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Accounts of absolute and conditional convergence always say that $$ \sum_{n=0}^\infty a_n = \sum_{n=0}^\infty a_{\sigma(n)} $$ if the series converges absolutely, if $\sigma$ is any bijection from the set of indices to itself.

Fubini's theorem tells us that $$ \sum_{n=0}^\infty\sum_{m=0}^\infty a_{n,m} = \sum_{m=0}^\infty\sum_{n=0}^\infty a_{n,m} $$ if we have absolute convergence. (Tonelli's theorem tells us that these are equal if all terms are nonnegative, regardless of whether the series converges or not.)

But in this answer, I explained how to rearrange $$ \sum_{n=0}^\infty\sum_{m=0}^\infty a_{n,m} $$ into $$ \sum_{n=0}^\infty \sum_{k=0}^n a_{k,n-k}. $$ What known theorems apply to this case, and where is this case mentioned in the literature?

How exotic does the class of rearrangements of series get? What other kinds of rearrangements don't fit instantly into one of the cases covered by the two categories above that are covered by standard results?

(I might post my own answer to this.)

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  • $\begingroup$ I've posted my own answer, but other arguments for this result may also be of interest, so don't hesitate to add those. $\endgroup$ Jun 7, 2013 at 22:09
  • $\begingroup$ @MinimusHeximus : I really don't think you should have more than two "$\sum$"s here. $\endgroup$ Jun 8, 2013 at 15:46

3 Answers 3

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Fubini's theorem gives us more than equality of the two iterated sums, one of which is $$ \sum_{n=0}^\infty {\Huge(} \underbrace{{}\qquad\sum_{m=0}^\infty a_{n,m}\qquad{}}_{\begin{smallmatrix}\text{$n$ remains constant} \\ \text{as $m$ goes through} \\ \text{the whole range.}\end{smallmatrix}} {\Huge)}. $$ Fubini's theorem tells us in addition that the two iterated sums are both equal to $$ \sum \{ a_{n,m} : (n,m) \in (\mathbb Z^{\ge0})^2 \}, $$ where that sum is defined as follows. Let $P=\{(n,m) : a_{n,m}\ge0\}$ and $N=\{(n,m) : a_{n,m}<0\}$. Then the sum is $$ \sup\left\{ \sum_{(n,m)\in A} a_{n,m} : A\subseteq P\text{ and } A\text{ is finite} \right\} - \sup\left\{ \sum_{(n,m)\in A} -a_{n,m} : A\subseteq N\text{ and } A\text{ is finite} \right\}. $$ Then the first proposition on rearrangements stated above tells us that this is equal to $$ \lim_{N\to\infty} \sum_{k=1}^N a_{n_k,m_k} $$ for every enumeration $(n_k,m_k)_{k=1}^\infty$ of $(\mathbb Z^{\ge0})^2$.

After that, observe that $$ \sum_{n=0}^\infty\left(\sum_{k=0}^n a_{k,n-k} \right) $$ is simply one such enumeration.

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It seems to me that you are alluding to the theory of unordered summation, which is discussed (for instance) in $\S$ 14.2.3 of these notes.

In particular it is discussed that unordered summation is equivalent to absolute convergence, and the application to double series is given as an exercise.

(In my opinion calling this phenomenon "Fubini's Theorem" is a bit fancier than necessary.)

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A couple of theorems (Theorems 1 & 2 described below) which are helpful in these kind of situations are given in [1]. In the first [1, p142, Theorem 4], the notion of rearrangement of an infinite series is extended to a partition of the series into finitely or infinitely many finite or infinite sub-series (where a sub-series is a series formed from a subsequence of the terms of the original series), and such sub-series are shown to sum to the original series provided the latter is absolutely convergent. A standard rearrangement of a series is thus a special case of such a partition, with every sub-series just a single term. The second theorem [1, p143, §90], which the author terms as 'Cauchy's Double Series Theorem', shows that under certain conditions the rows series, columns series, and diagonals series for a double series all have the same sum.

