Some Properties of Infinite Series

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Faculty of Technology and Science

Mattias Flygare

Some Properties of Infinite Series

Några egenskaper hos oändliga serier

Mathematics

Degree Project 15ECTS, Bachelor Level

Date: 2012-06-11

Supervisor: Viktor Kolyada Examiner: Håkan Granath

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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.

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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 nth partial sum of some function an, dened for all natural numbers n,

Sn=

n

X

r=1

ar, (2.1)

and is commonly written as

∞

X

n=1

a_n= lim

n→∞S_n. (2.2)

If the partial sum Sn has a nite limit S then we say that the series P^∞n=1an converges to S. If the limit 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 es- tablish some condition for convergence of the series. The following theorem states an obvious but important condition.

Theorem 2.1. If the series

∞

X

n=1

an

converges then

n→∞lim an = 0.

Proof. Since the series converges, the partial sums Sn−1and Snmust approach the same limit as n goes to innity. Thus

n→∞lim(S_n− Sn−1) = 0, but

n→∞lim(Sn− Sn−1) = lim

n→∞an, which proves the theorem.

We say that an → 0is a necessary condition for convergence of the series P^∞n=1an.

Denition 2.2. A series _∞

X

n=1

an

is said to be absolutely convergent if the series

∞

X

n=1

|an| converges.

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Denition 2.3. A series _∞ X

n=1

an

is said to be conditionally convergent if the series

∞

X

n=1

|an| diverges but the series

∞

X

n=1

an

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

n=1

bn

is absolutely convergent and there exists a natural number N such that

|an| ≤ |bn|, ∀n ≥ N,

then the series _∞

X

n=1

a_n converges absolutely.

Proof. Let

Sn =

n

X

r=1

|ar|,

Tn=

n

X

r=1

|br|.

Assume that P^∞n=1|bn| converges and |an| ≤ |bn| for n ≥ N. By this assumption the limit

T = lim

n→∞T_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 Snis a bounded monotone sequence and thus it must have a limit which

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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). Let

∞

X

n=1

an (2.3)

be a convergent series and let the sequence {nk} be an arbitrary, strictly increasing, subsequence of the positive integers. Also let

b_k= a_n_k−1₊₁+ a_n_k−1₊₂+ · · · + a_n_k, ∀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

k=2

b_k (2.4)

is convergent, with the same sum as (2.3).

Proof. Let

Sn=

n

X

r=1

ar

be the nth partial sum of (2.3) and let

Tk=

k

X

r=2

br

be the kth partial sum of (2.4). Then we have that

T_k=

k

X

r=2

b_r= a₁+ a₂+ · · · + a_n_k= S_n_k

so it is clear that the sequence {Tk}is just a subsequence of {Sn}. Every subsequence of a convergent sequence with limit S is also convergent, and also has the limit S. The theorem is proved.

Example 2.6. We can now show that if

P = 1 − 1 2^s+ 1

3^s − 1

4^s+ · · · , (2.5)

and

Q = 1 + 1 2^s+ 1

3^s + 1

4^s + · · · , (2.6)

then

P =

1 − 1

2^s−1

Q (2.7)

for s > 1.

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Let s > 1, so that both (2.5) and (2.6) are convergent. Let

P_n=

n

X

r=1

(−1)ⁿ⁺¹1 r^s and

Q_n=

n

X

r=1

1 r^s be the nth partial sum of P and Q respectively, so that

lim

n→∞P_n = P and

n→∞lim Q_n= Q.

By theorem 2.5 we may now group terms without changing the order of appearance without aecting the convergence or sum. Then

P =

1 − 1

2^s

+ 1

3^s − 1 4^s

+ · · · , and

Q =

1 + 1

2^s

+ 1

3^s + 1 4^s

+ · · · . so that we can rewrite the (2n)th partial sums of P and Q as

P_2n=

n

X

r=1

1

(2r − 1)^s− 1 (2r)^s

and

Q2n=

n

X

r=1

1

(2r − 1)^s+ 1 (2r)^s

. For any n we have that

Q2n− P2n=

n

X

r=1

1

(2r − 1)^s+ 1 (2r)^s

−

n

X

r=1

1

(2r − 1)^s − 1 (2r)^s

=

n

X

r=1

1

(2r − 1)^s+ 1

(2r)^s − 1

(2r − 1)^s + 1 (2r)^s

=

n

X

r=1

2 (2r)^s = 2

2^s

n

X

r=1

1 r^s = 1

2^s−1Qn

or equivalently

P2n= Q2n− 1 2^s−1Qn. Letting n go to innity, we get the original statement,

P =

1 − 1

2^s−1

Q.

