§ 7. Infinite sums and products of real numbers

7. Infinite sums and products of real numbers

Since every point of $\mathbf{R}$ has a countable fundamental system of neighbourhoods (§ 1, no. 4, Corollary to Proposition 3), it follows that a family $(x_t)$ of finite real numbers is summable in $\mathbf{R}$ only if the set of indices t such that $x_t \neq 0$ is countable (Chapter III, § 5, no. 2, Corollary to Proposition 1). The study of summable families in $\mathbf{R}$ thus reduces essentially to the study of summable sequences. However, we shall later have to consider uncountable families $(x_t)$ of finite real numbers, whose terms are functions of a parameter $t$ ; it can happen that this family is summable for all $t$ , but that the (countable) set of indices $i$ such that $x_i \neq 0$ depends on $t$ . For this reason we shall not impose any hypothesis on the power of the index set in what follows.

1. Families of positive finite numbers summable in $\mathbf{R}$

Theorem 1

A family* $(x_i)$ of finite real numbers $\geqslant 0$ is summable in $\mathbf{R}$ if and only if the set of partial finite sums of the family is bounded above in $\mathbf{R}$ . If so, the least upper bound of this set is the sum of the family $(x_i)$ .

For each finite subset $H$ of the index set $I$ , let $s_H = \sum_{i \in H} x_i$ ; since the $x_i$ are $\geqslant 0$ , the relation $H \subset H'$ implies $s_H \leqslant s_{H'}$ . In other words, the mapping $H \to s_H$ is an increasing function on the directed set $\mathfrak{F}(I)$ of finite subsets of $I$ ; therefore (§ 5, no. 2, Corollary to Theorem 2) it has a finite limit if and only if it is bounded above.

Remark

Let $(H_\lambda)$ be a family of finite subsets of $I$ such that, for each finite subset $H$ of $I$ , there is an index $\lambda$ such that $H \subset H_\lambda$ ; then $(x_i)$ is summable if and only if the family of the $s_{H_\lambda}$ is bounded above in $\mathbf{R}$ . In particular, let $(x_n)$ be a sequence of finite real numbers $\geqslant 0$ , and for each integer $n$ let $s_n = \sum_{p=0}^n x_p$ ; then the sequence $(x_n)$ is summable in $\mathbf{R}$ if and only if, for one sequence of strictly increasing integers $(n_k)$ , the partial sequence $(s_{n_k})$ is bounded above in $\mathbf{R}$ .

Examples.* 1) For each number $q$ such that $0 \leqslant q < 1$ , the sequence $(q^n)$ ("geometric progression of ratio $q$ ") is summable in $\mathbf{R}$ , since

s_n = \frac{1 - q^{n+1}}{1 - q} \leqslant \frac{1}{1 - q}

the sum of this sequence is

\lim_{n \to \infty} s_n = \frac{1}{1 - q}

2) Let $a$ and $b$ be two numbers such that $0 \leqslant a < 1$ and $0 \leqslant b < 1$ ; then the family $(a^m b^n)_{(m, n) \in \mathbf{N} \times \mathbf{N}}$ is summable in $\mathbf{R}$ . For each finite subset of $\mathbf{N} \times \mathbf{N}$ is contained in a subset of the form $[0, p] \times [0, p]$ , and we have

\sum_{m=0}^p \sum_{n=0}^p a^m b^n = \left( \sum_{m=0}^p a^m \right) \left( \sum_{n=0}^p b^n \right) = \frac{1 - a^{p+1}}{1 - a} \cdot \frac{1 - b^{p+1}}{1 - b} \leqslant \frac{1}{(1 - a)(1 - b)}.

3) For each integer $p > 1$ , the sequence $(n^{-p})$ ( $n > 0$ ) is summable, since

s_{2^n + 1} - s_{2^n} = \sum_{k=1}^{2^n} (2^n + k)^{-p} < 2^n \cdot (2^n)^{-p}

and therefore, adding these inequalities together,

s_{2^n} < \frac{1}{1 - 2^{1-p}}.

4) The sequence $(1/n)$ ( $n > 0$ ) is not summable in $\mathbf{R}$ , since

s_{2^{n+1}} - s_{2^n} = \sum_{k=1}^{2^n} \frac{1}{2^n + k} > \frac{2^n}{2^{n+1}} = \frac{1}{2}

and therefore, adding these inequalities together,

s_{2^n} > n/2

so that the criterion of Theorem 1 is not satisfied.