Example applications of Theorem 1 are (i) proving Theorem 2 below, and (ii) formal proof of the reciprocal formula for the Riemann Zeta function (eg. this answer to question Riemann zeta function - Euler product formula).

Example applications of Theorem 2 are (i) showing that the Cauchy product of two absolutely convergent series is absolutely convergent with sum the product of the sums of the two series, (ii) proving that a real power series function is analytic throughout its open interval of convergence (cf. [2, p176, Theorem 8.4]), and (iii) showing that the order of summation of a non-negative double series in the extended real number system $\overline{\mathbb{R}}$ can be changed without changing the sum. Application (iii) is useful in measure theory.

A further application of Theorems 1 & 2 is Theorem 3 and its counterpart Theorem 4. Theorem 3 is useful in deriving basic properties of Outer Measure on $\mathbb{R}$ [3, Chap 2], eg. in proving countable subadditivity on $P(\mathbb{R})$ and countable additivity on the 'system of intervals'  $\mathcal{I} = \{$ all countable unions of intervals in $\mathbb{R}\}$.

Below, the abbreviation 'AC' is used for 'absolutely convergent' and the term 'double series' is used to mean an infinite series each of whose terms is the sum of an infinite series, and the term 'double sequence' used to mean a sequence indexed on $\mathbb{N} \times \mathbb{N}$. Theorems 1 & 2 apply within $\mathbb{R}$ and $\mathbb{C}$, whilst application (iii) of Theorem 2 works within the extended real number system $\overline{\mathbb{R}}$, and Theorems 3 & 4 apply to non-negative term series in $\mathbb{R}$.

Theorem 1 (Rearrangement of Series in the Extended Sense)

Suppose AC series $S = \sum_{n = 1}^{\infty} a_n$ is partitioned into a finite or infinite sequence of finite or infinite sub-series (ie. every term $a_n$ of $S$ appears within exactly one of the sub-series) :

$$ \begin{array}{cccccccc} a_1^{(1)} & + & a_2^{(1)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = r^{(1)}) & \hspace{3em} & \mbox{Sub-series 1} \\[1ex] \vdots & & \vdots & & & \vdots & \hspace{3em} & \vdots \\[1ex] a_1^{(k)} & + & a_2^{(k)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = r^{(k)}) & \hspace{3em} & \mbox{Sub-series k} \\[1ex] \vdots & & \vdots & & & \vdots & & \vdots \end{array} $$

Then $\sum_{k = 1}^{\infty} r^{(k)}$ is AC with sum $S$.

Proof

Firstly note any sub-series of $S$ must also be AC. For any sub-series which is finite extend it by zeros to an infinite series. Let us first consider the case where there are infinitely many sub-series. Let the sequence of partial sums for $S$ be $(s_n)$. Let $\epsilon > 0$ and using the absolute convergence of $S$ choose $m$ so that $|a_{m + 1}| + |a_{m + 2}| + \cdots < \epsilon$ (note this also implies $|s_m - S| < \epsilon$).

Choose $M$ so each of $a_1, \ldots, a_m$ appear within the first $M$ sub-series. Then select an $N$ such that all of $a_1, \ldots, a_m$ appear within the first $N$ columns of the first $M$ rows of the above grid.

Now fix a row $k \geq M$ and consider the series which is the sum of the first $k$ sub-series and let its sequence of partial sums be $(b_n)$. Then $\forall n \geq N$, within the expression $|b_n - s_m|$ only terms $a_i$ with $i > m$ are left after subtracting $s_m = a_1 + \cdots + a_m$ from $b_n$, so that certainly $|b_n - s_m| \leq |a_{m + 1}| + |a_{m + 2}| + \cdots < \epsilon$. But then $\lim_{n \rightarrow \infty} |b_n - s_m| \leq \epsilon$, ie. $|r^{(1)} + \cdots + r^{(k)} - s_m| \leq \epsilon$. But then :

$$|r^{(1)} + \cdots + r^{(k)} - S| \leq |r^{(1)} + \cdots + r^{(k)} - s_m| + |s_m - S| < 2\epsilon.$$

