Faculty of Technology and Science
Mattias Flygare
Some Properties of Infinite Series
Några egenskaper hos oändliga serier
Mathematics
Degree Project 15ECTS, Bachelor Level
Date:
Supervisor:
Examiner:
Karlstads universitet 651 88 Karlstad
Tfn 054-700 10 00 Fax 054-700 14 60
[email protected] www.kau.se
2012-06-11
Viktor Kolyada
Håkan Granath
Contents
1 Introduction
5
2 Innite Series
6
2.1
Convergence of Innite Series . . . . . . . . . . . . . . . . . . . . . . . .
6
2.2
Associativity of Innite Series . . . . . . . . . . . . . . . . . . . . . . . .
8
2.3
Commutativity of Innite Series
. . . . . . . . . . . . . . . . . . . . . .
3 Innite Products
12
16
3.1
Convergence of Innite Products
. . . . . . . . . . . . . . . . . . . . . .
16
3.2
Tests for Convergence
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.3
Absolute Convergence
. . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
3.4
Associativity and Commutativity . . . . . . . . . . . . . . . . . . . . . .
22
3.5
Summary and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
4 Further Properties of Innite Series
30
4.1
Riemann's Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
4.2
Toeplitz's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
4.3
Multiplication of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . .
42
5 Miscellaneous Examples and Problems
2
49
Abstract
The subject of innite series and the properties thereof are explored, showing
the theorems of Bernhard Riemann, Augustin Louis Cauchy, Otto Toeplitz, Franz
Mertens and Niels Henrik Abel, among others and also several standard and nonstandard examples and problems where these theorems are useful.
Sammanfattning
Oändliga serier och deras egenskaper utforskas med hjälp av satser av bland andra Bernhard Riemann, Augustin Louis Cauchy, Otto Toeplitz, Franz Mertens och
Niels Henrik Abel. Flertalet exempel och problem där dessa satser är användbara
presenteras och löses.
3
Acknowledgements
Thank you for all your help and dedication, my supervisor Viktor Kolyada!
Also I want to thank my wife Anna-Lena and my two sons Sixten and Alexander,
you are truly great!
4
1 Introduction
In this thesis the subject of innite series is explored.
The subject has a long and
widespread history ranging at least back to ancient Greece [5, p. 21] where Archimedes
used the method of exhaustion to calculate decimals of the irrational number
π.
The
history of innite series is not, however, the focus of this thesis, but the properties
of these series.
In particular it is of interest how series behave under the common
operations of addition and multiplication, operations that we can take for granted when
it comes to working with nite sums but, as we shall see, demands some thought and
some conditions to be met in order to make sense for innite series.
When considering innite series it also becomes interesting to consider the multiplication analogue to series, innite products. Indeed there is a close link between the two
concepts and this text strives to show this link and to display and prove some of the
analogous theorems.
The thesis shows the theorems of Bernhard Riemann (1826-1866), Augustin Louis
Cauchy (1789-1857), Otto Toeplitz (1881-1940), Franz Mertens (1840-1927) and Niels
Henrik Abel (1802-1829), among others and also several standard and non-standard
examples and problems where these theorems are useful.
The main contribution from the author of this text has been to gather the relevant
information and present the theorems and proofs in a coherent and understandable way,
and also to solve the problems and examples that are included.
The thesis follows the main direction of Fikhtengol'ts [3, p. 1-29] and taking some
parts from Hyslop [4, p. 93-107]. Also for support and additional information, Bartle
and Sherbert [1], Bromwich [2] and Markushevich [6] was used.
5
2 Innite Series
The sum of a nite number of terms have some properties that can not unconditionally
be carried over to innite series, or sums with an innite number of terms. The concept
of an innite series involves taking the limit of what is usually called the
sum of some function
an ,
dened for all natural numbers
Sn =
n
X
nth
partial
n,
ar ,
(2.1)
r=1
and is commonly written as
∞
X
an = lim Sn .
If the partial sum
to
S.
If the limit
(2.2)
n→∞
n=1
P∞
Sn has a nite limit S then we say that the series n=1 an converges
of Sn does not exist, we say that the series diverges.
2.1 Convergence of Innite Series
Before exploring the properties of associativity and commutativity we rst need to establish some condition for convergence of the series. The following theorem states an
obvious but important condition.
Theorem 2.1.
If the series
∞
X
an
n=1
converges then
lim an = 0.
n→∞
Proof.
Sn−1
Since the series converges, the partial sums
limit as
n
and
Sn
must approach the same
goes to innity. Thus
lim (Sn − Sn−1 ) = 0,
n→∞
but
lim (Sn − Sn−1 ) = lim an ,
n→∞
n→∞
which proves the theorem.
We say that
an → 0
Denition 2.2.
is a
A series
necessary condition
∞
X
for convergence of the series
an
n=1
is said to be absolutely convergent if the series
∞
X
|an |
n=1
converges.
6
P∞
n=1
an .
Denition 2.3.
A series
∞
X
an
n=1
is said to be conditionally convergent if the series
∞
X
|an |
n=1
diverges but the series
∞
X
an
n=1
converges.
Absolute convergence is a stronger property than conditional and every absolutely
convergent series also converges.
A well known test for convergence is the comparison test, sometimes referred to as
the direct comparison test.
Theorem 2.4 (The comparison test).
If the series
∞
X
bn
n=1
is absolutely convergent and there exists a natural number N such that
|an | ≤ |bn |,
then the series
∞
X
∀n ≥ N,
an
n=1
converges absolutely.
Proof.
Let
Sn =
Tn =
n
X
r=1
n
X
|ar |,
|br |.
r=1
P∞
Assume that
n=1
|bn |
converges and
|an | ≤ |bn |
for
n ≥ N.
By this assumption
the limit
T = lim Tn
n→∞
exists and is nite. We have
Sn − SN ≤ Tn − TN , ∀n ≥ N.
Since
Sn
and
Tn
are both monotonically increasing we see that
SN ≤ Sn ≤ Tn − TN + SN ≤ T + SN
or
0 ≤ Sn ≤ T + SN .
This shows that
Sn
is a bounded monotone sequence and thus it must have a limit which
proves the convergence of
Pn
r=1
|an |.
7
2.2 Associativity of Innite Series
The following theorem expresses the associativity of convergent series, analogous to the
property of associativity of nite sums.
Theorem 2.5 (The theorem of associativity for innite series).
∞
X
Let
an
(2.3)
n=1
be a convergent series and let the sequence {nk } be an arbitrary, strictly increasing,
subsequence of the positive integers. Also let
bk = ank−1 +1 + ank−1 +2 + · · · + ank , ∀k = 2, 3, ...
so that the new terms bk are arbitrary groups of the original terms without changing the
order of terms. Then the series
∞
X
bk
(2.4)
k=2
is convergent, with the same sum as
Proof.
(2.3).
Let
Sn =
n
X
ar
r=1
be the
nth
partial sum of (2.3) and let
Tk =
k
X
br
r=2
be the
k th
partial sum of (2.4). Then we have that
Tk =
k
X
br = a1 + a2 + · · · + ank = Snk
r=2
{Tk } is just a subsequence of {Sn }. Every subsequence
limit S is also convergent, and also has the limit S . The
so it is clear that the sequence
of a convergent sequence with
theorem is proved.
Example 2.6.
We can now show that if
P =1−
and
1
1
1
+ s − s + ··· ,
2s
3
4
1
1
1
+ s + s + ··· ,
s
2
3
4
1
P = 1 − s−1 Q
2
Q=1+
then
for
s > 1.
8
(2.5)
(2.6)
(2.7)
s > 1,
Let
so that both (2.5) and (2.6) are convergent. Let
Pn =
n
X
1
(−1)n+1 s
r
r=1
and
Qn =
be the
nth
partial sum of
P
and
Q
n
X
1
s
r
r=1
respectively, so that
lim Pn = P
n→∞
and
lim Qn = Q.
n→∞
By theorem 2.5 we may now group terms without changing the order of appearance
without aecting the convergence or sum. Then
P =
and
Q=
so that we can rewrite the
1
2s
1
1+ s
2
1−
(2n)th
+
n X
r=1
Q2n =
n X
r=1
For any
n
1
1
− s
3s
4
1
1
+ s
s
3
4
partial sums of
P2n =
and
+
P
+ ··· ,
+ ··· .
and
Q
1
1
−
(2r − 1)s
(2r)s
1
1
+
(2r − 1)s
(2r)s
as
.
we have that
n X
=
n
X
r=1
1
1
+
(2r − 1)s
(2r)s
n X
1
1
−
s
s
(2r
−
1)
(2r)
r=1
r=1
n X
1
1
1
1
=
+
−
+
(2r − 1)s
(2r)s
(2r − 1)s
(2r)s
r=1
Q2n − P2n =
−
n
2
2 X 1
1
=
= s−1 Qn
s
s
s
(2r)
2 r=1 r
2
or equivalently
P2n = Q2n −
Letting
n
1
Qn .
2s−1
go to innity, we get the original statement,
P =
1−
9
1
2s−1
Q.
It is important to note that theorem 2.5 does not work in reverse, so to speak, that
is when beginning from a convergent sequence and taking some or all terms and writing
them as the sum of some new terms, we do not have convergence guaranteed for the
new series.
Example 2.7.
A good example of this is the series
0 + 0 + · · · + 0 + · · · = (1 − 1) + (1 − 1) + · · · + (1 − 1) + · · ·
which is obviously convergent with the sum
0,
but when dropping the parentheses so
that we get a new series we have
1 − 1 + 1 − 1 + ··· + 1 − 1 + ···
which is divergent. This shows that convergence is only preserved when grouping the
terms of a convergent series and
not preserved when ungrouping
the terms of a con-
vergent series. There are however conditions under which the reverse of theorem 2.5 is
true, expressed in the next theorem.
Theorem 2.8.
∞
X
Suppose
(ank−1 +1 + ank−1 +2 + · · · + ank ) = (a1 + · · · + an1 ) + (an1 +1 + · · · + an2 ) + · · ·
(2.8)
k=2
is a convergent series for an arbitrary, strictly increasing, subsequence of the positive
integers {nk }. If the terms in each set of parentheses all have the same sign, where this
sign may vary from one set of parentheses to the next, then the series
∞
X
an = a1 + a2 + · · ·
(2.9)
n=1
obtained from dropping the parentheses in
as the original series.
