Thus by taking away a (sufficiently large) finite number of terms from the family $(x_i)$, every finite partial sum of the surviving subfamily has to be as close to $o$ as we please.

An immediate consequence of the first part of Theorem 1 is the following proposition:

This necessary condition for a family $(x_i)$ to be summable is by no means sufficient in general, even if $G$ is complete; we shall see many examples later (see Chapter IV, § 7).

For if Cauchy’s criterion is satisfied by a family $(x_i)_{i \in I}$, then it is trivially satisfied by every subfamily.

If $(x_i)_{i \in I}$ is summable, it follows therefore that the sum $\sum_{i \in J} x_i$ is defined for every subset J of I, finite or not : it is again called the partial sum of the family $(x_i)$, corresponding to the subset J of the index set. The set of partial sums of a summable family is evidently contained in the closure of the set of finite partial sums.

Thus if we have a summable family in a complete group, we can associate its terms arbitrarily into subfamilies and form the sum of each subfamily thus obtained; the family of these partial sums is again summable and its sum is equal to that of the given family.

Let $s = \sum_{i \in I} x_i$, and let $V$ be any closed neighbourhood of 0 in $G$. Then there is a finite subset $J_0$ of $I$ such that, for each finite subset $J$ of $I$ which contains $J_0$, we have $\sum_{i \in J} x_i \in s + V$. Let $K_0$ be the subset of $L$ consisting of indices $\lambda$ such that $J_\lambda = I_\lambda \cap J_0$ is not empty: $K_0$ is clearly finite. Let $K$ be any finite subset of $L$ which contains $K_0$; we shall show that $\sum_{\lambda \in K} s_\lambda \in s + V$, which will establish the theorem.

Now $s_\lambda$ is very close to a finite partial sum of $(x_i)$, whose indices all belong to $I_\lambda$; more precisely, given any symmetric neighbourhood $W$ of 0, there exists for each $\lambda \in K$ a finite subset $H_\lambda$ of $I_\lambda$, containing $J_\lambda$ and such that $s_\lambda - \sum_{i \in H_\lambda} x_i \in W$. Let $J = \sum_{\lambda \in K} H_\lambda$; then $J$ is a finite subset of $I$ containing $J_0$, and we have

by the associativity of finite sums. Hence by reason of the choice of $J_0$ and the $H_\lambda$, we have

where $n$ is the number of elements in $K$; this relation holds for each $W$, hence also $\sum_{\lambda \in K} s_\lambda \in s + V$, since $V$ (being closed) is the intersection of the neighbourhoods $V + nW$ [§ 3, no. 1, formula (1)].

We can thus write the associativity formula for sums:

(I)

which is valid whenever the family $(I_\lambda)$ is a partition of its union and the right-hand side is defined. In particular, if the index set is a product $I = L \times M$, and if the "double" family $(x_{\lambda \mu})_{(\lambda, \mu) \in L \times M}$ is summable, we have the formula of change of order of summation

(2)

It should be remarked that the left-hand side of (1) can have a meaning, without the right-hand side being defined. Consider for example the case where $I = LX \{1, 2\}$ and $L$ is infinite, and $I_\lambda$ consists of the two elements $(\lambda, 1)$ and $(\lambda, 2)$; if we take $x_{\lambda, 1} = a,\ x_{\lambda, 2} = -a$, where $a$ is any non-zero element of $G$, then all the partial sums corresponding to the $I_\lambda$ are zero, and therefore the left-hand side of (1) is defined and equal to 0, whereas the right-hand side of (1) has no meaning, as Proposition 1 shows.

Likewise, it can happen that the left-hand side of (2) is not defined but that each of the two "double sums" in (2) has a meaning; and the elements of $G$ which they represent need not be equal (see Chapter IV, § 7, Exercise 17).

Thus although it is always possible to "associate" the terms of a sum, it is not possible, on the other hand, to "dissociate" into their elements those terms of a sum which appear themselves as sums. Nevertheless, this operation is legitimate whenever the number of these "dissociable" terms is finite.

It is enough to prove the proposition when $L = (1,2)$; having done so, we then proceed by induction on the number of elements of $L$. Let $s_1 = \sum_{i \in I_1} x_i$ and $s_2 = \sum_{i \in I_2} x_i$. For each neighbourhood $V$ of the origin, there is a finite subset $J_1$ (resp. $J_2$) of $I_1$ (resp. $I_2$) such that, for each finite subset $H_1$ (resp. $H_2$) of $I_1$ (resp. $I_2$) containing $J_1$ (resp. $J_2$), we have $\sum_{i \in H_1} x_i \in S_1 + V$ (resp. $\sum_{i \in H_2} x_i \in s_2 + V$). If we put $J_0 = J_1 \cup J_2$, it follows that, for each finite subset $H$ of $I$ which contains $J_0$, we have $\sum_{i \in H} x_i \in S_1 + s_2 + V + V$, and the result follows.