But this is true for every $k \geq M$ and hence $\sum_{k = 1}^{\infty} r^{(k)} = S$, as required. To show this convergence is absolute we can apply what we have just proved to the AC series $\sum_{n = 1}^{\infty} |a_n|$, where we have a partition into the AC sub-series :

$$ \begin{array}{cccccccc} |a_1^{(1)}| & + & |a_2^{(1)}| & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = t^{(1)}) & \hspace{3em} & \mbox{Sub-series 1} \\[1ex] \vdots & & \vdots & & & \vdots & \hspace{3em} & \vdots \\[1ex] |a_1^{(k)}| & + & |a_2^{(k)}| & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = t^{(k)}) & \hspace{3em} & \mbox{Sub-series k} \\[1ex] \vdots & & \vdots & & & \vdots & & \vdots \end{array} $$

from which it follows that $\sum_{k = 1}^{\infty} t^{(k)} = \sum_{n = 1}^{\infty} |a_n|$. Now, applying the triangle inequality, we have $|r^{(k)}| = \left|\sum_{n = 1}^{\infty} a_n^{(k)}\right| \leq t^{(k)}, \forall k \geq 1$. Therefore by the comparison test $\sum_{k = 1}^{\infty} |r^{(k)}|$ is convergent, ie. $\sum_{k = 1}^{\infty} r^{(k)}$ is AC, as required.

In the case of a finite partition we can proceed as above except setting $M$ to be the total number of sub-series in the partition, and selecting $k = M$. Then we conclude $|r^{(1)} + \cdots + r^{(M)} - S| < 2\epsilon, \forall \epsilon > 0$, so that $r^{(1)} + \cdots + r^{(M)} = S$, as required. QED

Theorem 2 (Cauchy's Double Series Theorem)

Suppose we have a sequence of AC series :

$$ \begin{array}{cccccccc} a_1^{(1)} & + & a_2^{(1)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = r^{(1)}) & \hspace{3em} & \mbox{Series 1} \\[1ex] \vdots & & \vdots & & & \vdots & & \vdots \\[1ex] a_1^{(k)} & + & a_2^{(k)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = r^{(k)}) & \hspace{3em} & \mbox{Series k} \\[1ex] \vdots & & \vdots & & & \vdots & & \vdots \end{array} $$

such that the following condition is satisfied : if $b^{(k)}$ is the sum of absolute terms of the $k$th series, ie. $b^{(k)} = \sum_{n = 1}^{\infty} |a_n^{(k)}|$, then $\sum_{k = 1}^{\infty} b^{(k)}$ is convergent. Then :

  1. each individual column series $c^{(n)} = \sum_{k = 1}^{\infty} a_n^{(k)}$, for $n \in \mathbb{N}$, is AC.
  2. the series $\sum_{k = 1}^{\infty} r^{(k)}$ of row sums and the series $\sum_{n = 1}^{\infty} c^{(n)}$ of column sums are both AC and have the same sum.
  3. the term-by-term diagonal series and the diagonal sum based diagonal series are both AC series and have the same sum as (2).

Proof

Consider the term-by-term diagonal series $\sum_{n = 1}^{\infty} d_n$ obtained by taking a diagonal rearrangement of the individual terms in the above grid in the following fashion :

enter image description here

Given any $n$, there must be an $r$ such that all the terms $d_1, \ldots, d_n$ must be contained within the first $r$ rows, hence $|d_1| + \cdots + |d_n| \leq b^{(1)} + \cdots + b^{(r)} \leq \sum_{k = 1}^{\infty} b^{(k)}$, and so series $\sum_{n = 1}^{\infty} d_n$ is AC, and thus can be partitioned into sub-series in any manner, by theorem (1) above.