Proof.
(2.8)
is also convergent and has the same sum
Let
Tk =
k
X
(ank−1 +1 + ank−1 +2 + · · · + ank )
r=2
be the
k th
partial sum of (2.8) and let
Sn =
n
X
ar
r=1
be the
nth
partial sum of (2.9). Then
so that for such an
r
ar
is always the same sign when
we have
Snk−1 ≤ Sr ≤ Snk or Snk−1 ≥ Sr ≥ Snk .
But
Tk−1 = Snk−1 and Tk = Snk
10
nk−1 ≤ r ≤ nk ,
so we have
Tk−1 ≤ Sr ≤ Tk or Tk−1 ≥ Sr ≥ Tk .
(2.10)
lim Tk = T but by (2.10), then we also
k→∞
T , so we see that (2.9) converges and has the same sum as (2.8).
Since the series (2.8) converges we know that
have that
lim Sr =
r→∞
Now that we know something about the associativity of convergent series we can
state and prove the next test for convergence.
Theorem 2.9 (The alternating series test - Leibniz criterion).
Let
{an }
be a decreasing sequence of strictly positive numbers and let
lim an = 0.
n→∞
Then the alternating series
∞
X
(−1)n+1 an
n=1
is convergent.
Proof.
Let
br = a2r−1 − a2r
so that
S2n =
n
2n
X
X
br .
(−1)r+1 ar =
r=1
r=1
Since
a2r−1 ≥ a2r
it follows that
br
is positive and so the partial sum
S2n
monotonically increasing. Now
let
cr = −(a2r − a2r+1 )
so that
S2n =
2n
X
(−1)r+1 ar = a1 +
r=1
Since
cr
cr − a2n .
r=1
is negative, we have that
S2n ≤ a1 ,
Thus the subsequence
number
n−1
X
S ∈ R.
S2n
∀n ∈ N.
is a bounded monotone sequence so it converges to some
This means that for each
>0
that
there exists a natural number
1
2
>0
N1
such
|S2n − S| ≤
Also, since
an
is strictly decreasing, for each
such that
|a2n+1 | ≤
11
1
.
2
there exists a natural number
N2
Let
N = max{N1 , N2 }
so that for
n≥N
we have
|S2n+1 − S| = |S2n + a2n+1 − S| ≤ |S2n − S| + |a2n+1 | ≤
1
1
+ = .
2
2
Therefore, both partial sums of an odd number of terms and partial sums of an even
number of terms are within
of
S
n. Since is arbitrary, Sn
n+1
(−1)
an converges.
n=1
for suciently large
must
P∞
therefore converge which means that the series
2.3 Commutativity of Innite Series
In the previous section, the importance of not rearranging the order of appearance of
terms has repeatedly been mentioned. We now explore the conditions under which the
terms of a convergent series may be rearranged without making it divergent or changing
its sum. The conditions are expressed in the following two theorems, the second of which
being the most important one.
Theorem 2.10.
Let
a1 + a2 + · · · + an + · · ·
(2.11)
be a convergent series with all terms non-negative, so that
an ≥ 0,
n = 1, 2, ....
Also let the sequence
{nk }
be an arbitrary rearrangement of the positive integers. Then the series
an1 + an2 + · · · + ank + · · ·
is convergent, and with the same sum as
Proof.
For any given positive integer
Then the
nth
(2.11).
k , let N
partial sum
Sn =
be the largest of the integers
n
X
ar
r=1
and the
k th
partial sum
Tk =
k
X
anr
r=1
satisfy the inequality
Tk ≤ SN .
Since
Sn
(2.12)
is increasing and converges to a limit, say
Tk ≤ S.
12
S,
we have
n1 , n2 , ..., nk .
Tk
The terms of (2.12) are also all non-negative so
is also increasing. By the comparison
test we have (absolute) convergence of the series (2.12) with sum
T
satisfying
T ≤ S.
By the exact same arguments, the series (2.11) can be seen as rearrangement of (2.12),
so we have
S ≤ T,
and so it follows that we have
S=T
and the theorem is proved.
We have seen that under some conditions we may remove parenthesis around terms
and also rearrange the order of appearance of terms.
The next theorem generalizes
this to show that under some conditions we can remove parenthesis around an innite
number of terms, and then also rearrange the terms of the resulting series.
In other
words, under some conditions, we can take two iterated series and make a double series
that converges to the same sum.
Theorem 2.11.
Let
a11
a21
a31
..
.
a12
a22
a32
..
.
···
···
···
a13
a23
a33
..
.
(2.13)
···
be an innite matrix with all ajk the same sign. Assume that ajk are arranged into a
sequence {br }. Then the series
∞
X
br
r=1
converges if and only if:
1. for any j the series
∞
X
ajk
k=1
converges,
2. the iterated series
∞ X
∞
X
ajk
j=1 k=1
converges.
Moreover, in this case
∞
X
r=1
br =
∞ X
∞
X
j=1 k=1
13
ajk .
Proof.
First assume that
Then for all
J
K
and
ajk
are non-negative. Assume
there exists an
R
J X
K
X
r=1 br converges with sum
B.
such that
R
X
ajk ≤
br ≤ B,
r=1
j=1 k=1
and thus it follows that for any
P∞
j,
∞
X
ajk
k=1
converges. Making
K→∞
we get
J X
∞
X
ajk ≤ B
j=1 k=1
and thus
∞ X
∞
X
ajk
j=1 k=1
converges, and its sum
A
satises
A ≤ B.
Now assume that
Then, for each
R
P∞
k=1
ajk
converges for all
(2.14)
j
and that
we can always nd a natural number
PR
r=1 br is contained in the partial sum
PN PN
j=1
∞ X
∞
X
k=1
ajk .
N
P∞ P∞
j=1
k=1
ajk
converges.
such that all the terms of
Thus, if
ajk = A
j=1 k=1
then
R
X
br ≤ A, ∀R ∈ N,
r=1
and so
∞
X
br
r=1
converges to a sum
B
and
B ≤ A.
(2.15)
Equations (2.14) and (2.15) together show that
A=B
and the theorem is proved for non-negative
ajk .
bounds for the partial sums, holds for non-positive
Similar reasoning, but with lower
ajk ,
so the theorem is proved.
Theorem 2.12 (The theorem of commutativity for absolutely convergent series).
a1 + a2 + · · · + an + · · ·
be an absolutely convergent series. Also let the sequence
{nk }
14
Let
(2.16)
be an arbitrary rearrangement of the positive integers. Then the series
an1 + an1 + · · · + ank + · · ·
is convergent, and with the same sum as
Proof.
(2.16).
The series (2.16) is absolutely convergent with sum
gence of the series
(2.17)
∞
X
S
and thus we have conver-
|an |.
(2.18)
n=1
Since (2.18) has only non-negative terms so according to theorem 2.10 any rearrangement
of the series converges to the same sum. Therefore the series (2.17) is also absolutely
convergent with the sum
T.
Let
p1 , p2 , ...
denote the positive terms of (2.16) and
q1 , q2 , ...
denote the negative terms. Now let
P =
∞
X
pn
(2.19)
|qn |
(2.20)
n=1
and
Q=
∞
X
n=1
be the sums of two series with non-negative terms. The sum
S
can be written as
S = P − Q,
so that now, any rearrangement of terms in (2.16) induces corresponding rearrangements
of the terms of (2.19) and (2.20), but according to theorem 2.10, these rearrangements
have no eect to the sums of
sum
T.
P
and
Q
and thus it has no eect to the sum rearranged
Thus
T = P − Q = S,
and thus the theorem is proved.
The condition of absolute convergence of the series thus guarantees that any rearrangement of terms is also convergent and the sum is unchanged.
If the series is
not absolutely convergent but still conditionally convergent we have another remarkable
result stated in a theorem formulated by Riemann which will be presented in section
4.1.
15
3 Innite Products
Before moving further, the concept of innite products will be explored.
3.1 Convergence of Innite Products
Suppose that
cn
is any real function of
n,
dened for all positive integral numbers
n.
Then the product
Pn := c1 c2 . . . cn
can be written with the use of the symbol
Π,
Pn =
(3.1)
signifying the product of the factors
n
Y
cn ,
as
cr .
(3.2)
Pn as n goes to innity is nite
converges to the limit P and we write
and non-zero we say that the innite
r=1
If the limit of
product
∞
Y
cn = P.
(3.3)
n=1
When
Pn does not tend to a nite real number we say that the innite product diverges.
cn > 0, we have that
!
n
n
Y
X
ln(Pn ) = ln
cr =
ln cr
(3.4)
For
r=1
r=1
P∞
Pn tends to zero, the series r=1 ln cr tends to −∞. This
Pn tends to zero, the innite product diverges to zero [4].
which tells us that if
we say that, when
is why
Convergence to zero is however possible. An innite product converges to zero if and
only if a
nite
and non-zero number of factors are zero, so that if we were to remove
these factors the product would converge to a non-zero number.
Proposition 3.1.
If the product
∞
Y
cn
n=1
is convergent, then
cn → 1.
Proof.
Since the product is convergent there is only a nite number of factors that are
nth partial
Pn and Pn−1
Pn
zero. Therefore we can remove these factors and consider the
product
where all zero-factors have been removed. Then the limits of
tend to the
same number. Hence
lim
n→∞
Pn
= lim cn = 1.
n→∞
Pn−1
16
It follows from proposition 3.1 that, if the product converges, there exists a natural
number
N
such that
∀n ≥ N,
cn > 0,
(3.5)
We may now write
∞
Y
N
−1
Y
cn =
n=1
cn
n=1
∞
Y
cn .
It is clear that to study the convergence or divergence of
the same properties for the product
(3.6)
n=N
Q∞
n=N cn .
