CHAPTER 6
INFINITE SERIES OF REAL NUMBERS
6.1 INTRODUCTION
P
DEFINITION. Let S = 1
1 an be a series of real numbers.
(1) The partial sums of S of order n are the numbers de¯ned, for all n 2 N
by
n
X
sn :=
ak :
k=1
(2) S is said to converge if and only if the sequence of partial sums fsn g
converges to some s 2 R as n ! 1; i.e. for ² > 0 there is an N 2 N such
that n ¸ N implies that jsn ¡ sj < ². In this case we shall write
s=
1
X
ak ;
k=1
P
and call s the sum or value of the series 1
k=1 ak :
(3) S is said to diverge if and only if the sequence of partial sums fsn g does
not converge as n ! 1. When sn diveges to 1 as n ! 1, we shall also
write
1
X
ak = 1:
k=1
Example. Decimal expansion
P1
Example. k=1 2¡k = 1:
P1
Example. k=1 (¡1)k diverges.
P
Example. Harmonic series 1
k=1
1
k
diverges.
Theorem.
[Divergence test] If the sequence fak g does not converge to 0, then the
P
series 1
k=1 ak diverges.
Theorem. [Telescopic series] If fak g is a convergent sequence, then
1
X
(ak ¡ ak+1 ) = a1 ¡ lim an :
n!1
k=1
P
k
Theorem. [Geometric series] The series 1
k=1 x converges if and only if jxj < 1, in
which case
1
X
x
xk =
:
1¡x
k=1
Typeset by AMS-TEX
1
2
Theorem. [Cauchy Criterior] Let fak g be a real sequence. Then the series
converges if and only if for any ² > 0 there is an N 2 N such that
m>n¸N
implies j
m
X
P1
k=1
ak
ak j < ²:
k=n
P1
P1
Theorem. Let fak g and fbk g be real sequence. If k=1 ak and k=1 bk are convergent
series, then
1
1
1
X
X
X
(ak + bk ) =
ak +
bk
k=1
and
k=1
1
X
cak = c
k=1
k=1
1
X
ak
k=1
for all c 2 R.
6.2 SERIES WITH NONNEGATIVE TERMS
P
Theorem. Suppose that ak ¸ 0 for k ¸ N . Then the series 1
k=1 ak converges if and
only if the sequence ofPpartial sums fsn gis bounded, i.e. if and only if there is a ¯nite
number M such that j nk=1 ak j · M for all n.
Theorem.
P1 [Integral test]. Suppose that f : [1; 1) ! R is positive decreasin on [1; 1).
Then R k=1 f (k) converges if and only if f is improperly integrable on [1; 1),i.e. if and
1
only if 1 f (x)dx < 1:
Example.
P1
1
k=1 1+k2 :
Corollary. [p-series test] The series
P1
1
k=1 kp
converges if and only if p > 1.
Theorem. [Comparison test]. Supposre that 0 · ak · bk for large k.
P
P1
(1) If P1
b
<
1,
the
ak < 1:
k
k=1
Pk=1
1
(2) If 1
a
=
1,
the
k=1 k
k=1 bk = 1:
Example.
P1
3k
k=1 k2 +1
q
log k
k :
Theorem. [Limit comparison test]. Suppose that ak ¸ 0 and bk > 0 for large k and
L = limn!1 abnn exists as an extended real number.
P1
P1
(1) If < L < 1, then
a
converges
if
and
only
if
k
k=1
k=1 bk converges.
P1
P1
(2) If L = 0 and P
b
converges,
then
a
convergea.
k
k=1 k
Pk=1
1
(3) If L = 1 and 1
b
diverges
,
then
a
k
k=1
k=1 k diverges.
3
Example.
P1
k=1
k
p
:
4k4 +k2 +5k
Example. Let ak ! 0 as k ! 1. Prove that
P
1
k=1 jak j converges.
P1
k=1
sin jak j converges if and only if
6.3 ABSOLUTE CONVERGENCE
P1
DEFINITION. Let S = k=1 ak be an in¯nite series.
P
(1) S is said to converge absolutely if and only if 1
k=1 jak j < 1:
(2) S is said to converge conditionally if and only if S converges but not
absolutely.
P1
Remark. A series k=1 ak converges absolutely if and only if for every ² > 0
there is N 2 N such that
m>n¸N
implies
m
X
jak j · ²:
n
Remark. If
versely.
P1
k=1
ak converges absolutely then
P1
k=1
ak converges, but not con-
Remark. Let x 2 R and fxk g be a real seuquence.
(1) If lim supk!1 xk < x, then xk < x for large k.
(2) If lim supk!1 xk > x, then xk > x for in¯nite many k.