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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 convergent series. There are however conditions under which the reverse of theorem 2.5 is true, expressed in the next theorem.

Theorem 2.8. Suppose

∞

X

k=2

(an_k−1+1+ an_k−1+2+ · · · + an_k) = (a1+ · · · + an₁) + (an₁+1+ · · · + an₂) + · · · (2.8) 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

n=1

a_n= a₁+ a₂+ · · · (2.9)

obtained from dropping the parentheses in (2.8) is also convergent and has the same sum as the original series.

Proof. Let

Tk =

k

X

r=2

(an_k−1+1+ an_k−1+2+ · · · + an_k) be the kth partial sum of (2.8) and let

S_n=

n

X

r=1

a_r

be the nth partial sum of (2.9). Then ar is always the same sign when nk−1≤ r ≤ nk, so that for such an r we have

Sn_k−1≤ Sr≤ Sn_k or Sn_k−1≥ Sr≥ Sn_k. But

Tk−1= Sn_k−1 and Tk = Sn_k

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so we have

Tk−1≤ Sr≤ Tk or Tk−1≥ Sr≥ Tk. (2.10) Since the series (2.8) converges we know that lim

k→∞Tk = T but by (2.10), then we also have that lim

r→∞Sr= T, so we see that (2.9) converges and has the same sum as (2.8).

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 {a_n}

be a decreasing sequence of strictly positive numbers and let

n→∞lim an = 0.

Then the alternating series

∞

X

n=1

(−1)ⁿ⁺¹an

is convergent.

Proof. Let

b_r= a_2r−1− a2r

so that

S2n=

2n

X

r=1

(−1)^r+1ar=

n

X

r=1

br. 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

r=1

(−1)^r+1ar= a1+

n−1

X

r=1

cr− a2n. Since cr is negative, we have that

S2n≤ a1, ∀n ∈ N.

Thus the subsequence S2n is a bounded monotone sequence so it converges to some number S ∈ R. This means that for each > 0 there exists a natural number N1 such that

|S2n− S| ≤ 1 2

Also, since an is strictly decreasing, for each > 0 there exists a natural number N2

such that

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Let

N = max{N1, N2} so that for n ≥ N we have

|S2n+1− S| = |S2n+ a2n+1− S| ≤ |S2n− S| + |a2n+1| ≤ 1 2 +1

2 = .

Therefore, both partial sums of an odd number of terms and partial sums of an even number of terms are within of S for suciently large n. Since is arbitrary, Sn must therefore converge which means that the series P^∞_n=1(−1)ⁿ⁺¹a_n converges.

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

a_n ≥ 0, n = 1, 2, ....

Also let the sequence

{nk}

be an arbitrary rearrangement of the positive integers. Then the series

a_n₁+ a_n₂+ · · · + a_n_k+ · · · (2.12) is convergent, and with the same sum as (2.11).

Proof. For any given positive integer k, let N be the largest of the integers n1, n₂, ..., n_k. Then the nth partial sum

Sn=

n

X

r=1

ar

and the kth partial sum

Tk =

k

X

r=1

an_r

satisfy the inequality

Tk≤ SN.

Since Sn is increasing and converges to a limit, say S, we have Tk≤ S.

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The terms of (2.12) are also all non-negative so Tk 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 a12 a13 · · · a₂₁ a₂₂ a₂₃ · · · a31 a32 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

r=1

b_r converges if and only if:

1. for any j the series

∞

X

k=1

ajk

converges,

2. the iterated series _∞

X

j=1

∞

X

k=1

ajk

converges.

Moreover, in this case

∞

X

r=1

br=

∞

X

j=1

∞

X

k=1

ajk.