Q∞
n=1 cn it is sucient to study
When it comes to convergence, a nite
number of factors can always be disregarded, which is why for the rest of this text we
shall assume that we have already disregarded all these bad factors so that
all values of
Since
cn
cn > 0
for
n.
tends to
1
for convergent products it is convenient to rewrite the product
in the following manner. Let
an : N → R
an = cn − 1
so that
Pn =
n
Y
(3.7)
(1 + ar ).
(3.8)
r=1
The above assumption that
all
n.
cn > 0
for all values of
n
now becomes that
an > −1
for
Proposition 3.1 may now be equivalently restated as
Proposition 3.2.
If the product
∞
Y
(1 + an )
n=1
is convergent, then
an → 0.
Proof.
Since, according to proposition 3.1,
cn
tends to
1 and since an = cn − 1 the proof
is complete.
3.2 Tests for Convergence
The following theorem is useful for testing products where
Theorem 3.3.
an
always has the same sign.
If
1. an > −1, ∀n ∈ N,
2. an has the same sign for all n,
P
Q∞
then the series ∞
n=1 an and the product
n=1 (1 + an ) converge or diverge together.
17
Proof.
an ≥ 0.
The proof is divided into two cases. First, assume that
P∞
sum of the series
n=1
an
The
nth
partial
is written as
Sn =
n
X
ar .
r=1
we have that
Pn = (1 + a1 )(1 + a2 ) · · · (1 + an ) ≥ 1 +
n
X
n
X
ar >
r=1
ar = Sn .
r=1
Since
1 + x ≤ ex ,
∀x ≥ 0
it follows that
(1 + a1 )(1 + a2 ) · · · (1 + an ) ≤ ea1 ea2 · · · ean ,
so we get that
Sn < Pn ≤ eSn .
Since
Pn
an ≥ 0
for all
n
it follows that
Sn
and
Pn
are both monotonically increasing.
is thus a bounded monotone sequence and must converge to a limit. The proof is
complete.
Now assume that
−1 < an ≤ 0.
Then since
1 + x ≤ ex ,
−1 < x ≤ 0
we have, by the same arguments as before,
0 < Pn ≤ eSn .
Since
Sn ≤ 0
it follows that if
P∞
n=1
an
diverges, then
Pn
tends to zero, that is, the
innite product diverges to zero.
On the other hand, suppose
a natural number
N = N ,
P∞
n=1
an
converges. Then, for every
>0
there exists
such that
−<
∞
X
ar ≤ 0.
r=N
We also know that
(1 + aN )(1 + aN +1 ) · · · (1 + an ) ≥ 1 +
n
X
ar ,
r=N
so that for
n≥N
we have
n
∞
X
X
Pn
= (1 + aN )(1 + aN +1 ) · · · (1 + an ) ≥ 1 +
ar ≥ 1 +
ar > 1 − .
PN −1
r=N
So
Pn
PN −1 has a lower bound and, since
−1 < an ≤ 0,
r=N
it is monotonely decreasing which
means that it has a nite and non-zero limit. This in turn of course means that
a nite non-zero limit which is to say that the innite product
Thus the proof is complete.
18
Q∞
n=1 (1 + an )
Pn
has
converges.
In many cases it is needed to have tests that work regardless of the sign of
an .
The
following theorem proves to be useful for some applications.
Theorem 3.4.
P
P∞
2
If the series ∞
n=1 an converges, then the series
n=1 an and the product
Q∞
n=1 (1 + an ) converge or diverge together.
P
2
Proof. Since the series ∞
n=1 an converges, there exists a natural number N such that
a2n ≤
1
,
4
n≥N
an ≤
1
,
2
n ≥ N.
or
For
−1 < x ≤ 1
we have the Taylor expansion
ln(1 + x) = x −
so for
n≥N
x2
x3
x4
+
−
+ ···
2
3
4
we get
2 3 4
2
a a a an
a3n
a4n
+
−
+ · · · ≤ n + n + n + · · ·
| ln(1 + an ) − an | = −
2
3
4
2
3
4
a2n
2
2 2
a2n 2
≤
1 + |an | + |an | + · · · ≤
1 + |an | + |an | + · · · =
2
3
4
2
∞
2 X
2
2
a
a
1
a
1
r
= n
|an | = n ·
< n·
= a2n .
2 r=0
2 1 − |an |
2 1 − 12
P∞
a2n
P∞
|ln(1 + an ) − an | which
in turn gives the convergence of the series
n=1 {ln(1 + an ) − an }. This is equivalent
to saying that Pn − Sn tends to a nite limit. Thus, if Sn has a nite limit then Pn
must have a nite limit. Conversely, if Sn diverges as n goes to innity, so does Pn .
The convergence of
Example 3.5.
n=1
thus gives the convergence of
P∞
An example of when theorem 3.4 is useful is the innite product
∞ Y
n=2
Here,
n=1
an = (−1)n n1
1
1 + (−1)
n
n
.
(3.9)
so
∞
X
a2n =
n=2
∞
X
1
,
n2
n=2
(3.10)
which is a convergent series. We can therefore use theorem 3.4. The series
∞
X
n=2
an =
∞
X
(−1)n
n=2
1
n
(3.11)
which is convergent by Leibniz criterion (theorem 2.9). We then know, by theorem 3.4,
that the product (3.9) is convergent.
19
Example 3.6.
If we instead consider the product
∞ Y
n=2
we see that
an = (−1)n √1n
1
1 + (−1)n √
n
.
(3.12)
so the series
∞
X
a2n =
n=2
∞
X
1
n
n=2
(3.13)
is divergent. This means that we cannot use theorem 3.4 for this product. Instead we
can observe that the convergence of the product (3.12) is equivalent to the convergence
of the series
∞
X
1
ln 1 + (−1) √
n
n=1
and that for
−1 < an ≤ 1
n
(3.14)
we have
ln(1 + x) = x −
x2
+ O(x3 ).
2
(3.15)
Then we get
1
1
1
1
+O
= (−1)n √ −
ln 1 + (−1)n √
,
n
n 2n
n3/2
(3.16)
where we can see that the second of these three series is divergent while the other two are
convergent. This is enough to know that the sum (3.14) is divergent which is equivalent
to the divergence of the product (3.12).
3.3 Absolute Convergence
Denition 3.7.
The product
∞
Y
(1 + an )
n=1
is said to be absolutely convergent if the product
∞
Y
(1 + |an |)
n=1
is convergent.
We now explore the link between absolute convergence of series and products.
Theorem 3.8.
If the series
∞
X
an
n=1
is absolutely convergent, then the series
∞
X
ln(1 + an )
n=1
is also absolutely convergent.
20
Proof.
n≥N
Since
P∞
n=1
|an |
is convergent, there exists a natural number
we have
|an | ≤
First we assume that
0 ≤ an ≤
N
such that for
1
.
2
1
2 . In similar fashion as in the proof of theorem 3.4 we
have
| ln(1 + an )| = ln(1 + |an |) < 2|an |
Now assume that
− 12 ≤ an < 0.
Then we have, for all
| ln(1 + an )| = − ln(1 − |an |) = ln
So, for all values of
n≥N
1
1 − |an |
∀n ≥ N.
n ≥ N,
= ln 1 +
|an |
1 − |an |
≤
|an |
≤ 2|an |.
1 − |an |
we have
| ln(1 + an )| ≤ 2|an |
so according to the comparison test (theorem 2.4), the series
and thus the series
P∞
n=1
ln(1 + an )
P∞
n=1
| ln(1+an )| converges
is absolutely convergent.
From this we can now deduce another theorem.
Theorem 3.9.
If the product
∞
Y
(1 + |an |)
n=1
is convergent then the product
∞
Y
(1 + an )
n=1
is also convergent.
Proof.
Since
|an |
obviously never changes sign we can use theorem 3.3 which says that
the product
∞
Y
(1 + |an |)
n=1
and the series
∞
X
|an |
n=1
converge or diverge together. Since the product converges, then so does the series. By
theorem 3.8 we see that since
P∞
n=1
the series
an
∞
X
is absolutely convergent we have convergence of
| ln(1 + an )|
n=1
which implies the convergence of
∞
X
ln(1 + an ).
n=1
21
This is equivalent with the convergence of the product
∞
Y
(1 + an ),
n=1
and the proof is complete.
Comparing theorem 3.9 with denition 3.7 we see that the theorem simply states
that every product that is absolutely convergent is also convergent.
3.4 Associativity and Commutativity
The properties of associativity and commutativity of innite products are similar to
those of innite series and the following two theorems are the analogues to the corresponding theorems for series.
Theorem 3.10 (The theorem of associativity for innite products).
Let
P = (1 + a1 )(1 + a2 ) · · · (1 + an ) · · ·
(3.17)
be a convergent innite product, and let
c1 c2 · · · ck · · ·
(3.18)
be a new product obtained by grouping the factors in
changing the order, where
(3.17)
in arbitrary groups without
ck = (1 + ank−1 +1 ) · · · (1 + ank ), ∀k = 1, 2, 3, ...
and where n1 = 1 and {nk } is some arbitrary subsequence of the positive integers such
that nk < nk+1 for all k . Then the new product (3.18) is convergent and
c1 c2 · · · ck · · · = P.
Proof.
Let
Rk =
k
Y
cr
r=1
be the
k th
partial product of (3.18) and let
Pn =
n
Y
(1 + ar )
r=1
be the
nth
partial product of the product
P.
ln Rk =
Then, if the partial sum
k
X
ln cr
r=1
has a nite limit, the product (3.18) converges. Since
k
X
ln ck =
r=1
and since
to
P.
k
X
k
X
ln (1 + anr−1 +1 ) · · · (1 + anr ) =
ln(1 + ank ) = ln Pnk
r=1
lim Pn → P ,
n→∞
r=1
then
lim (ln Pnk ) → ln P
k→∞
22
and it follows that
Rk
also converges
An important note on theorem 3.10 is that, similarly to the corresponding theorem
for series, the converse of the theorem is not true. That is, if we have a product that
is convergent and we expand any number of factors into two or more new factors, and
by doing so we may now end up with a divergent product. Under some circumstances
however, the factors can be expanded into new factors while preserving the convergence
as well as the value of convergence, which is stated by the following theorem.
Theorem 3.11.