5) Let $(I_n)$ be a sequence of mutually disjoint non-empty open intervals, all contained in an interval of finite length $l$ . The sum of the lengths of a finite number of intervals of this family is $\leq l$ ( $§ 1$ , no. 5) and therefore the family of lengths of the $I_n$ is summable in $\mathbf{R}$ , and its sum is $\leq l$ .

Theorem 2 (Principle of comparison)

Let* $(x_i)_{i \in I}$ and $(y_i)_{i \in I}$ be two families of finite real numbers $\geq 0$ , such that $x_i \leq y_i$ for all $i$ . If $(y_i)$ is summable in $\mathbf{R}$ then so is $(x_i)$ , and we have $\sum_i x_i \leq \sum_i y_i$ ; if in addition there is an index $x$ such that $x_x < y_x$ , then $\sum_i x_i < \sum_i y_i$ .

The hypothesis implies that,

\sum_{i \in H} x_i \leq \sum_{i \in H} y_i,

for every finite subset

H

I

and the first part of the theorem follows from this; the inequality relating the sums follows from the principle of extension of inequalities (

§ 5

, no. 2, Theorem 1). If

x_x < y_x

, then

\sum_i x_i = x_x + \sum_{i \neq x} x_i < y_x + \sum_{i \neq x} y_i = \sum_i y_i.

This theorem provides the most commonly used criterion for deciding whether or not a sequence $(x_n)$ of real numbers $\geq 0$ is summable in $\mathbf{R}$ : we try to compare the given sequence with a simpler sequence $(y_n)$ of which we already know whether it is summable or not. If there exists a finite number $a > 0$ such that $x_n \leq a y_n$ for all $n$ from a certain point onwards, and if $(y_n)$ is summable, then $(x_n)$ is summable; if on the other hand there is a finite number $b > 0$ such that $x_n \geq b y_n$ for all $n$ from a certain point onwards, and if $(y_n)$ is not summable in $\mathbf{R}$ , then $(x_n)$ is not summable in $\mathbf{R}$ . We shall see in a later volume how such comparison sequences may be obtained in the cases which arise most frequently.

Example 1

Let $a$ be a finite real number $> 0$ , and consider the sequence $\left( \frac{a^n}{n!} \right)$ ; let $n_0$ be the smallest integer such that $a < n_0$ . Then for each $n \geq n_0$ we have

\frac{a^n}{n!} \leq \frac{a^{n_0}}{n_0!} \cdot \left( \frac{a}{n_0} \right)^{n-n_0};

since

q = \frac{a}{n_0} < 1

, the sequence

(q^{n-n_0})

is summable, and therefore so is the sequence

\left( \frac{a^n}{n!} \right)

Example 2

Let $(a_n)$ be a summable sequence of positive numbers. Since

\lim_{n \to \infty} a_n = 0,

there exists an integer

n_0

such that

a_n \leq 1

whenever

n \geq n_0

. Consequently, for each

n \geq n_0

we have

a_n^2 \leq a_n

, and therefore the sequence

(a_n^2)

is summable in

\mathbf{R}

. So is the sequence

(a_n^p)

for each integer

p > 1

Example 3

Let $a$ and $b$ be two numbers $< 1$ ; then

\frac{1}{a^m + b^n} \leq \frac{1}{2(\sqrt{a})^m (\sqrt{b})^n}

and hence the family

\left( \frac{1}{a^m + b^n} \right)

is summable in

\mathbf{R}

Corollary

Let* $(x_i)_{i \in I}$ be a summable family of finite numbers $\geq 0$ in $\mathbf{R}$ . If $H$ is any subset of $I$ , we have

\sum_{i \in H} x_i \leq \sum_{i \in I} x_i,

and equality holds only if

x_i = 0

for all

i \in \complement H

2. Families of finite numbers of arbitrary sign summable in $\mathbf{R}$

Theorem 3

Let* $(x_i)_{i \in I}$ be a family of finite real numbers; then the following statements are equivalent:

a) The family $(x_i)$ is summable in $\mathbf{R}$ . b) The family $(|x_i|)$ is summable in $\mathbf{R}$ . c) The set of finite partial sums of the family $(x_i)$ is bounded in $\mathbf{R}$ .