For each $n$, $c^{(n)}$ is a sub-series of AC series $\sum_{n = 1}^{\infty} d_n$, and hence is AC itself, thus establishing (1).

Both $\{c^{(n)} : n \in \mathbb{N}\}$ and $\{r^{(k)} : k \in \mathbb{N}\}$ are infinite partitions of $\sum_{n = 1}^{\infty} d_n$ into infinite sub-series and so by theorem (1) both are AC with sum $\sum_{n = 1}^{\infty} d_n$, thus establishing (2).

Let $\sum_{n = 1}^{\infty} e_n$ be the diagonal sum based diagonal series (so for example the 2\textsuperscript{nd} term of that series would be $a_1^{(2)} + a_2^{(1)}$). Then, $\{e^{(n)} : n \in \mathbb{N}\}$ is an infinite partition of $\sum_{n = 1}^{\infty} d_n$ into finite sub-series and so by theorem (1) it is AC with sum $\sum_{n = 1}^{\infty} d_n$, thus establishing (3). QED

Example Applications of Theorem 2

(i) Show that the Cauchy product of two AC series is AC with sum equal to the product of the sums of the two series

Let $\sum_{k = 1}^{\infty} a_k$ and $\sum_{n = 1}^{\infty} b_n$ be two AC series with sums $A, B$ respectively. Let $\sum_{i = 1}^{\infty} c_i$ be their Cauchy product, ie. $\forall i, c_i = a_1 b_i + a_2 b_{i - 1} + \cdots + a_i b_1$. Then $\sum_{i = 1}^{\infty} c_i$ is the diagonal sum based series (ie. the series $\sum_{n = 1}^{\infty} e_n$ in the proof of Theorem 2 above) for the following double series : $$ \begin{array}{c@{\hspace{1em}}|@{\hspace{1em}}c@{\hspace{2em}}c@{\hspace{2em}}c@{\hspace{2em}}c@{\hspace{2em}}c@{\hspace{9em}}c} & b_1 & b_2 & b_3 & \cdots & b_n \\[1ex] \hline \rule{0ex}{4ex} a_1 & a_1b_1 & a_1b_2 & a_1b_3 & \cdots & & \mbox{(sum = $r^{(1)}$)} \\[2ex] a_2 & a_2b_1 & a_2b_2 & a_2b_3 & \cdots & & \mbox{(sum = $r^{(2)}$)} \\[2ex] a_3 & a_3b_1 & a_3b_2 & a_3b_3 & \cdots & & \mbox{(sum = $r^{(3)}$)} \\[2ex] \vdots &&&& \ddots \\[2ex] a_k &&&&& a_k b_n & \mbox{(sum = $r^{(k)}$)} [2ex] \end{array} $$

for which the sum $v^{(k)}$ of absolute terms of the $k$\textsuperscript{th} row is $|a_k| \sum_{n = 1}^{\infty} |b_n|$, so that $\sum_{k = 1}^{\infty} v^{(k)}$ converges to $(\sum_{k = 1}^{\infty} |a_k|)(\sum_{n = 1}^{\infty} |b_n|) = A B$.

Thus the conditions of Theorem 2 apply and then by part (3) of that theorem we have that $\sum_{i = 1}^{\infty} c_i$ is AC with sum equal to $\sum_{k = 1}^{\infty} r^{(k)}$. But $r^{(k)} = a_k \sum_{n = 1}^{\infty} b_n = a_k B$, and hence $\sum_{i = 1}^{\infty} c_i = \sum_{k = 1}^{\infty} r^{(k)} = B \sum_{k = 1}^{\infty} a_k = BA$, as required. QED

(ii) Prove that a real power series function is analytic throughout its open interval of convergence (cf. [2, p176, Theorem 8.4])