Let
P = c1 c2 · · · ck · · ·
be a convergent innite product, and let
(1 + a1 )(1 + a2 ) · · · (1 + an ) · · ·
be a product of new factors made from the original product in such a way that
(1 + ank−1 +1 ) · · · (1 + ank ) = ck , ∀k = 1, 2, 3, ...
and where n1 = 1 and {nk } is some arbitrary subsequence of the positive integers such
that nk < nk+1 for all k . If for each k , anr has the same sign for nk−1 + 1 ≤ nr ≤ nk ,
then the product
∞
Y
(1 + an )
(3.19)
n=1
also converges to the limit P .
Proof.
Each group of factors keeps the same sign then each groups sub-factors are greater
than one or less than one. This means that the series obtained from taking the logarithm
of the product
P
can be written as a series of grouped terms,
[ln(1 + a1 ) + · · · ln(1 + an1 )] + [ln(1 + an1 +1 ) + · · · ln(1 + an2 )] + · · ·
with parentheses around each group's terms. Applying theorem 2.8 we get that the series
obtained by dropping the parentheses is also convergent and their sum is the same. This
series, when the parentheses are dropped, is exactly the series obtained by taking the
logarithm of the product (3.19), and so this product converges, and to
P.
We now explore the property of commutativity.
Theorem 3.12 (The theorem of commutativity for innite products).
∞
Y
(1 + an )
Let
(3.20)
n=1
be an absolutely convergent product. Then the product obtained when arbitrarily rearranging the order of the factors in (3.20) is convergent and with the same sum.
Proof.
By hypothesis, the product
∞
Y
(1 + |an |)
n=1
23
is convergent and thus, by theorem 3.3 the series
∞
X
|an |
n=1
is convergent. By theorem 3.8 the series
∞
X
ln(1 + an )
n=1
is thus also absolutely convergent.
According to theorem 2.12 we may rearrange any
number of terms in this series, and thus the corresponding factors in the product. This
proves the theorem.
3.5 Summary and Examples
We have seen that an absolutely convergent product is still absolutely convergent when
reordering it's factors. When expanding a product into new factors, the conditions of
theorem 3.11 must however always be remembered, in particular the condition that each
new sub-factor in each group should have the same sign.
An example of how such a
problem can occur can be illustrated by studying the innite product representation of
sin x.
the trigonometric function
The derivation of this expression is not included in this
text but can be found for instance in [2, p. 184-186].
Example 3.13.
In this example, we consider the convergence of the innite product
∞ Y
x2
sin x = x
1− 2 2 .
n π
n=1
(3.21)
Also we will try to conrm the factorisation
x
x x x
1+
··· 1 −
1+
···
sin x = x 1 −
π
π
kπ
kπ
and study its convergence.
In this example
an = −
x2
n2 π 2
which converges absolutely, so the product (3.21) is absolutely convergent. Observe that
1−
x2
x x =
1
−
1
+
n2 π 2
nπ
nπ
so that we can write the expanded product
sin x = x
∞ Y
1−
n=1
Written in this way, the two factors
in the
nth
product
Qn =
n Y
r=1
(1 −
1−
x x 1+
.
nπ
nπ
x
nπ ) and
(1 +
x x
1+
rπ
rπ
24
x
nπ ) always appear together
nth
and since the
product of the original expression
Pn =
n Y
1−
r=1
is equal to
Qn
for all
n,
x2
2
r π2
then the convergence is also equal for these two products. Now
we rewrite the product so that we have
f (x) = x
∞
Y
n=1
where
n+1 2
(−1)n x
1 + n+1 π
2
!
x
x x x
1+
··· 1 −
1+
···
=x 1−
π
π
kπ
kπ
n+1
2 , in other words the sequence
denotes the integer part of
Now we have
(3.22)
1, 1, 2, 2, ....
(−1)n x
an = n+1 π
2
which is convergent due to the Leibniz criterion, but not absolutely convergent. Also,
according to theorem 3.10, if the product, obtained by factoring together the factors of
another convergent product, is also convergent, then they converge to the same value.
Thus,
sin x
can be written as an absolutely convergent product (3.21), as well as a
product (3.22) which is not absolutely convergent.
Example 3.14.
Knowing the product for
innite product representation of
cos x,
cos x =
sin x
which is given by
∞ Y
1−
n=1
we now show that
∞
Y
m=1
cos
of equation (3.21) and also using the
4x2
(2n − 1)2 π 2
sin x
x
=
,
2m
x
,
(3.23)
x 6= 0.
(3.24)
Equation (3.23) gives us
∞
Y
x
x2
cos m =
1−
2
2
[2m−1 (2n − 1)] π 2
n=1
so that
∞
Y
,
∞ Y
∞
Y
x
x2
cos m =
1−
2
2
[2m−1 (2n − 1)] π 2
m=1
m=1 n=1
It is clear that for every
0<
so there for any
!
x
x
there exists an
1−
N
!
.
(3.25)
such that
!
x2
< 1, ∀n, m ≥ N,
2
[2m−1 (2n − 1)] π 2
there are only a nite number of factors that does not satisfy this
inequality. Therefore we can assume that we have
0<
1−
−π < x < π ,
so that
!
x2
2
[2m−1 (2n − 1)] π 2
25
< 1, ∀n, m ∈ N.
Consider the logarithm of the left hand side of (3.25),
∞
Y
x
ln
cos m
2
m=1
!
=
∞ X
∞
X
!
x2
ln 1 −
2
[2m−1 (2n − 1)] π 2
m=1 n=1
where
∞ X
∞
X
amn ,
m=1 n=1
!
x2
amn = ln 1 −
=
2
[2m−1 (2n − 1)] π 2
.
Now we form the matrix
a11
a21
a31
a12
a22
a32
a13
a23
a33
···
···
···
.
.
.
.
.
.
.
.
.
···
which explicitly is the matrix
ln 1 −
ln 1 −
ln 1 −
2
2
ln 1 − [3]x2 π2
ln 1 − [5]x2 π2
2
2
ln 1 − [6]x2 π2
ln 1 − [10]x2 π2
2
2
ln 1 − [12]x2 π2
ln 1 − [20]x2 π2
x2
[1]2 π 2 x2
[2]2 π 2 x2
[4]2 π 2
.
.
.
Now we consider
.
.
.
.
.
.
∞
Y
sin x
x2
=
1− 2
x
[r] π 2
r=1
···
···
(3.26)
···
···
!
and the logarithm
ln
sin x
x
=
∞
X
ln 1 −
r=1
= ln 1 −
!
x2
2
[r] π 2
!
x2
2
[1] π 2
=
∞
X
br =
r=1
+ ln 1 −
Now we can see that the sequence
{br }
+ ln 1 −
2
[2] π 2
where
br = ln 1 −
!
x2
x2
2
[r] π 2
x2
2
[3] π 2
!
+ ···
!
.
is some arrangement of the terms of the matrix
(3.26) and that the terms of the matrix are all of negative sign. Since the series
converges then, according to theorem 2.11,
∞
X
amn
m=1
converges for all
n,
the iterated series
∞ X
∞
X
n=1 m=1
26
amn
P∞
r=1 br
converges and that
∞
Y
x
ln
cos m
2
m=1
!
=
∞ X
∞
X
amn =
n=1 m=1
∞
X
br = ln
r=1
sin x
x
,
and then equation (3.24) follows.
We have shown that (3.21) and (3.23) imply (3.24), but similarly (3.23) and (3.24)
also imply (3.21). The preceding proof is designed to show the use of the innite product
representation of
sin x
cos x,
and
however a more direct proof is also possible:
First note that
x
x
x
x
x
cos = 22 sin 2 cos cos 2
2
2
2
2
2
sin x = 2 sin
so that by induction we get, for any
n,
n
x Y
x
cos k ,
2n
2
sin x = 2n sin
k=1
or, for
x 6= 0,
n
Y
cos
k=1
x
x
sin x
sin x
2n
=
=
·
.
2k
2n sin 2xn
x
sin 2xn
Since
lim
n→∞
we get
∞
Y
cos
m=1
x
2n
=1
sin 2xn
x
sin x
=
, x 6= 0,
m
2
x
and the proof is done.
Example 3.15.
Consider the innite product
∞
Y
√
n
a,
a > 0.
n=1
For
a = 1 we have convergence to 1 since all factors are 1.
ln
∞
Y
√
n
∞
X
a=
ln a
a=
n=1
n=1
Then
√
n
ln
Now let
a 6= 1.
Then we have
∞
X
1
ln a.
n
n=1
is simply a real constant, negative or positive depending on if a is smaller or
greater than
1.
Thus we have a divergent series so the product must also diverge.
Example 3.16.
Now consider
∞
Y
√
n2
a,
a > 0.
n=1
Then we have
ln
∞
Y
√
n2
n=1
a=
∞
X
ln
√
n2
n=1
27
a=
∞
X
1
ln a.
n2
n=1
a>0
This series is convergent for all
and since
∞
X
π2
1
=
2
n
6
n=1
we have that
∞
Y
√
n2
π2
a=a
/6
,
a > 0.
n=1
Example 3.17.
Also consider the series
∞
Y
√
n
n.
(3.27)
n=1
In this case, if the product converges, it is the same even if the product begins at
We have
∞
Y
√
n
ln
n=
∞
X
√
n
ln
∞
∞
X
X
1
1
ln n ≥
ln 2.
n
n
n=2
n=2
n=
n=2
n=2
n = 2.
(3.28)
Since the series on the right hand side of (3.28) diverges we also have divergence for the
product (3.27).
Example 3.18.
Now we show that
∞
Y
n
(1 + x2 ) =
n=0
The
nth
product is
Pn =
1
,
1−x
n Y
1 + x2
|x| < 1.
r
(3.29)
r=0
and for
n=1
we have
P1 = (1 + x) 1 + x
2
2
3
=1+x+x +x =
3
X
r
x =
r=0
Assume that
Pn =
2n+1
X−1
2
2X
−1
xr .
r=0
xr
r=0
for some
n ≥ 1.