This follows immediately from the condition for convergence, with respect to a filter, of a function which takes its values in a product space (Chapter I, § 1, no. 6, Corollary 1 to Proposition 10); in effect, for each finite subset $J$ of $I$, we have

If $J$ is any finite subset of the index set, then $f \left( \sum_{i \in J} x_i \right) = \sum_{i \in J} f(x_i)$, and the image under $f$ of a convergent filter base is a convergent filter base (Chapter I, § 7, no. 4, Corollary 1 to Proposition 9).

For $x \to -x$ and $x \to nx$ are continuous homomorphisms of $G$ into $G$; on the other hand, if $(x_i)$ and $(y_i)$ are summable, then the family $(x_i, y_i)$ is summable in $G \times G$, and since $(x, y) \to x + y$ is a continuous homomorphism of $G \times G$ into $G$, we deduce (6).

For the series whose general term is $x_n - y_n$ has all its terms zero from a certain index onwards.

This corollary may be put in the form that we may alter arbitrarily a finite number of terms of a convergent series without it ceasing to be convergent. In particular, if $y_n = 0$ for $n < m$, and $y_n = x_n$ for $n \geq m$, we see that the series whose general term is $y_n$ converges if and only if the series whose general term is $x_n$ converges; its sum is denoted by $\sum_{n=m}^{\infty} x_n$ and is called the residue of index $m$ of the series $(x_n)$. Since

If a sequence $(x_n)_{n \in I}$ has as index set an infinite subset $I$ of $\mathbf{N}$, and if $\varphi$ denotes the strictly order-preserving bijection of $\mathbf{N}$ onto $I$, then the series defined by the sequence $(x_{\varphi(n)})_{n \in \mathbf{N}}$ is called, by abuse of language, the series defined by the sequence $(x_n)_{n \in I}$; if this series is convergent, its sum is denoted by $\sum_{n \in I}^{\infty} x_n$. It is immediately verified that this series converges if and only if the series whose general term is $(z_n)$ converges, where $z_n = x_n$ if $n \in I$ and $z_n = 0$ if $n \in \mathbf{C} I$.

It is important to notice that if the series defined by a sequence $(x_n)_{n \in \mathbf{N}}$ converges, there can exist infinite subsets $I$ of $\mathbf{N}$ such that the series defined by the subsequence $(x_n)_{n \in I}$ does not converge (see Exercise 5 and Chapter IV, § 7).

Propositions 4 and 5 extend immediately to series, and we leave their formulation in this case to the reader.

Proposition 8 (Restricted associativity of series). Let $(k_n)$ be a strictly increasing sequence of integers $\geqslant 0$. If the series whose general term is $x_n$ converges, and if we put $u_n = \sum_{p=k_{n-1}}^{k_n-1} x_p$, then the series whose general term is $u_n$ is convergent, and we have $\sum_{n=0}^{\infty} u_n = \sum_{n=0}^{\infty} x_n$.

For the sequence of partial sums of the series $(u_n)$ is a subsequence $(s_{k_n-1})$ of the sequence $(s_n)$ of partial sums of the series $(x_n)$.

If the sequence is summable, then clearly the series is commutatively convergent. To prove the reverse implication we shall argue by reductio ad absurdum and suppose that the series $(x_n)$ is commutatively convergent but that the sequence $(x_n)$ is not summable. The image of the filter $\Phi$ under the mapping $H \to s_H$ therefore cannot be a Cauchy filter base in $G$, otherwise this filter base would converge, since by hypothesis it has a cluster point (Chapter II, § 3, no. 2, Proposition 5, Corollary 2). Hence there is a neighbourhood $V$ of 0 in $G$ such that, for each finite subset $J$ of $\mathbf{N}$, there is a finite subset $H$ of $\mathbf{N}$ which does not meet $J$ and is such that $\sum_{n \in H} x_n \notin V$. Hence, by induction, we can define a partition of $\mathbf{N}$ into finite subsets $H_k$ ($k \in \mathbf{N}$) such that $\sum_{n \in H_k} x_n \notin V$ for an infinite number of indices $k$. Clearly there is a permutation $\sigma$ of $\mathbf{N}$ such that for each $k$, the values of $n$ for which $\sigma(n) \in H_k$ are consecutive. If $\sigma$ is such a permutation, then the series whose general term is $x_{\sigma(n)}$ cannot be convergent, and we have a contradiction.

If the group $G$ is written multiplicatively, the infinite product defined by a sequence $(x_n)$ of points of $G$ (or infinite product whose general term is $x_n$, or even product $x_n$ if there is no possibility of confusion) is defined to be the pair formed by the sequence $(x_n)$ and the sequence of partial products

The infinite product is said to converge if the sequence $(p_n)$ converges, and the limit of this sequence is denoted by $\prod_{n=0}^{\infty} x_n$ (or $\prod_{n=0}^{\infty} x_n$ by abuse of notation). We leave to the reader the task of transcribing into multiplicative notation the properties of series which we have established.