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Proof. First assume that ajkare non-negative. Assume P^∞r=1br converges with sum B.

Then for all J and K there exists an R such that

J

X

j=1 K

X

k=1

a_jk≤

R

X

r=1

b_r≤ B, and thus it follows that for any j,

∞

X

k=1

ajk

converges. Making K → ∞ we get

J

X

j=1

∞

X

k=1

ajk≤ B

and thus _∞

X

j=1

∞

X

k=1

ajk

converges, and its sum A satises

A ≤ B. (2.14)

Now assume that P^∞k=1ajk converges for all j and that P^∞j=1

P∞

k=1ajk converges.

Then, for each R we can always nd a natural number N such that all the terms of PR

r=1br is contained in the partial sum P^Nj=1

PN

k=1ajk. Thus, if

∞

X

j=1

∞

X

k=1

ajk= A

then R

X

r=1

b_r≤ A, ∀R ∈ N,

and so _∞

X

r=1

br

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. Similar reasoning, but with lower bounds for the partial sums, holds for non-positive ajk, so the theorem is proved.

Theorem 2.12 (The theorem of commutativity for absolutely convergent series). Let a₁+ a₂+ · · · + a_n+ · · · (2.16) be an absolutely convergent series. Also let the sequence

{nk}

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be an arbitrary rearrangement of the positive integers. Then the series

an₁+ an₁+ · · · + an_k+ · · · (2.17) is convergent, and with the same sum as (2.16).

Proof. The series (2.16) is absolutely convergent with sum S and thus we have convergence of the series

∞

X

n=1

|an|. (2.18)

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

n=1

p_n (2.19)

and

Q =

∞

X

n=1

|qn| (2.20)

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 P and Q and thus it has no eect to the sum rearranged sum T . 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.

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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 := c1c2. . . cn (3.1)

can be written with the use of the symbol Π, signifying the product of the factors cn, as

Pn=

n

Y

r=1

cr. (3.2)

If the limit of Pn as n goes to innity is nite and non-zero we say that the innite product converges to the limit P and we write

∞

Y

n=1

cn = P. (3.3)

When Pn does not tend to a nite real number we say that the innite product diverges.

For cn> 0, we have that

ln(Pn) = ln

n

Y

r=1

cr

!

=

n

X

r=1

ln cr (3.4)

which tells us that if Pn tends to zero, the series P^∞_r=1ln cr tends to −∞. This is why we say that, when Pn tends to zero, the innite product diverges to zero [4].

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

n=1

cn

is convergent, then

cn→ 1.

Proof. Since the product is convergent there is only a nite number of factors that are zero. Therefore we can remove these factors and consider the nth partial product Pn

where all zero-factors have been removed. Then the limits of Pn and Pn−1tend to the same number. Hence

n→∞lim P_n Pn−1

= lim

n→∞c_n= 1.

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It follows from proposition 3.1 that, if the product converges, there exists a natural number N such that

cn> 0, ∀n ≥ N, (3.5)

We may now write

∞

Y

n=1

cn =

N −1

Y

n=1

cn

∞

Y

n=N

cn. (3.6)

It is clear that to study the convergence or divergence of Q^∞n=1cnit is sucient to study the same properties for the product Q^∞_n=Nc_n. 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 cn> 0for all values of n.

Since cn tends to 1 for convergent products it is convenient to rewrite the product in the following manner. Let

a_n: N → R

a_n= c_n− 1 (3.7)

so that

Pn=

n

Y

r=1

(1 + ar). (3.8)

The above assumption that cn> 0for all values of n now becomes that an > −1for all n. Proposition 3.1 may now be equivalently restated as

Proposition 3.2. If the product

∞

Y

n=1

(1 + an) is convergent, then

an→ 0.

Proof. Since, according to proposition 3.1, cntends to 1 and since an = c_n− 1the proof is complete.

3.2 Tests for Convergence

The following theorem is useful for testing products where analways has the same sign.

Theorem 3.3. If 1. an > −1, ∀n ∈ N,

2. an has the same sign for all n,

then the series P^∞n=1an and the product Q^∞n=1(1 + an)converge or diverge together.