Let $I_1$ be the set of all $i \in I$ such that $x_i \geq 0$ , and $I_2$ the set of all $i \in I$ such that $x_i < 0$ . The family $(x_i)_{i \in I}$ *[resp. $(|x_i|)_{i \in I}$ ] is summable if and only if each of the families $(x_i)_{i \in I_1}$ and $(x_i)_{i \in I_2}$ *[resp. $(|x_i|)_{i \in I_1}$ and $(|x_i|)_{i \in I_2}$ ] is summable (Chapter III, § 5, no. 3, Propositions 2 and 3). Now it comes to the same thing to say that $(x_i)_{i \in I_1}$ is summable, or that $(|x_i|)_{i \in I_1}$ is summable, or that the set of finite partial sums of the family $(x_i)_{i \in I_1}$ is bounded (no. 1, Theorem 1); and the same is true with $I_1$ replaced by $I_2$ . The theorem follows immediately.

Theorem 3 shows that the summability in $\mathbf{R}$ of a family of finite real numbers depends only on the summability of the family of their absolute values.

We recall (Chapter III, § 5, no. 5, Proposition 6) that if $(x_i)$ and $(y_i)$ are two summable families of finite real numbers, then the family $(x_i + y_i)$ is summable and $\sum (x_i + y_i) = \sum x_i + \sum y_i$ . Moreover, if $(x_i)$ is a summable family of finite real numbers and $a$ is any finite real number, then the family $(ax_i)$ is summable in $\mathbf{R}$ , and we have $\sum ax_i = a \cdot \sum x_i$ .

3. Product of two infinite sums

Proposition 1

If the families* $(x_\lambda)_{\lambda \in L}$ and $(y_\mu)_{\mu \in M}$ of finite real numbers are summable in $\mathbf{R}$ , then so is the family $(x_\lambda y_\mu)_{(\lambda, \mu) \in L \times M}$ *and we have

(1)

\sum_{(\lambda, \mu) \in L \times M} x_\lambda y_\mu = (\sum_{\lambda \in L} x_\lambda) (\sum_{\mu \in M} y_\mu).

Every finite subset of $L \times M$ is contained in a finite subset of the form $H \times K$ , where $H$ is a finite subset of $L$ and $K$ is a finite subset of $M$ . By hypothesis, there exists a number $a > 0$ such that $\sum_{\lambda \in H} |x_\lambda| \leq a$ and $\sum_{\mu \in K} |y_\mu| \leq a$ for all finite subsets $H$ and $K$ of $L$ and $M$ respectively; therefore

\sum_{(\lambda, \mu) \in H \times K} |x_\lambda y_\mu| = (\sum_{\lambda \in H} |x_\lambda|) (\sum_{\mu \in K} |y_\mu|) \leq a^2,

and this shows that the family

(x_\lambda y_\mu)

is summable in

\mathbf{R}

, by Theorems 1 and 3. By associativity we can write [Chapter III, § 5, no. 3, formula (2)]

\sum_{(\lambda, \mu) \in L \times M} x_\lambda y_\mu = \sum_{\lambda \in L} (\sum_{\mu \in M} x_\lambda y_\mu) = \sum_{\lambda \in L} x_\lambda (\sum_{\mu \in M} y_\mu) = (\sum_{\lambda \in L} x_\lambda) (\sum_{\mu \in M} y_\mu);

hence the result.

4. Families multipliable in $\mathbf{R}^*$

In the multiplicative group $\mathbf{R}^*$ of finite non-zero real numbers, a family $(x_i)_{i \in I}$ can be multipliable only if $\lim x_i = 1$ with respect to the filter of complements of finite subsets of $I$ (Chapter III, § 5, no. 2, Proposition 1). In particular there can be only a finite number of indices $i$ such that $x_i < 0$ . We may therefore limit ourselves to considering only families $x_i$ all of whose terms are strictly positive; it is then convenient to put $x_i = 1 + u_i$ , where the $u_i$ are subject to the inequalities $-1 < u_i < +\infty$ for all $i$ . Since each point of $\mathbf{R}^*$ has a countable fundamental system of neighbourhoods, the set of indices $i$ such that $u_i \neq 0$ is countable, if the family $(1 + u_i)$ is multipliable in $\mathbf{R}^*$ .