By definition a function $f : I \rightarrow \mathbb{R}$, where $I$ is a non-zero width interval in $\mathbb{R}$, is \emph{analytic} at $x_0 \in I$ if there is a neighborhood $U$ of $x_0$ in $\mathbb{R}$ such that within $U \cap I$, $f(x)$ equals a power series function about $x_0$, ie. for some $a_0, a_1, \ldots \in \mathbb{R}$, $f(x) = \sum_{n = 0}^{\infty} a_n (x - x_0)^n, \forall x \in U \cap I$. It is readily shown that such an $f$ is analytic at $x_0$ iff there exists a neighborhood $U$ of $x_0$ in $\mathbb{R}$ such that $f$ is infinitely differentiable on $U \cap I$ and $f$ equals its Taylor Series expansion about $x_o$ on $U \cap I$, ie. $$ f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x - x_0)^n, \hspace{2em} \forall x \in U \cap I. $$

If we have a power series function $f(x) = \sum_{n = 0}^{\infty} a_n x^n$ with radius of convergence $R$, then with $I = U = (-R, R)$, $f$ is clearly analytic at $0$. However it does not immediately follow that $f$ is analytic \emph{throughout} $I$.

To prove $f$ is analytic at an arbitrary point $x_0 \in I$, we can use Theorem 2 to prove it equals its Taylor Series expansion in a neighborhood $U \subseteq I$ of $x_0$, namely $U = (x_0 - \lambda, x_0 + \lambda)$, where $\lambda = R - |x_0|$.

From the general properties of power series functions, $f(x)$ is infinitely differentiable in $I = (-R, R)$ (and hence in $U \cap I = U \subseteq I$) with : $$ f^{(n)}(x) = \sum_{k = n}^{\infty} \frac{k!}{(k - n)!} a_k x^{k - n}, \hspace{2em} \forall n \geq 0 \mbox{ and } |x| < R. $$

Thus setting $a = x_0$, and $b = x - x_0$ the Taylor Series expansion of $f(x)$ about $x_0$ is : \begin{eqnarray} g(x) & = & \sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!} (x - x_0)^n \label{eq:taylor-series} \tag{1} \\ & = & \sum_{n = 0}^{\infty} \left( \sum_{k = n}^{\infty} \binom{k}{n} a_k x_0^{k - n} \right) (x - x_0)^n \nonumber \\ & = & \sum_{n = 0}^{\infty} \;\; \sum_{k = n}^{\infty} \binom{k}{n} a_k a^{k - n} b^n \nonumber \\ \Rightarrow \hspace{3em} g(x) & = & \sum_{n = 0}^{\infty} \;\; \sum_{k = n}^{\infty} a_n^{(k)}, \label{eq:g-double-series} \tag{2} \end{eqnarray}

where $a_n^{(k)} = \binom{k}{n} a_k a^{k - n} b^n$ is defined $\forall n, k \in \mathbb{W}$ with $n \leq k$ ($\mathbb{W}$ = set of non-negative integers). We show this double series converges and equals $f(x), \; \forall x \in U$, and hence that $f$ is analytic at $x_0$.

Now for $x \in U$, $x = a + b$ is within the open interval $(-R, R)$ of convergence of $f(x)$, so : \begin{eqnarray} f(x) & = & \sum_{k = 0}^{\infty} a_k (a + b)^k \nonumber \\ & = & \sum_{k = 0}^{\infty} a_k \left( \sum_{n = 0}^{k} \binom{k}{n} a^{k - n} b^n \right) \nonumber \\ & = & \sum_{k = 0}^{\infty} \;\; \sum_{n = 0}^{k} \binom{k}{n} a_k a^{k - n} b^n \nonumber \\ \Rightarrow \hspace{2em} f(x) & = & \sum_{k = 0}^{\infty} \;\; \sum_{n = 0}^{k} a_n^{(k)} \label{eq:f-double-series} \tag{3} \end{eqnarray}