Pn+1
Then


=
r
2n+1
X−1
x
·2n+1
2n+1
X−1
X−1
n+1
2n+1
r
1+x
=
x +
xr · x2
=
r=0
n+1
2
=
X−1
r=0
r=0
n+1
2
xr +
X−1
n+1
x2
+r
r=0
=
A
X
r=0
xr ,
r=0
where
A = 2n+1 − 1 + 2n+1 = 2 · 2n+1 − 1 = 2n+2 − 1.
28
(3.30)
Thus, by induction and (3.30), we have that
Pn =
2n+1
X−1
xr
r=0
for all
n
and since
Pn
is a subsequence of the partial sum
Sn =
n
X
xr
r=0
and since
∞
X
xr =
n=0
1
,
1−x
we have also shown the equality (3.29).
29
|x| < 1,
4 Further Properties of Innite Series
After having taken the detour around innite products, we now explore some more
properties of addition and multiplication of innite series.
4.1 Riemann's Theorem
Before stating Riemann's theorem we rst state the following lemma of non-commutativity
for conditionally convergent series.
Lemma 4.1.
Let
∞
X
an
(4.1)
bn
(4.2)
n=1
be a conditionally convergent series, let
∞
X
n=1
be the series formed from the positive terms of
let
∞
X
cn
(4.1),
arranged in the same order, and
(4.3)
n=1
be the series formed from the absolute values of the negative terms of
the same order. Then the series (4.2) and (4.3) both diverge.
Proof.
Let
n
X
Pn =
(4.1),
arranged in
ar ,
r=1
Qj =
j
X
br
r=1
and
Rk =
k
X
cr ,
r=1
be the
nth
sum of (4.1), (4.2) and (4.3) respectively, where
terms in the
n
rst terms of (4.1) and
k
j
is the number of positive
is the number of negative terms in the
n
rst
terms of (4.1). Thus we have that
Pn = Qj − Rk
and because
Pn
converges as
n
goes to innity we have that
diverge together. Now let
Pn∗ =
n
X
Qj
and
Rk
converge or
|ar |,
r=1
so that we also have
Pn∗ = Qj + Rk .
30
(4.4)
The series (4.1) is not absolutely convergent so
hand side of (4.4) also diverges as
n
Pn∗
diverges, which means that the right
goes to innity.
Since
Qj
and
Rk
converge or
diverge together this shows that both (4.2) and (4.3) diverge.
We can now state and prove the remarkable theorem of Riemann.
Theorem 4.2 (Riemann's theorem).
Let
∞
X
an
(4.5)
n=1
be a conditionally convergent series. Then the terms of (4.5) can be rearranged to give
a new series that converges to ±∞ or any real number A.
Proof.
According to lemma 4.1 the series
∞
X
bn
(4.6)
n=1
made from the positive terms of (4.5) without changing the order of terms, and
∞
X
cn
(4.7)
n=1
made from the absolute value of the negative terms of (4.5) without changing the order
of terms, both diverge. Thus, starting from any position in the series, we can choose
enough terms to make their sum exceed any real number. Now select enough terms, let
us say
k1
terms, from the series (4.6) so that
b1 + · · · + bk1 > 1.
Then subtract the rst term of (4.7). Next, select enough terms
k2 − k1
so that
bk1 +1 + · · · + bk2 > 2.
and subtract the next term from (4.7).
Repeat this procedure so that for each
n
we
have
bkn +1 + · · · + bkn+1 > n + 1.
We now have the series
(b1 + · · · + bk1 ) − c1 + (bk1 +1 + · · · + bk2 ) − c2 + · · ·
that diverges to innity. Since all terms in each parenthesis have the same sign we can,
according to theorem 2.8, drop the parenthesis without changing the convergence of
the series. The series after having dropped the parenthesis is simply a rearrangement
of (4.5), so we will thus have a rearrangement of the series (4.5) that diverges to
Similarly we can rearrange the series to diverge to
Now we will try to nd a rearrangement to a series that converges to
select enough terms, say
j1
terms, from (4.6) so that
b1 + · · · + bj 1 > A
31
∞.
−∞.
A ∈ R.
First
Note that if
A<0
we can simply select zero terms in this rst step. Next we select
k1
terms from (4.7) to subtract so that
(b1 + · · · + bj1 ) − (c1 + · · · + ck1 ) < A.
We repeat this process indenitely so that all the terms of (4.5) eventually appear in
the new series. Now suppose that each time we add or subtract a group of terms we
never use more terms than are exactly needed to make the sum greater or less than
Thus for each
n≥2
A.
we have
|(b1 + · · · + bj1 ) − (c1 + · · · + ck1 ) + · · · + (bjn−1 +1 + · · · + bjn ) − A| < bjn
and
|(b1 + · · · + bj1 ) − (c1 + · · · + ck1 ) + · · ·
+ (bjn−1 +1 + · · · + bjn ) − (ckn−1 +1 + · · · + ckn ) − A| < ckn
so the partial sum never deviates from
A with more than the last term added in the last
group of terms in the series. Since (4.5) is convergent we must have that
lim bjn = 0
n→∞
and
lim ckn = 0
n→∞
so it follows that the series
(b1 + · · · + bj1 ) − (c1 + · · · + ck1 ) + · · ·
+ (bjn−1 +1 + · · · + bjn ) − (ckn−1 +1 + · · · + ckn ) + · · ·
is convergent, with sum
A
and according to theorem 2.8 it is still convergent to the
same sum after dropping the parenthesis, which makes it a rearrangement of the series
(4.5).
Example 4.3.
Consider the Taylor expansion
ln(1 + x) = x −
and let
x=1
x3
x2
+
− ···
2
3
so that
∞
X
1 1
(−1)n+1
ln(2) = 1 − + − · · · =
,
2 3
n
n=1
which is the
alternating harmonic series.
Pn =
Let
n
X
(−1)r+1
r=1
be its
nth
sum and let
1−
(4.8)
r
1 1 1 1 1
− + − − + ···
2 4 3 6 8
32
(4.9)
be the series obtained by rearranging the terms of (4.8) in such a way that a positive
term is always followed by two negative, without changing the internal order of the
positive and negative terms respectively. Now let
Qn
be the
nth
sum of this new series.
Then we have
Q3n
X
n 1
1
1
1
1
=
=
=
−
−
−
2r − 1 4r − 2 4r
2(2r − 1) 4r
r=1
r=1
n 1
1X
1
P2n
=
−
=
.
2 r=1 (2r − 1) 2r
2
n X
Also we have
1
4n
Q3n−1 = Q3n +
and
Q3n−2 = Q3n +
1
1
+
4n − 2 4n
so we have that
lim Q3n = lim Q3n−1 = lim Q3n−2 =
n→∞
n→∞
n→∞
Thus the series (4.9) is convergent, with sum
ln 2
.
2
ln 2
2 .
We can now develop a formula for a more general rearrangement of the alternating
harmonic series.
Example 4.4.
First we introduce a constant called
the EulerMascheroni constant,
dened as being the limiting dierence between the harmonic series and the natural
logarithm,
n
X
1
γ = lim
n→∞
r=1
r
!
− ln n .
(4.10)
The existence of this limit is proved by considering the sequence
γn =
n
X
1
r=1
r
− ln n.
Since
γn − γn+1 =
n
X
1
− ln n −
r
= − ln 1 −
r=1
n+1
X
r=1
1
n+1
1
+ ln(n + 1) = ln
r
−
γ1 = 1 > ln 2 = ln
γn > ln
n+1
n
−
1
1
1
>
−
= 0,
n+1
n+1 n+1
the sequence is monotone decreasing. Now we note that
Assume that
n+1
n
33
1+1
1
,
.
1
=
n+1
for some
n.
Then we have
γn+1 = γn +
1
n+1
1
+
− ln
=
n+1
n
n+1
1
n+2
> ln 1 +
= ln
.
n+1
n+1
1
+ ln n − ln(n + 1) > ln
n+1
n+1
n
By induction we thus have that
γn > ln
so
γn
n+1
n
> 0, ∀n ∈ N,
(4.11)
is monotone decreasing and is bounded below. Therefore
also see immediately that this limit is between
0 and 1.
γn
has a limit. We can
In fact, the value of the constant
has been found to be approximately
γ = 0, 57721566490...
From (4.10) we can now deduce the following formula
Hn = 1 +
where
n
1 1
1
+ + · · · + = ln n + γ + n
2 3
n
is a function depending on
n
such that
lim n = 0,
n→∞
and
Hn
is the
nth
partial sum of the harmonic series. Now consider the series



1
1
1
1 1


− − − ··· − +
1 + + · · · +
3
2p − 1 2 4
2q 

|
{z
} |
{z
}
p terms
q terms


 1
1
1
1
1
1


+
+
+ ··· +
−
−
− ··· −  + ···
4p − 1 2q + 2 2q + 4
4q 
 2p + 1 2p + 3
|
{z
} |
{z
}
p terms
q terms
(p + q)nth partial sum
1
1 1
1
1
P(p+q)n = 1 + + · · · +
−
+ + ··· +
.
3
2pn − 1
2 4
2qn
|
{z
} |
{z
}
This series has the
n×p terms
n×q terms
Now we observe that
1 1
1
Hn
ln n γ
n
+ + ··· +
=
=
+ +
2 4
2n
2
2
2
2
and that
1+
1
1
1
ln n γ
n
+ ··· +
= H2n − Hn = ln(2n) + γ + 2n −
− −
=
3
2n − 1
2
2
2
2
ln n γ
n
= ln 2 +
+ + 2n − .
2
2
2
34
(4.12)
(p + q)nth
Now we rewrite the
partial sum as
Hqn
Hpn
−
=
= H2pn −
2
2
ln(pn) γ
pn
ln(qn) γ
qn
= ln 2 +
+ + 2pn −
−
+ +
=
2
2
2
2
2
2
r p
pn
qn
= ln 2
+ 2pn −
−
.
q
2
2
P(p+q)n
This implies the formula


r 
p
1
1 1
1
1


− − − ··· − +
ln 2
= 1 + + · · · +
q
3
2p − 1 2 4
2q 

|
{z
} |
{z
}
p terms
q terms


 1
1
1
1
1
1


+
−
+
+ ··· +
−
− ··· −  + ···
4p − 1 2q + 2 2q + 4
4q 
 2p + 1 2p + 3
{z
} |
{z
}
|
p terms
q terms
We can now check the result of example 4.3 by using this formula. In that example,
p=1
and
q = 2.