Theorem 4

The family* $(1 + u_i)$ is multipliable in $\mathbf{R}^*$ if and only if the family $(u_i)$ is summable in $\mathbf{R}$ . Lemma. (i) If $(a_i)_{1 \leq i \leq p}$ is a finite sequence of numbers $> 0$ , then

\prod_{i=1}^p (1 + a_i) \geq 1 + \sum_{i=1}^p a_i.

(ii) If also

a_i < 1

for each

i

, then

\prod_{i=1}^p (1 - a_i) \geq 1 - \sum_{i=1}^p a_i.

These inequalities are clear if $p = 1$ , and are proved by induction on $p$ . If

\prod_{i=1}^{p-1} (1 + a_i) \geq 1 + \sum_{i=1}^{p-1} a_i,

we have

\prod_{i=1}^p (1 + a_i) \geq (1 + a_p) \left( 1 + \sum_{i=1}^{p-1} a_i \right) = 1 + \sum_{i=1}^p a_i + a_p \cdot \sum_{i=1}^{p-1} a_i \geq 1 + \sum_{i=1}^p a_i.

Again, if

\prod_{i=1}^{p-1} (1 - a_i) \geq 1 - \sum_{i=1}^{p-1} a_i,

we have

\prod_{i=1}^p (1 - a_i) \geq (1 - a_p) \left( 1 - \sum_{i=1}^{p-1} a_i \right) = 1 - \sum_{i=1}^p a_i + a_p \cdot \sum_{i=1}^{p-1} a_i \geq 1 - \sum_{i=1}^p a_i.

Having proved the lemma, we note that if the family $(1 + u_i)$ is multipliable then so are the families $(1 + u_i^+)$ and $(1 - u_i^-)$ , since $\mathbf{R}^*$ is a complete group (Chapter III, § 5, no. 3, Proposition 2); and conversely, if the families $(1 + u_i^+)$ and $(1 - u_i^-)$ are multipliable, then so is $(1 + u_i)$ (Chapter III, § 5, no. 3, Proposition 3). We need therefore consider only the cases in which all the $u_i$ are $\geq 0$ and in which they are all $\geq 0$ .

Suppose first that $u_i \geq 0$ for all $i$ . If the family $(1 + u_i)$ is multipliable, then for each $\varepsilon > 0$ there is a finite subset $J$ of the index set in $I$ such that, for each finite subset $H$ of $I$ disjoint from $J$ , we have $1 \leq \prod_{i \in H} (1 + u_i) \leq 1 + \varepsilon$ ; by (2) it follows that $\sum_{i \in H} u_i \leq \varepsilon$ , which shows that $(u_i)$ is summable in $\mathbf{R}$ by virtue of Cauchy’s criterion (Chapter III, § 5, no. 2, Theorem 1).

Conversely, suppose that $(u_i)$ is summable in $\mathbf{R}$ . For each $\varepsilon$ such that $0 < \varepsilon < 1$ , there is a finite subset $J$ of $I$ such that, for each finite subset $H$ of $I$ disjoint from $J$ , we have $0 \leq \sum_{i \in H} u_i \leq \varepsilon$ . By (3), we have therefore $\prod_{i \in H} (1 - u_i) \geq 1 - \varepsilon$ ; but since $1 + u \leq \frac{1}{1 - u}$ for every number $u$ such that $0 \leq u < 1$ , it follows that

1 \leq \prod_{i \in H} (1 + u_i) \leq \frac{1}{1 - \varepsilon},

and this shows that $(1 + u_i)$ is multipliable (Cauchy’s criterion).

The proof is similar when all the $u_i$ are $\leq 0$ . To show that $(u_i)$ is summable if $(1 + u_i)$ is multipliable, use formula (2) and the inequality $1 - u \leq \frac{1}{1 + u}$ ( $0 \leq u < 1$ ); to show that $(1 + u_i)$ is multipliable if $(u_i)$ is summable, use formula (3).