which is the sum of a double series as in Theorem 2, with each row series being a finite sum, so that the grid of terms has all zeros above the main diagonal : $$ \begin{array}{c@{\hspace{1em}}c@{\hspace{1em}}c@{\hspace{1em}}c@{\hspace{1em}}c@{\hspace{8em}}c} a_0^{(0)} &&&&& \mbox{(sum = $r^{(0)}$)} \\[2ex] a_0^{(1)} & a_1^{(1)} &&&& \mbox{(sum = $r^{(1)}$)} \\[2ex] a_0^{(2)} & a_1^{(2)} & a_2^{(2)} &&& \mbox{(sum = $r^{(2)}$)} \\[2ex] \vdots & & & \ddots && \vdots \\[2ex] a_0^{(k)} & a_1^{(k)} & \cdots & & a_k^{(k)} & \mbox{(sum = $r^{(k)}$)} \\[2ex] \vdots &&&&& \vdots \end{array} $$

For the condition of Theorem 2 to be satisfied we require $\sum_{k = 1}^{\infty} b^{(k)}$ to be convergent, where $b^{(k)}$ is the sum of absolute terms for row $k$. But : \begin{eqnarray*} |a_n^{(k)}| & = & \binom{k}{n} |a_k| |a|^{k - n} |b|^n \\ \Rightarrow \hspace{2em} b^{(k)} & = & \sum_{n = 0}^{k} |a_n^{(k)}| = |a_k| \sum_{n = 0}^{k} \binom{k}{n} |a|^{k - n} |b|^n \\ \Rightarrow \hspace{2em} b^{(k)} & = & |a_k| ( |a| + |b| )^k \end{eqnarray*}

where, since $x \in U$, $|a| + |b| = |x_0| + |x - x_0| < |x_0| + \lambda = R$, so that $|a| + |b|$ is within the open interval of convergence $(-R, R)$ of power series function $f(x) = \sum_{n = 0}^{\infty} a_n x^n$. Hence $\sum_{k = 1}^{\infty} b^{(k)}$ is convergent. Thus by Theorem 2, if $c^{(n)}$ is the $n$\textsuperscript{th} column sum then $\forall n$ this sum is AC, and $\sum_{n = 1}^{\infty} c^{(n)} = \sum_{k = 1}^{\infty} r^{(k)}$. But for a given $x \in U$, equation (\ref{eq:f-double-series}) tells us $f(x) = \sum_{k = 1}^{\infty} r^{(k)}$ and equation (\ref{eq:g-double-series}) tells us $g(x) = \sum_{n = 1}^{\infty} c^{(n)}$, with the radius of the Taylor Series (\ref{eq:taylor-series}) no less than $\lambda$. QED

(iii) Show that the order of summation of a non-negative double series in the extended real number system $\overline{\mathbb{R}}$ can be changed and the sum within $\overline{\mathbb{R}}$ remains the same

Let $a_{k,n}$ be a double sequence in $[0, \infty]$. Since we are working within $\overline{\mathbb{R}}$, and every term is non-negative, so that partial sums are always increasing, the following two double series are always convergent within $\overline{\mathbb{R}}$, either to a non-negative real number or to $\infty$ : \begin{eqnarray} \sum_{k=1}^{\infty}\;\sum_{n=1}^{\infty} a_{k, n} & = & \sum_{k=1}^{\infty} r^{(k)}, \hspace{2em} \mbox{where $r^{(k)} =$ sum in $\overline{\mathbb{R}}$ of $k^{\mathrm{th}}$ row} \label{eq:double-series-1} \tag{4} \\ \sum_{n=1}^{\infty}\;\sum_{k=1}^{\infty} a_{k, n} & = & \sum_{n=1}^{\infty} c^{(n)}, \hspace{2em} \mbox{where $c^{(n)} =$ sum in $\overline{\mathbb{R}}$ of $n^{\mathrm{th}}$ column} \label{eq:double-series-2} \tag{5} \end{eqnarray}

We require to show that (\ref{eq:double-series-1}) and (\ref{eq:double-series-2}) are equal. Recall by 'double series' we mean a series which has the sums of infinite series as its terms. Note also a series of non-negative terms in $\overline{\mathbb{R}}$ can converge to $\infty$ in two ways : (a) if it contains $\infty$ as a term, or (b) all its terms are in $\mathbb{R}$ but the partial sums are unbounded. It must either converge to $\infty$ or converge to a real number.