Then, according to the formula, we get the sum
r !
√
1
1
ln 2
= ln 2 = ln 2
2
2
which is the same result as before. Also notable is the case when we have one positive
p = 1 and q = 4.
r !
1
ln 2
= ln 1 = 0.
4
term followed by four negative ones so that
Example 4.5.
In this case we get
Let
∞
X
an = ∞,
an ≥ 0, ∀n ∈ N,
n=1
Then we state that for any
S∈R
lim an = 0.
n→∞
there exists a sequence
∞
X
{n }
n an = S.
such that
|n | = 1
and
(4.13)
n=1
The proof of this statement follows and is similar to the proof of Riemann's theorem.
Since the series diverges to innity, it is always possible to start from any position in the
series and select enough terms to sum to any number. For simplicity we assume that
S≥0
and select the rst
k1
terms from the series such that
kX
1 −1
ar ≤ S
r=1
35
and
k1
X
ar ≥ S.
r=1
Now we set
1 ≤ n ≤ k1 .
n = 1,
Next we subtract the next
k2 − k1
terms of the series such that
k1
X
r=1
and
k1
X
kX
2 −1
ar −
ar ≥ S
r=k1 +1
k2
X
ar −
r=1
ar ≤ S,
r=k1 +1
and set
n = −1,
k1 + 1 ≤ n ≤ k2 .
Continue in the same fashion indenitely so that for each
|
kn
X
n
we will have
r ar − S| ≤ |akn |
r=1
and since
lim akn = 0
n→∞
we have convergence of the sequence when the terms are grouped. Since the terms in
each group have the same sign we can split the groups up to obtain the individual terms,
which shows that there exists a sequence
sum
Example 4.6.
We now propose that the following three statements are equivalent.
P∞
n=1
|an | < ∞,
(b)
∀{n }
such that
|n | = 1,
(c)
∀{δn }
such that
δn = 1
(a)
{n } such that the series (4.13) converges, with
S.
For all terms such that
P∞
or
an = 0
n=1 n an converges,
δn = 0,
P∞
n=1 δn an converges.
the value of
n
and
δn
do not matter. Therefore we can,
without loss of generality, assume that
an 6= 0, ∀n ∈ N.
To prove the equivalence of the three statements we rst show that both (b) and (c)
individually imply (a). Assume (b). Then we can select a sequence
n =
and
∞
X
n=1
n an =
|an |
an
∞
∞
X
X
|an |
an =
|an |
an
n=1
n=1
36
{n }
such that
converges, which implies (a).
Now assume (c). Then we can select the sequence
δn = 1, ∀n ∈ N.
so that the series
∞
X
δn an =
n=1
an
n=1
{δn }
converges. Now instead select a sequence
an > 0.
∞
X
Then the series
∞
X
where
δn = 0
if
an ≤ 0
and
δn = 1
if
δn an
n=1
is convergent by assumption and it is the series obtained from selecting only the positive
terms of the convergent series
∞
X
an ,
n=1
without changing the order of appearance of terms. Lemma 4.1 says that if
P∞
n=1
an
is conditionally convergent then the series of only its positive terms is divergent. Thus,
P∞
n=1
an
is not conditionally convergent but convergent which by denition means that
it is absolutely convergent, which implies (a).
Finally we will prove that (a) implies both (b) and (c) individually.
Then, if
|n | = 1
for all
n,
∞
X
|n an | =
|an |
n=1
n=1
so the series
∞
X
Assume (a).
P∞
n=1 n an is absolutely convergent and thus convergent, which implies (b).
Also, we have that if
δr ∈ {0, 1}
n
X
then
|δr ar | ≤
r=1
n
X
|ar |, ∀n ∈ N,
r=1
so by the comparison test we get that
Pn
r=1 δr ar is absolutely convergent and thus
convergent which implies (c).
This completes the proof of the equivalence of (a), (b) and (c).
4.2 Toeplitz's Theorem
In section 4.3 we will consider the multiplication between two series.
Since there are
innitely many terms that should be paired with each other it is not automatically
apparent what such multiplication means.
In particular, the order of appearance of
terms in the resulting series is of importance. To be able to formulate some theorems
that describe this, we rst need the theorem of Toeplitz and some corollaries thereof.
Theorem 4.7 (Toeplitz's Theorem).
Let
tnm ,
1≤m≤n
37
be numbers that can be expressed by the triangular matrix
t11
t21
t31
..
.
tn1
..
.
t22
t32
..
.
tn2
..
.
t33
..
.
tn3
..
.
..
.
···
···
tnn
..
.
..
.
such that for each xed m we have
lim tnm = 0.
(4.14)
n→∞
Also let the sum of the absolute values of each term in each row be bounded by a constant
K:
n
X
|tnm | ≤ K, ∀n ∈ N
(4.15)
m=1
Finally, let {an } be any sequence that converges to zero. Then the sequence {Sn } given
by
Sn = tn1 a1 + tn2 a2 + · · · + tnn an
also converges to zero.
Proof.
Since
an → 0,
for each
>0
there exists an
|an | <
N ∈N
such that
, ∀n > N.
2K
n > N we have that
|Sn | = tn1 a1 + · · · + tnN aN + tn(N +1) aN +1 + · · · + tnn an ≤
≤ |tn1 a1 + tn2 a2 + · · · + tnN aN | + tn(N +1) aN +1 + · · · + tnn an ≤
Thus for
≤ |tn1 a1 + tn2 a2 + · · · + tnN aN | + |tn(N +1) ||aN +1 | + · · · + |tnn ||an |
For
m>N
we have
|tnm ||am | < K
= ,
2K
2
which gives us that
|Sn | ≤ |tn1 a1 + tn2 a2 + · · · + tnN aN | + .
2
But (4.14) tells us that for any xed N there exists an M such that
|tn1 a1 + tn2 a2 + · · · + tnN aN | < , ∀n > M.
2
This means that for n > M we have
|Sn | < and since
is arbitrarily small it follows that
lim Sn = 0.
n→∞
38
Corollary 4.8.
Suppose the coecients of one row of the triangular matrix in Toeplitz's
theorem
Tn = tn1 + · · · + tnn
satisfy the condition
lim Tn = 1
n→∞
as well as the conditions (4.14) and
the limit zero, we have that
(4.14)
of Toeplitz's theorem. Now, instead of having
lim an = a.
n→∞
Then
lim Sn = lim (tn1 a1 + tn2 a2 + · · · + tnn an ) = a.
n→∞
Proof.
The sequence
n→∞
{an − a}
converges to zero and since
Tn a = tn1 a + · · · tnn a
we have that
Sn = tn1 a1 + tn1 (a1 − a) + · · · + tnn an + tnn (an − a) =
= Tn a + tn1 (a1 − a) + · · · + tnn (an − a) = Tn a + Rn
where
Rn = tn1 (a1 − a) + · · · + tnn (an − a).
By Toeplitz's theorem,
lim Rn = 0
n→∞
and since
lim Tn a = a
n→∞
we have proved the corollary.
The limit of arithmetic means, or Cesàro means, is then just a special case of corollary
4.8, stated below as a new corollary.
Corollary 4.9.
If
lim an = a
n→∞
then the limit of the arithmetic means satises
a1 + · · · an
lim
= a.
n→∞
n
Proof. Choose
1
1
1
Tn = tn1 + · · · + tnn = + + · · · + .
n}
|n n {z
n terms
Then we have the condition
lim Tn = 1
n→∞
and so
lim Sn = lim (tn1 a1 + tn2 a2 + · · · + tnn an ) =
n→∞
n→∞
39
a1 + a2 + · · · + an
= a.
n
Corollary 4.10.
Suppose that
lim an = lim bn = 0,
n→∞
n→∞
and that
|b1 | + |b2 | + · · · + |bn | ≤ K, ∀n ∈ N,
and let
Sn = a1 b1 + a2 b2 + · · · + an bn .
Then
lim Sn = 0.
n→∞
Proof.
Let each column of the triangular matrix tnm in Toeplitz's theorem be the terms
of the sequence
{bn },
so that we have
b1
b2
b3
b1
b2
b1
.
.
.
.
.
.
.
.
.
..
bn
bn−1
bn−2
···
b1
.
.
.
.
.
.
.
.
.
···
.
.
.
.
..
.
Then all conditions of Toeplitz's theorem are satised and we have
lim Sn = 0.
n→∞
There is now one nal useful corollary connected with Toeplitz's theorem.
Corollary 4.11.
Suppose
lim an = a,
lim bn = b,
n→∞
and let
Sn =
n→∞
a1 bn + a2 bn−1 + · · · + an b1
.
n
Then
lim Sn = ab.
n→∞
Proof.
Let
cn = an − a, dn = bn − b,
so that
lim cn = 0,
n→∞
lim dn = 0,
n→∞
and
aj bk = (aj − a)(bk − b) + aj b + abk − ab = cj dk + aj b + abk − ab, ∀j, k ∈ {1, ..., n}.
40
Also, since
dn
converges, any term of the sequence is less than some number
K,
and so
|dn | |dn−1 |
|d1 |
nK
+
+ ··· +
≤
= K.
n
n
n
n
Now we rewrite
Sn =
Sn
as
dn−1
d1
dn
+ c2
+ · · · + cn
c1
n
n
n
+
a1 + a2 + · · · an
b1 + b2 + · · · bn
b+a
− ab.
n
n
(4.16)
The rst term on the right hand side of (4.16) now satises the conditions of corollary
4.10 and so it goes to zero as
n
goes to innity. The second and third terms of (4.16)
satisfy the conditions of corollary 4.9, so the limit becomes
lim Sn = 0 + ab + ab − ab = ab,
n→∞
which completes the proof.
We can now use corollary 4.8 to prove the
StolzCesàro theorem
which can be seen
as an analogue to l'Hopital's rule, since it states that if the limit of the fraction between
the two sequences rate of change exists and is nite, then it is equal to the limit of the
fraction between the terms of the sequence.