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Proof. The proof is divided into two cases. First, assume that an≥ 0. The nth partial sum of the series P^∞_n=1an is written as

Sn=

n

X

r=1

ar. we have that

Pn = (1 + a1)(1 + a2) · · · (1 + an) ≥ 1 +

n

X

r=1

ar>

n

X

r=1

ar= Sn. Since

1 + x ≤ e^x, ∀x ≥ 0 it follows that

(1 + a1)(1 + a2) · · · (1 + an) ≤ eâ¹eâ²· · · eâⁿ, so we get that

S_n< P_n≤ e^Sⁿ.

Since an ≥ 0 for all n it follows that Sn and Pn are both monotonically increasing.

Pn 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 ≤ e^x, −1 < x ≤ 0 we have, by the same arguments as before,

0 < P_n≤ e^Sⁿ.

Since Sn≤ 0it follows that if P^∞_n=1a_n diverges, then Pn tends to zero, that is, the innite product diverges to zero.

On the other hand, suppose P^∞n=1a_n converges. Then, for every > 0 there exists a natural number N = N, such that

− <

∞

X

r=N

a_r≤ 0.

We also know that

(1 + a_N)(1 + a_{N +1}) · · · (1 + a_n) ≥ 1 +

n

X

r=N

a_r,

so that for n ≥ N we have P_n

PN −1

= (1 + a_N)(1 + a_{N +1}) · · · (1 + a_n) ≥ 1 +

n

X

r=N

a_r≥ 1 +

∞

X

r=N

a_r> 1 − .

So _P^P_{N −1}ⁿ has a lower bound and, since −1 < an ≤ 0, it is monotonely decreasing which means that it has a nite and non-zero limit. This in turn of course means that Pn has a nite non-zero limit which is to say that the innite product Q^∞_n=1(1 + an)converges.

Thus the proof is complete.

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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. If the series P^∞n=1a²_nconverges, then the series P^∞n=1anand the product Q∞

n=1(1 + an)converge or diverge together.

Proof. Since the series P^∞n=1a²_n converges, there exists a natural number N such that

a²_n ≤1

4, n ≥ N or

a_n≤1

2, n ≥ N.

For −1 < x ≤ 1 we have the Taylor expansion

ln(1 + x) = x −x² 2 +x³

3 −x⁴ 4 + · · · so for n ≥ N we get

| ln(1 + an) − a_n| =

−a²_n 2 +a³_n

3 −a⁴_n 4 + · · ·

≤

a²_n 2

+

a³_n 3

+

a⁴_n 4

+ · · ·

≤ a²_n 2

1 +2

3|an| +2

4|a²_n| + · · ·

≤ a²_n 2

1 + |an| + |an|²+ · · ·

=

= a²_n 2

∞

X

r=0

|an|^r= a²_n 2 · 1

1 − |an| < a²_n 2 · 1

1 − ¹₂ = a²_n.

The convergence of P^∞_n=1a²_nthus gives the convergence of P^∞_n=1|ln(1 + a_n) − a_n|which in turn gives the convergence of the series P^∞n=1{ln(1 + an) − an}. This is equivalent to saying that Pn− S_n 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. Example 3.5. An example of when theorem 3.4 is useful is the innite product

∞

Y

n=2

1 + (−1)ⁿ1 n

. (3.9)

Here, an= (−1)^{n 1}_n so

∞

X

n=2

a²_n=

∞

X

n=2

1

n², (3.10)

which is a convergent series. We can therefore use theorem 3.4. The series

∞

X

n=2

a_n=

∞

X

n=2

(−1)ⁿ1

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.

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Example 3.6. If we instead consider the product

∞

Y

n=2

1 + (−1)ⁿ 1

√n

. (3.12)

we see that an = (−1)^{n 1}^√_n so the series

∞

X

n=2

a²_n =

∞

X

n=2

1

n (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

n=1

ln

1 + (−1)ⁿ 1

√n

(3.14) and that for −1 < an≤ 1 we have

ln(1 + x) = x −x²

2 + O(x³). (3.15)

Then we get ln

1 + (−1)ⁿ 1

√n

= (−1)ⁿ 1

√n− 1 2n + O

1 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

n=1

(1 + a_n) is said to be absolutely convergent if the product

∞

Y

n=1

(1 + |an|) is convergent.