(3) If xk ! x as k ! 1 , then lim supk!1 xk = x.
Theorem. [Root test]. Let ak 2 R and
r = lim supk!1 jak j1=k .
P
(1) If r < 1, then P1
k=1 ak converges absolutely.
(2) If r > 1, then 1
k=1 ak diverges.
Theorem. [Ratio test]. Let ak 2 R with ak 6
= 0 for large k and suppose that r =
jak+1 j
limk!1 jak j exists as an extended real number.
P
(1) If r < 1, then P1
k=1 ak converges absolutely.
(2) If r > 1, then 1
k=1 ak diverges.
Corollary. Let ak 2 R with ak 6
= 0 for large k .
P
jak+1 j
(1) If r = lim supk!1 jak j < 1, then 1
k=1 ak converges absolutely.
P1
jak+1 j
(2) If r = lim inf k!1 jak j > 1, then k=1 ak diverges.
Remark.The Root and Ratio test are inconclusive when r = 1.
4
P
P1
DEFINITION. A series 1
k=1 bk is called a rearrangement of
k=1 ak if and only
there is a 1-1 function f : N ! N such that bf (k) = ak for all k.
P1
P1
P1
Theorem.
If
a
converges
absolutely
and
b
is
any
rearramgement
of
k
k
k=1
k=1
k=1 ak ,
P1
then k=1 bk converges and
1
1
X
X
ak =
bk :
k=1
k=1
P1
Theorem. [Riemann].
P1 Let x 2 R. If k=1 ak is conditionally convergent, then there is
a rearrangement of k=1 ak that converges to x.
6.4 ALTERNATING SERIES
gk2N and fbk gk2N be real sequences, and for each
Theorem. [Abel's formula]. Let fakP
pair of integers n ¸ m ¸ 1 set An;m = nm ak . Then
n
X
ak bk = An;m bn ¡
k=m
n¡1
X
Ak;m (bk+1 ¡ bk )
k=m
for all integers n ¸ m ¸ 1.
Theorem.
Pn [Dirichlet's test]. Let ak ; bb 2 R for all k. If the sequence of partial sums
sn = 1 ak is bounded and bk & 0 as k ! 1, then
1
X
ak bk
k=1
converges.
Corollary. [Alternating series test]. If ak & 0 as k ! 1, then
1
X
k=1
converges.
Example.
Example.
P1
2
P1
(¡1)k = log k:
k=1
sin(kx)=k.
6.5 ESTIMATION OF SERIES
(¡1)k ak
5
Theorem. Suppose that f : [1; 1) ! R is positive and decreasing on [1; 1). Then
Z n
n
X
f (n) ·
f (k) ¡
f (x)dx · f (1); for n 2 N:
1
k=1
Moreover, if
P1
k=1
f (k) converges, then
Z 1
n
1
X
X
0·
f (k) +
f (x)dx ¡
f (k) · f (n);
n
k=1
k=1
for all n 2 N:
Example.
P1
k=1
2
ke¡k 10¡3 .
Pn
1
k=1 k
+ log n + Cn for some Cn 2 (0; 1):
P1
Pn
Theorem. Sup[pose that ak & 0 as k ! 1. If s = k=1 (¡1)k ak and sn = k=1 (¡1)k ak ,
the
0 · jsn ¡ sj · an+1
Example.
for all n 2 N.
Example. ® > 0 and
P1
k
2
k=1 (¡1) k=(k
+ ®), 10¡2 .
P1
P1
Theorem. Suppose that k=1 ak converges absolutely and s is the value of k=1 jak j,
(1) If there exists x 2 (0; 1) and N 2 N such that
jak j1=k · x
for all k ¸ N , then
0·s¡
n
X
jak j ·
k=1
xn+1
1¡x
for all n ¸ N .
(2) If there exists x 2 (0; 1) and N 2 N such that
jak+1 j
·x
jak j
for all k ¸ N , then
0·s¡
n
X
k=1
for all n ¸ N .
Example.
P1
k2k
k=1 (3k2 +k)k :
6.6 ADDITIONAL TEST
jak j ·
jaN jxn¡N +1
1¡x
6
Theorem. [Logarithmic test]. Suppose that ak 6
= 0 fot large k and
log(1=ak )
k!1
log k
p = lim
exists as
an extended real number. If p > 1 then
P1
, then k=1 jak j diverges.
P1
k=1
ak converges absolutely. If p < 1
Theorem. [Raabe's test]. Suppose that there is a constant C and a parameter p such
that
ak+1
p
j
j·1¡
ak
k+C
P1
for large k. If p > 1 then k=1 ak converges absolutely.