 1) diverges.
n
n=1
Note. For series of all positive or all negative terms,
convergence () absolute convergence.
Theorem (9.1.2). If a series in R is absolutely convergent, then it is
convergent.
Proof. We use the Cauchy Criterion 3.7.4. Let ✏ > 0 be given.
P
P
Since |xn| converges, 9 M (✏) 2 N 3 m > n M (✏) =) (for
xn)
|sm
sn| = |xn+1 + · · · + xm|  |xn+1| + · · · + |xm| < ✏ .
|
{z
}
X
from
|xn|
P
Since ✏ is arbitrary,
xn converges.
154
⇤
9.1. ABSOLUTE CONVERGENCE
155
Remark. A conditionally convergent series must contain infinitely many
positive and negative terms, For such a series, we can make a rearrangement
of terms that converges to any c 2 R. To do this, take positive terms until the
sum is greater than c, negative terms until it is less than c, and so on.
P
Theorem (9.1.5 — Rearrangement Theorem). Let
P xn be
Pan absolutely
convergent series in R. Then any rearrangement
yn of
xn converges
to the same value.
P
Proof. Suppose xn = x. Let ✏ > 0 be given. 9 N 2 N 3 for n, q > N,
q
X
|x sn| < ✏ and
|xk | < ✏.
k=N +1
Let M 2 N 3 all of x1, . . . , xn are contained as summands in
tM = y1 + · · · + yM
Thus, for m
M , tm sn is the sum of a finite number of terms xk with
k > N . So, for some q > N ,
q
X
|tm sn| 
|xk | < ✏.
k N +1
Thus, if m
|tm
M,
x| = |tm
Since ✏ is arbitrary,
P
sn + sn
yn = x.
x|  |tm
sn| + |sn
x| < ✏ + ✏ = 2✏.
⇤
156
9. INFINITE SERIES
9.2. Tests for Absolute Convergence
Theorem (9.2.1 — Limit Comparison Test, II). Suppose (xn) and (yn)
xn
are nonzro real sequences and suppose r = lim
exists in R..
yn
P
P
(a) If r 6= 0,
xn is absolutely convergent ()
yn is absolutely
convergent.
P
P
(b) If r = 0 and
yn is absolutely convergent, then
xn is absolutely
convergent.
Proof. This result follows immediately from the Limit comparison test. ⇤
Theorem (Theorem 9.2.2 — Root Test). Let (xn) be a sequence in R.
(a) If 9 r < 1 and K 2 N 3 |xn|1/n  r for n
P
then
xn converges absolutely.
(b) If 9 K 2 N 3 |xn|1/n
P
then
xn diverges.
1 for n
K,
K,
Proof.
(a) |xn|1/n  r for n K =) |xn|  rn for n K.
P n
P
Since
r converges for r < 1, |xn| converges by comparison.
(b) |xn|1/n
1 for n
K =) |xn|
1 for n
the series diverges by the n’th-term test.
K =)
⇤
9.2. TESTS FOR ABSOLUTE CONVERGENCE
157
1/n
Corollary
(9.2.3).
Let
(x
)
be
a
sequence
in
R
and
suppose
r
=
lim|x
|
n
n
P
exists. Then
xn is absolutely convergent when r < 1 and divergent when
r > 1.
Proof. If r < 1, 9 r1 with r <Pr1 < 1 and K 2 N 3 for n > K,
|xn|1/n  r1. Applying the root test,
xn converges absolutely.
If r > 1, 9 K 2 R 3 |xn|1/n > 1 for n
P
xn diverges by the n’th-term test.
K =)
⇤
Example.
(3) Suppose xn = (np)1/n =)
p
ln xn = ln(np)1/n = ln n =)
n
✓ 1◆
⇣ p ln n ⌘
p· n
lim(ln xn) = lim
= lim
= 0 =)
n
1
| {z }
L’Hopital’s Rule
lim (ln xn) = 0 =) lim eln x = e0 = 1.
n!1
n!1
Thus, for all p-series, absolutely convergent and divergent,
1 1/n
1
1
lim p
= lim p 1/n = = 1.
n!1 n
n!1 (n )
1
158
9. INFINITE SERIES
Theorem (9.2.4 — Ratio Test). Let (xn) be a sequence of nonzero real
numbers.
xn+1
(a) If 9 r 2 R with 0 < r < 1 and K 2 N 3
 r for n K,
xn
P
then
xn is absolutely convergent.
xn+1
(b) If 9 K 2 N 3
1 for n K,
xn
P
then
xn diverges.
Proof.
xn+1
(a) If
 r 8 n K, by induction
xn
|xK+m|  |xK |rm for m 2 N.
P
P n
Thus,
P for n K, the terms of |xn| are dominated by the terms of |xK | r ,
so |xn| converges by comparison.
xn+1
(b) If
1 for n K, by induction
xn
|xK+m| |xK | for m 2 N
P
and so
xn diverges by the n’th-term test.
⇤
Corollary (9.2.5). Let (xn) be a nonzero sequence in R and suppose
P
xn+1
r = lim
exists in R. Then
xn converges absolutely for r < 1 and
xn
diverges for r > 1.
Example.
(4) For any p-series, absolutely convergent or divergent,
1
⇣ n ⌘p
(n+1)p
p
r = lim
=
lim
=
1
= 1.
1
n
+
1
np