We now explore the link between absolute convergence of series and products.

Theorem 3.8. If the series

∞

X

n=1

a_n is absolutely convergent, then the series

∞

X

n=1

ln(1 + an) is also absolutely convergent.

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Proof. Since P^∞n=1|an| is convergent, there exists a natural number N such that for n ≥ N we have

|an| ≤ 1 2.

First we assume that 0 ≤ an ≤ ¹₂. In similar fashion as in the proof of theorem 3.4 we have

| ln(1 + an)| = ln(1 + |an|) < 2|an| ∀n ≥ N.

Now assume that −¹₂ ≤ a_n< 0. Then we have, for all n ≥ N,

| ln(1 + an)| = − ln(1 − |a_n|) = ln

1

1 − |an|

= ln

1 + |an| 1 − |an|

≤ |an|

1 − |an| ≤ 2|an|.

So, for all values of n ≥ N we have

| ln(1 + an)| ≤ 2|an|

so according to the comparison test (theorem 2.4), the series P^∞n=1| ln(1+a_n)|converges and thus the series P^∞n=1ln(1 + an)is absolutely convergent.

From this we can now deduce another theorem.

Theorem 3.9. If the product

∞

Y

n=1

(1 + |an|) is convergent then the product

∞

Y

n=1

(1 + an) is also convergent.

Proof. Since |an| obviously never changes sign we can use theorem 3.3 which says that the product

∞

Y

n=1

(1 + |a_n|)

and the series _∞

X

n=1

|a_n|

converge or diverge together. Since the product converges, then so does the series. By theorem 3.8 we see that since P^∞n=1an is absolutely convergent we have convergence of

the series _∞

X

n=1

| ln(1 + a_n)|

which implies the convergence of

∞

X

n=1

ln(1 + an).

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This is equivalent with the convergence of the product

∞

Y

n=1

(1 + an), 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 + a_n) · · · (3.17) be a convergent innite product, and let

c1c2· · · ck· · · (3.18)

be a new product obtained by grouping the factors in (3.17) in arbitrary groups without changing the order, where

c_k = (1 + a_n_k−1₊₁) · · · (1 + a_n_k), ∀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

c1c2· · · ck· · · = P.

Proof. Let

R_k =

k

Y

r=1

c_r be the kth partial product of (3.18) and let

Pn=

n

Y

r=1

(1 + ar)

be the nth partial product of the product P . Then, if the partial sum

ln Rk=

k

X

r=1

ln cr

has a nite limit, the product (3.18) converges. Since

k

X

r=1

ln ck =

k

X

r=1

ln(1 + an_r−1+1) · · · (1 + an_r) =

k

X

r=1

ln(1 + an_k) = ln Pn_k

and since lim

n→∞Pn→ P, then lim

k→∞(ln Pn_k) → ln P and it follows that Rk also converges to P .

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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 = c1c2· · · 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 + a_n_k−1₊₁) · · · (1 + a_n_k) = c_k, ∀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

n=1

(1 + an) (3.19)

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 + a_n₁)] + [ln(1 + a_n₁₊₁) + · · · ln(1 + a_n₂)] + · · ·

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). Let

∞

Y

n=1

(1 + an) (3.20)

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

∞

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is convergent and thus, by theorem 3.3 the series

∞

X

n=1

|an| is convergent. By theorem 3.8 the series

∞

X

n=1

ln(1 + an)

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 the trigonometric function sin x. 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

sin x = x

∞

Y

n=1

1 − x²

n²π²

. (3.21)

Also we will try to conrm the factorisation sin x = x

1 − x π

1 + x π

· · · 1 − x

kπ

1 + x kπ

· · · and study its convergence.

In this example

a_n= − x² n²π²

which converges absolutely, so the product (3.21) is absolutely convergent. Observe that

1 − x² n²π² =

1 − x nπ

1 + x

nπ

so that we can write the expanded product

sin x = x

∞

Y

n=1

1 − x

nπ

1 + x

nπ

.