In Chapter V (§ 4), the topological study of the group $\mathbf{R}^*$ will enable us to give another criterion for multipliability of a family in $\mathbf{R}^*$ with the help of the logarithmic function; in a later volume we shall establish the equivalence of this criterion with the one above, using the differential properties of the logarithm.

5. Summable families and multipliable families in $\overline{\mathbf{R}}$

On the interval $[0, + \infty]$ of $\overline{\mathbf{R}}$ , addition is an associative and commutative law of composition (§ 4, no. 3); hence the notion of a summable family of numbers in this interval is defined (Chapter III, § 5, no. 1, Remark 3).

Proposition 2

Every family* $(x_i)$ of real numbers $\geq 0$ is summable in $\overline{\mathbf{R}}$ .

For the mapping $H \to s_H$ of the directed set $\mathfrak{F}(I)$ into $\overline{\mathbf{R}}$ is increasing; hence (§ 5, no. 2, Theorem 2) has a limit.

The same reasoning shows that every family of real numbers $\leq 0$ is summable in $\overline{\mathbf{R}}$ .

Similarly, multiplication is an associative and commutative law of composition on each of the intervals $[0, 1]$ and $[1, + \infty]$ of $\overline{\mathbf{R}}$ , and therefore the notion of a multipliable family is defined on each of these intervals.

Proposition 3

Every family $(1 + u_i)$ [resp. $(1 - u_i)$ ] of numbers $\geq 1$ (resp. $\geq 0$ and $\leq 1$ ) is multipliable in $\overline{\mathbf{R}}$ .

Same proof as for Proposition 2.

Corollary

The product $\prod (1 + u_i)$ [resp. $\prod (1 - u_i)$ ] of numbers $\geq 1$ (resp. $> 0$ and $\leq 1$ ) is equal to $+ \infty$ (resp. 0) if and only if $\sum u_i = + \infty$ .

For if $\sum u_i$ is finite then $\prod (1 + u_i)$ and $\prod (1 - u_i)$ are in $\mathbf{R}^*$ , and conversely (no. 4, Theorem 4).

Remark

The theorem of associativity (Chapter III, § 5, no. 3, Theorem 2) remains valid when G is replaced by $\overline{\mathbf{R}}$ and the $x_i$ are assumed to be $\geq 0$ . This is clear if $\sum_{i \in I} x_i$ is finite. Suppose, on the contrary, that $\sum_{i \in I} x_i = + \infty$ . Then for each finite $a > 0$ there is a finite subset H of I such that $\sum_{i \in H} x_i \geq a$ . Let K be a finite subset of L such that $H \subset \bigcup_{\lambda \in K} I_\lambda$ ; then since $s_\lambda \geq \sum_{i \in I_\lambda \cap H} x_i$ for all $\lambda \in K$ , we have

\sum_{\lambda \in K} s_\lambda \geq \sum_{i \in H} x_i \geq a,

which shows that

\sum_{\lambda \in L} s_\lambda = + \infty

. We leave to the reader the task of formulating the analogous proposition for multipliable families of numbers in

[0, 1]

or in

[1, + \infty]

6. Infinite series and infinite products of real numbers

A series of finite real numbers is simply said to be convergent if it is convergent in $\mathbf{R}$ .

Definition 1

A series of finite real numbers is said to be absolutely convergent if the series of absolute values of its terms is convergent.

Proposition 4

A series of finite real numbers is commutatively convergent if and only if it is absolutely convergent.

This follows from Chapter III, § 5, no. 7, Proposition 9 and from Theorem 3 of no. 2.

In other words, if $(u_n)$ is a sequence of finite real numbers, it comes to the same thing to say that the series whose general term is $u_n$ is commutatively convergent, or that it is absolutely convergent, or that the sequence $(u_n)$ is summable in $\mathbf{R}$ . All the properties of summable families proved in Chapter III, § 5 therefore apply to absolutely convergent series. In particular, if the series whose general term is $u_n$ is absolutely convergent, then the sum $\sum_{n \in H} u_n$ exists for all subsets $H$ of $\mathbf{N}$ ; and, if $(H_p)$ is a partition of $\mathbf{N}$ , we have

\sum_{n=0}^{\infty} u_n = \sum_p \left( \sum_{n \in H_p} u_n \right)

(associativity of absolutely convergent series).