Theorem 4.12 (The StolzCesàro theorem).
Let {xn } and {yn } be two sequences that
satisfy the following:
1. x0 = y0 = 0,
2. yn > yn−1 for all n ∈ N,
3. lim yn = ∞,
n→∞
xn −xn−1
n→∞ yn −yn−1
4. lim
= a.
Then
lim
n→∞
Proof.
We will choose the values of
tnm
4.8. Let
tnm =
xn
= a.
yn
so that they satisfy the conditions of corollary
ym − ym−1
yn
so that we have the triangular matrix
1
y1
y2
y1
y3
.
.
.
y1
yn
.
.
.
y2 −y1
y2
y2 −y1
y3
.
.
.
y2 −y1
yn
.
.
.
y3 −y2
y3
.
.
.
y3 −y2
yn
.
.
.
..
.
···
···
41
yn −yn−1
yn
.
.
.
..
.
Now it becomes more clear that for all
m
lim tnm =
n→∞
and that the sum of the
nth
we have
ym − ym−1
=0
yn
row is
y1
y2 − y1
yn − yn−1
yn
+
+ ··· +
=
=1
yn
yn
yn
yn
so that both conditions of corollary 4.8 are satised.
Now let
an =
xn − xn−1
yn − yn−1
so that
lim an = a,
n→∞
and let
bn =
n
X
tnk ak =
k
n
X
yk − yk−1
k
n
·
yn
X xk − xk−1
xk − xk−1
=
.
yk − yk−1
yn
k
The right hand side is a telescopic sum so we have
xn
xn − x0
=
.
yn
yn
bn =
Corollary 4.8 states that, when the conditions are satised, if
lim an = a
n→∞
then
lim bn = a,
n→∞
which proves the theorem.
4.3 Multiplication of Series
Now we consider the multiplication of two innite series. Consider the two series
∞
X
an
(4.17)
bn .
(4.18)
n=1
and
∞
X
n=1
Writing the multiplication of these two series
∞
X
an ×
n=1
∞
X
bn
it is not completely obvious how this should be interpreted. If we consider the
nth
(4.19)
n=1
mth
and
partial sums of the two above series
Am =
m
X
r=1
42
ar
(4.20)
and
n
X
Bn =
br ,
(4.21)
r=1
and the product of these two sums can be written as
Cmn = Am × Bn =
m
X
ar ×
r=1
n
X
r=1
br =
mn
X
ajr bkr
(4.22)
r=1
{jr } is an arbitrarily arranged sequence of the natural numbers
1, ..., m each occurring n times, and {kr } is an arbitrarily arranged sequence of the
natural numbers 1, ..., n each occurring m times. In words, we now have a new nite
sum with m × n terms, each term a multiplication between one of the terms from (4.20)
where the sequence
and one of the terms from (4.21) and where these terms appear in an arbitrary order.
Theorem 4.13 (Cauchy's Theorem).
Let
∞
X
A=
an
(4.23)
bn ,
(4.24)
n=1
and
∞
X
B=
n=1
be two absolutely convergent series. Then the series formed from taking the product of
each term of (4.23) with each term from (4.24), in any order is absolutely convergent
with sum AB .
Proof.
An arbitrary arrangement of the resulting series can be written as
∞
X
ajn bkn = aj1 bk1 + aj2 bk2 + aj3 bk3 + ...
(4.25)
n=1
Now let
n
X
A∗n =
|ar |
r=1
and
Bn∗ =
n
X
|br |
r=1
be the partial sums of absolute values of the terms of each series.
limit as
n
They both have a
goes to innity since the series are absolutely convergent, so now let
A∗ = lim A∗n
n→∞
and
B ∗ = lim Bn∗ .
n→∞
Then
A∗n ≤ A∗ , Bn∗ ≤ B ∗ , ∀n ∈ N.
43
Now consider the partial sum
n
X
|ajr bkr |
r=1
and let
N
be the largest of the integers
n
X
j1 , k1 , ..., jn , kn .
Then we have
|ajr bkr | ≤ A∗n Bn∗ ≤ A∗ B ∗
r=1
for all
n,
which shows that the series (4.25) converges absolutely. Thus any rearrange-
ment (see theorem 2.12) and regrouping (see theorem 2.5) of terms of (4.25) also converges and to the same number. Rearrange (4.25) so that
∞
X
ajn bkn = a1 b1 + (a1 b2 + a2 b2 + a2 b1 ) + · · ·
n=1
in such a way so the
nth
partial sum of the series can be written as
An B n =
n
X
an ×
r=1
The sequence
An B n
clearly has the limit
AB
n
X
bn .
r=1
and the theorem is proved.
When ordering the terms of the resulting series in a diagonal fashion, meaning that
if we think of the terms of the series as an
n×m
matrix and we take each term of the
resulting series to be the sum of each diagonal of this matrix, so that
AB = a1 b1 + (a1 b2 + a2 b1 ) + (a1 b3 + a2 b2 + a3 b1 ) + · · · =
∞ X
n
X
ar bn−r+1
(4.26)
n=1 r=1
it is exactly in the form that Cauchy rst represented the product of two innite series.
When given in this way it is sometimes therefore referred to as the product of the series
A
and
B
in
Cauchy's form .
A very practical application of this form can be seen when
multiplying two power series with each other. Then all the sub-terms of each parenthesis
will have the same power of
x so the resulting series will automatically be in the form of
a power series. This is sometimes referred to as term-by-term multiplication of power
series by Cauchy's rule. Note that the resulting power series is true only for all
x where
both of the original series are uniformly convergent.
Example 4.14.
In this example, we show that
∞
X
1
=
nxn−1 = 1 + 2x + 3x2 + · · ·
(1 − x)2
n=1
|x| < 1
(4.27)
with the help of multiplication of power series by Cauchy's rule. Note rst that
∞
X
1
=
xn−1 = 1 + x + x2 + · · ·
(1 − x) n=1
|x| < 1,
and that the coecients of this power series are all ones. This means that the product
between the series and itself is a power series that has coecient that are the number
of terms in each parenthesis of the Cauchy form (4.26). This means it is exactly (4.27),
and the equality is shown.
44
Example 4.15.
A more general result that contains the previous example is that if
An =
n
X
ar
r=0
is the
(n + 1)th
partial sum of the series
∞
X
an ,
(4.28)
n=0
then
∞
1 X
an xn .
An x =
1 − x n=0
n=0
∞
X
n
This can be seen by observing that, the terms if the parenthesis in (4.26) are the
rst terms of (4.28) multiplied with ones. This means that they are exactly
Example 4.16.
n+1
An .
One way to show the validity of the addition formula for the expo-
nential function
ex ey = ex+y ,
is by using Cauchy's form. Take the series
ex =
∞
X
1 n
x ,
n!
n=0
and multiply the right hand side by itself, substituting
we have the series
1 + (x + y) +
x for y
1
1 2
x + xy + y 2
2
2
in one of the series. Then
+ ···
(4.29)
where the terms are now ordered so that the sum of the exponents of
term equals the number of terms in each parenthesis (minus one).
x
and
y
in each
This is analogous
to multiplication of power series by Cauchy's rule, but for power series of two dierent
variables. Now we see that the
n
X
r=0
=
(n + 1)th
n
n 1
n!
1 X
1 X n r n−r
xr y n−r =
xr y n−r =
x y
.
r!(n − r)!
n! r=0 r!(n − r)!
n! r=0 r
Now we recall that
n X
n
r=0
so that the
term of this series can be written as
(n + 1)th
r
xr y n−r = (x + y)n
term of (4.29) can be written as
1
(x + y)n
n!
so that we have
∞
X
1 n
x
n!
n=1
!
∞
X
1 n
y
n!
n=1
!
45
=
∞
X
n=1
(x + y)n = ex+y .
We can now state two theorems that generalize the theory of multiplication between
two series. Absolute convergence for both series involved guarantees us that the product
converges regardless of the order of terms, which should come as no surprise since the
order of terms can completely alter the convergence and sum of conditionally convergent
series. When we specify in what order the new terms are to appear and how they are
to be grouped, however, the product of two series can make sense even when they are
not absolutely convergent, particularly in the sense of multiplication in Cauchy's form.
This is expressed by the next theorem by Merten's.
Theorem 4.17 (Merten's theorem).
Let
∞
X
A=
an
n=1
and
∞
X
B=
bn
n=1
be two convergent series. Suppose that at least one of the series is absolutely convergent.
Then the product in Cauchy's form
a1 b1 + (a1 b2 + a2 b1 ) + (a1 b3 + a2 b2 + a3 b1 ) + · · ·
is convergent and with sum AB .
Proof.
Assume that
∞
X
bn
(4.30)
n=1
is absolutely convergent and let
Rn =
n
X
|br |
r=1
be the
nth
sum of the absolute values of the terms in
B.
Then we have that
Rn ≤ K, ∀n ∈ N.
Also let
An =
n
X
an , Bn =
r=1
be the
nth partial sums
nth term of the
The
of
A
and
B
n
X
(4.31)
bn ,
r=1
respectively.
series
a1 b1 + (a1 b2 + a2 b1 ) + (a1 b3 + a2 b2 + a3 b1 ) + · · ·
can be written as
Sn =
n
X
ar bn−r+1
r=1
so that the series (4.32) can be written as
S1 + S2 + · · · + Sn + · · ·
46
(4.32)
with the
n'th
partial sum
Tn = S1 + S2 + · · · + Sn .
bk
Ordering the terms of (4.33) by the
(4.33)
terms, we see that
Tn = b1 An + b2 An−1 + · · · + bn A1 .
Since
An
converges to
A,
the
nth
partial sum diers from
A
by some number
αn
which
has the property
lim αn = 0.
n→∞
This means that we can rewrite
Tn
as
Tn = ABn − γn
where
γn = b1 αn + b2 αn−1 + · · · + bn α1 .
Since
bn → 0, αn → 0
and because of equation (4.31), we can use corollary 4.10 and get
that
lim γn = 0,
n→∞
which means that
lim Tn = AB.
n→∞
This concludes the proof of the theorem.