Written in this way, the two factors (1 − _nπ^x) and (1 +_nπ^x )always appear together in the nth product

Qn=

n

Y

r=1

1 − x

rπ

1 + x

rπ

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and since the nth product of the original expression

Pn=

n

Y

r=1

1 − x²

r²π²

is equal to Qn for all n, 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

1 + (−1)ⁿx

_n+1

2 π

!

= x 1 − x

π

1 + x

π

· · · 1 − x

kπ

1 + x

kπ

· · · (3.22)

where ⁿ⁺¹₂

denotes the integer part of ⁿ⁺¹₂ , in other words the sequence 1, 1, 2, 2, ....

Now we have

a_n= (−1)ⁿx

_n+1

2 π

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 sin x of equation (3.21) and also using the innite product representation of cos x, which is given by

cos x =

∞

Y

n=1

1 − 4x² (2n − 1)²π²

, (3.23)

we now show that _∞

Y

m=1

cos x

2^m = sin x

x , x 6= 0. (3.24)

Equation (3.23) gives us

cos x 2^m =

∞

Y

n=1

1 − x²

[2^m−1(2n − 1)]²π²

! , so that

∞

Y

m=1

cos x 2^m =

∞

Y

m=1

∞

Y

n=1

1 − x²

[2^m−1(2n − 1)]²π²

!

. (3.25)

It is clear that for every x there exists an N such that

0 < 1 − x²

[2^m−1(2n − 1)]²π²

!

< 1, ∀n, m ≥ N,

so there for any x there are only a nite number of factors that does not satisfy this inequality. Therefore we can assume that we have −π < x < π, so that

0 < 1 − x² !

< 1, ∀n, m ∈ N.

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Consider the logarithm of the left hand side of (3.25),

ln

∞

Y

m=1

cos x 2^m

!

=

∞

X

m=1

∞

X

n=1

ln 1 − x²

[2^m−1(2n − 1)]²π²

!

=

∞

X

m=1

∞

X

n=1

amn, where

a_mn= ln 1 − x²

[2^m−1(2n − 1)]²π²

! . Now we form the matrix

a₁₁ a₁₂ a₁₃ · · · a21 a22 a23 · · · a₃₁ a₃₂ a₃₃ · · · ... ... ... ···

which explicitly is the matrix ln

1 − _[1]^x2²π²

ln

1 − _[3]^x2²π²

ln

1 − _[5]^x2²π²

· · · ln

1 − _[2]^x2²π²

ln

1 − _[6]^x2²π²

ln

1 − _[10]^x2²π²

· · · ln

1 − _[4]^x2²π²

ln

1 − _[12]^x2²π²

ln

1 − _[20]^x2²π²

· · ·

... ... ... · · ·

(3.26)

Now we consider

sin x

x =

∞

Y

r=1

1 − x² [r]²π²

!

and the logarithm

ln sin x x

=

∞

X

r=1

ln 1 − x² [r]²π²

!

=

∞

X

r=1

br=

= ln 1 − x² [1]²π²

!

+ ln 1 − x² [2]²π²

!

+ ln 1 − x² [3]²π²

! + · · · where

br= ln 1 − x² [r]²π²

! .

Now we can see that the sequence {br} 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 P^∞_r=1br

converges then, according to theorem 2.11,

∞

X

m=1

amn

converges for all n, the iterated series

∞

X

n=1

∞

X

m=1

a_mn

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converges and that

ln

∞

Y

m=1

cos x 2^m

!

=

∞

X

n=1

∞

X

m=1

amn=

∞

X

r=1

br= ln 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 and cos x, however a more direct proof is also possible:

First note that

sin x = 2 sinx 2cosx

2 = 2²sin x 2²cosx

2cos x 2² so that by induction we get, for any n,

sin x = 2ⁿsin x 2ⁿ

n

Y

k=1

cos x 2^k, or, for x 6= 0,

n

Y

k=1

cos x

2^k = sin x

2ⁿsin₂^x_n = sin x x ·

x 2ⁿ

sin₂^x_n. Since

n→∞lim

x 2ⁿ

sin₂^xn

= 1 we get

∞

Y

m=1

cos x

2^m = sin x

x , x 6= 0, and the proof is done.

