Case One

Theorem. Let xn≥0x_n \ge 0 for n∈N+n \in \mathbb{N}^+. Then

∑n≥1xn=sup⁡{∑n∈Exn:E⊆N+∧E is finite }. \sum_{n \ge 1} x_n = \sup \left\{ \sum_{n \in E} x_n : E \subseteq \mathbb{N}^+ \wedge E \text{ is finite } \right\}.

Proof. First,

∑n≥1xn=lim⁡k→∞∑n∈[k]xn≤sup⁡{∑n∈Exn:E⊆N+∧E is finite }. \sum_{n \ge 1} x_n = \lim_{k \rightarrow \infty} \sum_{n \in [k]} x_n \le \sup \left\{ \sum_{n \in E} x_n : E \subseteq \mathbb{N}^+ \wedge E \text{ is finite } \right\}.

Second, for each E⊆N+E \subseteq \mathbb{N}^+ such that EE is finite, there exists an k∈N+k \in \mathbb{N}^+ that is larger than all numbers in EE. Hence,

∑n≥1xn≥∑n∈[k]xn≥∑n∈Exn. \sum_{n \ge 1} x_n \ge \sum_{n \in [k]} x_n \ge \sum_{n \in E} x_n.

Taking supremum over the RHS finishes the proof.

□\square

Theorem. Let xn≥0x_n \ge 0 for n∈N+n \in \mathbb{N}^+, σ\sigma be a permutation of N+\mathbb{N}^+, i.e., an one-to-one mapping of N+\mathbb{N}^+ onto itself, and yn=xσ(n)y_n = x_{\sigma(n)} for n∈N+n \in \mathbb{N}^+. Then

∑n≥1xn=∑n≥1yn. \sum_{n \ge 1} x_n = \sum_{n \ge 1} y_n.

Proof. First, for each k∈N+k \in \mathbb{N}^+, there exits a k′∈N+k' \in \mathbb{N}^+, such that {x1,…,xk}⊆{y1,…,yk′}\{ x_1, \ldots, x_k \} \subseteq \{ y_1, \ldots, y_{k'} \}. Therefore,

∑n∈[k]xn≤∑n∈[k′]yn≤∑n≥1yn. \sum_{n \in [k]} x_n \le \sum_{n \in [k']} y_n \le \sum_{n \ge 1} y_n.

Taking k→∞k \rightarrow \infty proves ∑n≥1xn≤∑n≥1yn\sum_{n \ge 1} x_n \le \sum_{n \ge 1} y_n.

Second, for each k∈N+k \in \mathbb{N}^+, let k′=max⁡i∈[k]σ(i)k' = \max_{i \in [k] } \sigma(i). We have {y1,…,yk}⊆{x1,…,xk′}\{ y_1, \ldots, y_k \} \subseteq \{ x_1, \ldots, x_{k'} \}, and

∑n≥1xn≥∑n∈[k′]xn≥∑n∈[k]yn. \sum_{n \ge 1} x_n \ge \sum_{n \in [k']} x_n \ge \sum_{n \in [k]} y_n.

Taking k→∞k \rightarrow \infty proves ∑n≥1xn≥∑n≥1yn\sum_{n \ge 1} x_n \ge \sum_{n \ge 1} y_n.

□\square

Case Two

Consider a matrix A=(ai,j)A = (a_{i,j}) with countable rows and columns.

Suppose that for each row i∈N+i \in \mathbb{N}^+, the following limit exits

xi=∑j≥1ai,j, x_i = \sum_{j \ge 1} a_{i, j},

and that the limit of the sum of xix_i exits

s=∑i≥1xi. s = \sum_{i \ge 1} x_i.

Suppose that for each column j∈N+j \in \mathbb{N}^+, the following limit exits

yj=∑i≥1ai,j. y_j = \sum_{i \ge 1} a_{i,j}.

Moreover, ∀ϵ>0\forall \epsilon > 0, ∃N∈N+\exists N \in \mathbb{N}^+, s.t., if n>Nn > N, for all m∈N+m \in \mathbb{N}^+, we have

∣∑j=1m∑i=1nai,j−∑j=1myj∣≤ϵ. \left| \sum_{j = 1}^m \sum_{i = 1}^n a_{i,j} - \sum_{j = 1}^m y_j \right| \le \epsilon.

then the limit of ∑j≥1yj\sum_{j \ge 1} y_j exits, and

∑i≥1xi=∑j≥1yj. \sum_{i \ge 1} x_i = \sum_{j \ge 1} y_j.

Proof: For each m∈N+m \in \mathbb{N}^+,

∣s−∑j=1myj∣≤∣s−∑i=1nxi∣+∣∑i=1nxi−∑i=1n∑j=1mai,j∣+∣∑i=1n∑j=1mai,j−∑j=1myj∣=∣s−∑i=1nxi∣+∣∑i=1n(xi−∑j=1mai,j)∣+∣∑i=1n∑j=1mai,j−∑j=1myj∣ \begin{aligned} \left| s - \sum_{j = 1}^m y_j \right| &\le \left| s - \sum_{i = 1}^n x_i \right| + \left| \sum_{i = 1}^n x_i - \sum_{i = 1}^n \sum_{j = 1}^m a_{i,j} \right| + \left|\sum_{i = 1}^n \sum_{j = 1}^m a_{i,j} - \sum_{j = 1}^m y_j \right| \\ &= \left|s - \sum_{i = 1}^n x_i \right| + \left| \sum_{i = 1}^n \left( x_i - \sum_{j = 1}^m a_{i,j} \right) \right| + \left|\sum_{i = 1}^n \sum_{j = 1}^m a_{i,j} - \sum_{j = 1}^m y_j \right| \end{aligned}

First, we choose large enough nn, such that

∣s−∑i=1nxi∣+∣∑i=1n∑j=1mai,j−∑j=1myj∣≤2ϵ. \left|s - \sum_{i = 1}^n x_i \right| + \left|\sum_{i = 1}^n \sum_{j = 1}^m a_{i,j} - \sum_{j = 1}^m y_j \right| \le 2 \epsilon.

Fix an nn, we can choose large enough mm, such that

∣∑i=1n(xi−∑j=1mai,j)∣≤ϵ, \left| \sum_{i = 1}^n \left( x_i - \sum_{j = 1}^m a_{i,j} \right) \right| \le \epsilon,

Therefore,

∣s−∑j=1myj∣≤3ϵ. \left| s - \sum_{j = 1}^m y_j \right| \le 3 \epsilon.

□\square