As we have already remarked in Chapter III, § 5, a series of real numbers can be convergent without being commutatively convergent, that is, without being absolutely convergent.

Example

Alternating series.* A series defined by a sequence $(u_n)$ of finite real numbers is called alternating if $u_n = (-1)^n v_n$ , where $v_n \geq 0$ for all $n$ . Let us show that a sufficient condition for the convergence of such a series is that the sequence $(v_n)$ decreases and has 0 as its limit. For if $s_n$ denotes

\sum_{p=0}^{n} u_p,

the hypothesis that

(v_n)

is decreasing implies that

s_{2n+1} \leq s_{2n+3} \leq s_{2n+2} \leq s_{2n}

for all

n \geq 0

. The sequence

(s_{2n})

[resp.

(s_{2n+1})

] is decreasing and bounded below (resp. increasing and bounded above) and therefore has a finite limit

a

(resp.

b

), and

b \leq a

; since

a - b = \lim_{n \to \infty} (s_{2n} - s_{2n+1}) = \lim_{n \to \infty} v_{2n+1} = 0,

the assertion is proved.

If we take for example $v_n = 1/n$ , the conditions above are satisfied, and therefore the series whose general term is $(-1)^n / n$ (the "alternating harmonic series") is convergent. We have seen in no. 1 that the series whose general term is $1/n$ (the "harmonic series") is not convergent, and thus the alternating harmonic series is not absolutely convergent.

We recall (Chapter III, § 5, no. 6, Proposition 7) that, if $(u_n)$ and $(v_n)$ are two convergent series of finite real numbers, then the series $(u_n + v_n)$ is convergent, and

\sum_{n=0}^{\infty} (u_n + v_n) = \sum_{n=0}^{\infty} u_n + \sum_{n=0}^{\infty} v_n;

also, if the series

(u_n)

is convergent then the series

(au_n)

is convergent for all finite real numbers

a

, and

\sum_{n=0}^{\infty} au_n = a \cdot \sum_{n=0}^{\infty} u_n

Finally, if the series $(u_n)$ and $(v_n)$ are convergent, and if $u_n \leq v_n$ for all $n$ , we have $\sum_{n=0}^{\infty} u_n \leq \sum_{n=0}^{\infty} v_n$ by the principle of extension of inequalities (§ 5, no. 2, Theorem 1).

It should be noted that, if we suppose the series $(v_n)$ to be convergent but not absolutely convergent, and if $|u_n| \leq |v_n|$ for each $n$ , we can by no means infer that the series $(u_n)$ is convergent, as is seen by taking $u_n = |v_n|$ .

An infinite product of finite non-zero real numbers is said simply to be convergent if it is convergent in $\mathbf{R}^*$ ; its value is then a finite non-zero real number.

Definition 2

An infinite product whose general factor is $1 + u_n$ is said to be absolutely convergent if the product whose general factor is $1 + |u_n|$ is convergent.

Proposition 5

An infinite product of finite real numbers is commutatively convergent if and only if it is absolutely convergent.

This follows from Chapter III, § 5, no. 7, Proposition 9, and from Theorem 4 above.

Moreover, the product whose general factor is $1 + u_n$ is absolutely convergent if and only if the series whose general term is $u_n$ is absolutely convergent.

A product of non-zero real numbers can be convergent without being commutatively convergent, i.e. without being absolutely convergent.

Example

If we take $u_{2n-1} = -1/n$ and $u_{2n} = 1/n$ for $n \geq 2$ , the product $(1 + u_n)$ is not absolutely convergent, since the series $(u_n)$ is not absolutely convergent; but since

\prod_{p=3}^{n} (1 + u_p) = \prod_{p=2}^{n} \left(1 - \frac{1}{p^2}\right),

and

\prod_{p=3}^{2n+1} (1 + u_p) = \left(1 - \frac{1}{n+1}\right) \prod_{p=2}^{n} \left(1 - \frac{1}{p^2}\right),

it follows from Theorem 4 that the product is convergent and that its value is

\prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2}\right).

Moreover, it should be observed that the convergence of the series whose general term is $u_n$ is neither necessary nor sufficient for the convergence of the product whose general factor is $1 + u_n$ (see Exercises 21 and 22).

Exercises

See the exercises for § 7.