Example 3.15. Consider the innite product

∞

Y

n=1

√n

a, a > 0.

For a = 1 we have convergence to 1 since all factors are 1. Now let a 6= 1. Then we have

ln

∞

Y

n=1

√n

a =

∞

X

n=1

ln√ⁿ a =

∞

X

n=1

1 nln a.

Then ln a 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

n=1

n2√

a, a > 0.

Then we have _∞

√ ^∞ √ ^∞ 1

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This series is convergent for all a > 0 and since

∞

X

n=1

1 n² = π²

6

we have that _∞

Y

n=1

n2√

a = a^π2^/⁶, a > 0.

Example 3.17. Also consider the series

∞

Y

n=1

√n

n. (3.27)

In this case, if the product converges, it is the same even if the product begins at n = 2.

We have

ln

∞

Y

n=2

√n

n =

∞

X

n=2

ln √ⁿ n =

∞

X

n=2

1 nln n ≥

∞

X

n=2

1

nln 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=0

(1 + x²ⁿ) = 1

1 − x, |x| < 1. (3.29)

The nth product is

Pn=

n

Y

r=0

1 + x²^r and for n = 1 we have

P1= (1 + x) 1 + x² = 1 + x + x²+ x³=

3

X

r=0

x^r=

2²−1

X

r=0

x^r. (3.30)

Assume that

P_n=

2ⁿ⁺¹−1

X

r=0

x^r for some n ≥ 1. Then

P_n+1=





2ⁿ⁺¹−1

X

r=0

x^r





1 + x²ⁿ⁺¹

=

2ⁿ⁺¹−1

X

r=0

x^r+

·2ⁿ⁺¹−1

X

r=0

x^r· x²ⁿ⁺¹ =

=

2ⁿ⁺¹−1

X

r=0

x^r+

2ⁿ⁺¹−1

X

r=0

x²ⁿ⁺¹^+r=

A

X

r=0

x^r, where

A = 2ⁿ⁺¹− 1 + 2ⁿ⁺¹= 2 · 2ⁿ⁺¹− 1 = 2ⁿ⁺²− 1.

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Thus, by induction and (3.30), we have that

P_n=

2ⁿ⁺¹−1

X

r=0

x^r for all n and since Pn is a subsequence of the partial sum

Sn =

n

X

r=0

x^r

and since _∞

X

n=0

x^r= 1

1 − x, |x| < 1, we have also shown the equality (3.29).

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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

n=1

an (4.1)

be a conditionally convergent series, let

∞

X

n=1

b_n (4.2)

be the series formed from the positive terms of (4.1), arranged in the same order, and

let _∞

X

n=1

cn (4.3)

be the series formed from the absolute values of the negative terms of (4.1), arranged in the same order. Then the series (4.2) and (4.3) both diverge.

Proof. Let

P_n=

n

X

r=1

a_r,

Qj=

j

X

r=1

br

and

R_k=

k

X

r=1

c_r,

be the nth sum of (4.1), (4.2) and (4.3) respectively, where j is the number of positive terms in the n rst terms of (4.1) and k is the number of negative terms in the n rst terms of (4.1). Thus we have that

P_n= Q_j− Rk

and because Pn converges as n goes to innity we have that Qj and Rk converge or diverge together. Now let

P_n^∗=

n

X

r=1

|ar|, so that we also have

P_n^∗= Qj+ Rk. (4.4)

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The series (4.1) is not absolutely convergent so Pn^∗ diverges, which means that the right hand side of (4.4) also diverges as n 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

n=1

an (4.5)

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

n=1

b_n (4.6)

made from the positive terms of (4.5) without changing the order of terms, and

∞

X

n=1

cn (4.7)

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 k1terms, 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

bk_n+1+ · · · + bk_n+1 > n + 1.

We now have the series

(b₁+ · · · + b_k₁) − c₁+ (b_k₁₊₁+ · · · + b_k₂) − 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 A ∈ R. First select enough terms, say j1 terms, from (4.6) so that

Some Properties of Infinite Series

Mattias Flygare