Merten's theorem shows that when multiplying two series in Cauchy's form, only one
of the series have to be absolutely convergent. This raises the question if it is possible
to have convergence when multiplying two conditionally convergent series in Cauchy's
form. The answer is yes, and the next theorem by Abel tells us what the sum of this
product is.
Theorem 4.18 (Abel's theorem).
Let
∞
X
A=
an
n=1
and
B=
∞
X
bn
n=1
be two convergent series. Suppose their product in Cauchy's form
C=
∞
X
n=1
cn =
∞ X
n
X
n=1 r=1
is convergent. Then
C = AB.
47
ar bn−r+1
Proof.
Let
An =
n
X
ar , Bn =
r=1
be the
nth
n
X
br ,
r=1
partial sums of their respective series and let
Sn = C1 + C2 + · · · + Cn .
Then we can rewrite
Sn
as
Sn = A1 Bn + A2 Bn−1 + · · · + An B1
and so by corollary 4.9 we have
C = lim
n→∞
C1 + C2 + · · · + Cn
.
n
By corollary 4.11 we have that
A1 Bn + A2 Bn−1 + · · · + An B1
.
n→∞
n
AB = lim
But we also have that
C1 + C2 + · · · + Cn = A1 Bn + A2 Bn−1 + · · · + An B1
and so
lim
n→∞
A1 Bn + A2 Bn−1 + · · · + An B1
C1 + C2 + · · · + Cn
= lim
n→∞
n
n
which also means that
C = AB.
48
(4.34)
5 Miscellaneous Examples and Problems
In this section some dierent problems regarding series are stated and solved.
Problem 5.1.
Does there exist a sequence
∞
X
an
{an }, an 6= 0,
and
n=1
converge? Can
Solution.
{an }
such that the series
∞
X
1
2a
n
n
n=1
with this property be positive?
Let
Then the series
an =
(−1)n
.
n
an =
∞
X
(−1)n
n
n=1
∞
X
n=1
converges, by Leibniz criterion.
Also we have that
1
(−1)n
1
=
= an ,
=
n
n2 an
n
n2 (−1)
n
which means that there exists an alternating series with the properties asked for. Now
assume that
an
is positive and that the series
∞
X
an
and
n=1
∞
X
1
n2 an
n=1
are both convergent. Thus the series
∞ X
an +
n=1
is also convergent. The
nth
1
n2 an
term of this series is
an +
1
n2 an
=
n2 a2n + 1
2nan
2
≥ 2
=
n2 an
n an
n
where we have used the inequality
a2 + b2 ≥ 2ab.
an
positive and the series
P∞
n=1
P∞
1
is absolutely convergent, so an + 2
n=1 an
n an is also
an + n21an is absolutely convergent for all n so we can
is positive and thus the series
use the direct comparison test of theorem 2.4. This shows that the series
∞ X
an +
n=1
1
2
n an
diverges which means that both of the series can not be absolutely convergent. Thus
the assumption of positive
an
leads to a contradiction and we can conclude that for
both series to converge, the term
an
can not be positive for all
conditionally convergent.
49
n
but must indeed be
Problem 5.2.
Let
an > 0
and
P∞
n=1
an < ∞.
∞
X
1
1− n
an
Prove that
< ∞.
n=1
Solution.
We will compare the series to another convergent series. Let
0<q<1
be a constant real number. Then the series
∞
X
qn
n=1
converges. Now let
{i1 , i2 , ...}, in < in+1
be the subsequence of the sequence of natural numbers such that
ain < q in , ∀n ∈ N.
Then
∞
X
1− i1n
<
ain
∞
X
q in −1
n=1
n=1
where the right hand side of the inequality is a convergent series. Now let
{j1 , j2 , ...}, jn < jn+1
be the subsequence of the sequence of natural numbers such that
ajn ≥ q jn , ∀n ∈ N.
Then
and since the
∞
∞
∞
X
X
ajn
ajn
1X
p
aj ,
≤
=
√
jn
jn a
q n=1 n
jn
q jn
n=1
n=1
n=1
P∞
P∞
series
n=1 an converges and an > 0 then also
n=1 ajn
∞
X
1− j1n
ajn
=
converges. In
conclusion, all the terms in the series
∞
X
1
1− n
an
n=1
are smaller than the terms of a convergent series.
Thus the series must converge, by
direct comparison.
Problem 5.3.
Let
an > 0
and
P∞
n=1
an < ∞.
∞
X
an
ln
ln(1
+ n)
n=1
50
Prove that
1
an
< ∞.
Solution.
The solution is similar to problem 5.2. Consider any
n
such that
1
.
n2
an >
Then we have that
an
ln
ln(1 + n)
n
Now consider all
1
an
an
an
ln n2 =
2 ln (n) < 2an .
ln(1 + n)
ln(1 + n)
<
such that
1
.
n2
an ≤
We will also use that
1
= 0, ∀ > 0,
x
number N such that for n ≥ N
1
1
/4
≤ 1.
an ln
an
lim x ln
x→0
so that there exists a natural
Then, for
n ≥ N,
we have
we have
an
ln
ln(1 + n)
1
an
3
/
1
an 4
· an/4 ln
=
ln(1 + n)
1
an
≤
3
an/4
≤
1
n2
3 /4
=
1
n
3/
2
which is convergent. Thus the series
∞
X
an
ln
ln(1
+ n)
n=1
1
an
converges.
Problem 5.4.
Let
an ≥ 0
and
∞
X
an = ∞
n=1
and study the convergence of the series
∞
X
an
.
1
+
an
n=1
Solution.
Let
bn =
so that
bn ≥ 0 .
Assume that
exists some subsequence
ank
an
(5.1)
an
,
1 + an
does not go to zero as
n
goes to innity. Then there
such that
lim ank = λ, λ 6= 0.
k→∞
Assume that
λ = ∞.
Then
lim bnk = 1,
k→∞
so in this case (5.1) can not be convergent. Now assume that
lim bnk =
k→∞
λ
6= 0,
1+λ
51
λ < ∞,
then
so also in this case (5.1) can not be convergent. Finally we assume that
lim an = 0.
n→∞
This implies that
lim bn = 0,
n→∞
so that we can use the limit comparison test which states that for two series
∞
X
an
and
n=1
with
an ≥ 0
and
bn ≥ 0
for all
n,
∞
X
bn
n=1
if the limit
lim
n→∞
an
bn
exists and is nite then the two series converge or diverge together. We have that
lim
n→∞
an
= 1 + an = 1.
bn
so since the rst series diverges, then so does the other.
In conclusion, under these
conditions, for all possible cases the series (5.1) diverges.
Problem 5.5.
Let
an ≥ 0
and
∞
X
an = ∞
n=1
and study the convergence of the series
∞
X
an
.
1
+
a2n
n=1
Solution.
This series may diverge or converge and examples of both will be given. For
instance, if
1
an = √ .
n
Then we have the series
√1
n
∞
X
n=1
1+
√1
n
√
√
√
√
∞
∞
∞
X
n
1
2 X
n
1
2 X1
= +
+
> +
+
2 =
n+1
2
3
n+1
2
3
n
n=1
n=3
n=3
so it diverges. On the other hand, if
an = n2 ,
then we have
∞
X
∞
X
n2
1
<
4
2
1
+
n
n
n=1
n=1
which converges.
52
Problem 5.6.
Let
an ≥ 0
and
∞
X
an = ∞
n=1
and study the convergence of the series
∞
X
an
.
1
+
nan
n=1
Solution.
It may converge or diverge. If
an =
we have the series
∞
X
1
n
n=1
1 + n n1
∞
X
1
2n
n=1
=
k∈N

1
an =
0
which diverges. On the other hand, if
then
1
n
and
if
n = 2k
if
n 6= 2k
∞
∞
X
X 1
an
=
1 + nan
1 + 2k
n=1
k=1
which converges.
Problem 5.7.
Let
an ≥ 0
and
∞
X
an = ∞
n=1
and study the convergence of the series
∞
X
an
.
1 + n2 an
n=1
Solution.
∞
X
∞
∞
X
X
an
an
1
,
<
=
2a
2a
2
1
+
n
n
n
n
n
n=1
n=1
n=1
so the series converges.
Problem 5.8.
Let
a1 = 1
and
an+1 =
and prove that the series
an
, ∀n ∈ N,
1 + nan
∞
X
an
n=1
converges.
53
Solution.
We consider the reciprocal
λn+1 =
1
1 + nan
1
=
=
+n
an+1
an
an
and see that
λn+1 − λn = n
which means that we have
n
X
(λk+1 − λk ) =
k=1
n
X
k.
k=1
The left hand side is a telescopic sum so we get the equality
λn+1 − λ1 =
Since
λ1 = 1
n(n + 1)
.
2
we get that
λn+1 =
and going back to
an
we thus have
an+1 =
so that
an =
We also have that for
n(n + 1) + 2
,
2
n≥2
2
,
n(n + 1) + 2
2
, ∀n ∈ N.
n(n − 1) + 2
1
2
≤ ,
n−1
n
which gives us the inequality
an =
and so
2
4
4
≤ 2
< 2 , ∀n ≥ 2,
n(n − 1) + 2
n +2
n
∞
X
an <
n=2
which proves that the series converges.
54
∞
X
4
n2
n=2
References
[1] R.G. Bartle and D.R. Sherbert.
[2] T.J.I'A. Bromwich.
Introduction to Real Analysis.
Wiley, 2000.
An Introduction to the Theory of Innite Series.
AMS Chelsea
Publishing Series. American Mathematical Society, 2005.
[3] G.M. Fikhtengol'ts.
Innite Series: Ramications.
Pocket mathematical library.
Gordon and Breach, 1970.
[4] J.M. Hyslop.
Innite Series.
Dover Books on Mathematics Series. Dover Publica-
tions, 2006.
An Episodic History of Mathematics: Mathematical Culture Through
Problem Solving. MAA Textbooks. Mathematical Association of America, 2010.
[5] S.G. Krantz.
[6] A.I. Markushevich.
Innite Series.
Survey of recent East European mathematical
literature. Heath, 1967.
55