The proof falls into 2 cases :

Case $\sum_{k=1}^{\infty} r^{(k)} \in \mathbb{R}$ :

Then every $r^{(k)} \in \mathbb{R}$ and thus every $a_{k,n} \in \mathbb{R}$, so we are working within $\mathbb{R}$.

Then the Cauchy Double Series Theorem applies to the double series (\ref{eq:double-series-1}) so that every column sum $c^{(n)}$ converges within $\mathbb{R}$, and $\sum_{n=1}^{\infty} c^{(n)} = \sum_{k=1}^{\infty} r^{(k)}$ within $\mathbb{R}$ (and hence within $\overline{\mathbb{R}}$), which is the required result.

Case $\sum_{k=1}^{\infty} r^{(k)} = \infty$ :

In this case the condition for the Cauchy Double Series Theorem does not apply to (\ref{eq:double-series-1}). However if it were the case that $\sum_{n=1}^{\infty} c^{(n)} \in \mathbb{R}$ then every $c^{(n)} \in \mathbb{R}$ and every $a_{k,n} \in \mathbb{R}$ and we are working within $\mathbb{R}$. Then the Cauchy Double Series Theorem applies to the transpose of $(a_{k,n})$ to give $\sum_{k=1}^{\infty} r^{(k)}$ (= sum of col sums of the transpose series) $= \sum_{n=1}^{\infty} c^{(n)}$ (= sum of row sums of transpose series) $\in \mathbb{R}$ - a contradiction. Hence $\sum_{n=1}^{\infty} c^{(n)} \notin \mathbb{R}$ and thus $\sum_{n=1}^{\infty} c^{(n)} = \infty = \sum_{k=1}^{\infty} r^{(k)}$, again giving the required result. QED

Theorem 3

Suppose we have a sequence of non-negative term series in $\mathbb{R}$ whose rows $r^{(k)}$ converge in $\mathbb{R}$ :

$$ \begin{array}{cccccccc} a_1^{(1)} & + & a_2^{(1)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = r^{(1)}) \\[1ex] \vdots & & \vdots & & & \vdots & \hspace{3em} \\[1ex] a_1^{(k)} & + & a_2^{(k)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & (\mbox{sum} = r^{(k)}) \\[1ex] \vdots & & \vdots & & & \vdots \end{array} $$

with $\sum_{n = 1}^{\infty} d_n$ the term-by-term diagonal series as indicated in Fig. 1 above. Then $\sum_{k = 1}^{\infty} r^{(k)}$ is convergent iff $\sum_{n = 1}^{\infty} d_n$ is convergent, and in this case these sums are equal.

Proof

$\Rightarrow.$ Since we are dealing with non-negative term series, absolute convergence is the same as convergence. Hence Cauchy Double Series Theorem part (3) implies the result.

$\Leftarrow.$ The above row series form a partition into sub-series of the diagonal series $\sum_{n = 1}^{\infty} d_n$ in the extended sense. Since the latter series is AC, Theorem 1 implies its sum equals the sum of sums of these sub-series, ie. $\sum_{n = 1}^{\infty} d_n = \sum_{k = 1}^{\infty} r^{(k)}$. QED

Theorem 4

Suppose we have a sequence of non-negative term series in $\mathbb{R}$ whose columns $c^{(n)}$ converge in $\mathbb{R}$ :

$$ \begin{array}{cccccccc} a_1^{(1)} & + & a_2^{(1)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & \hspace{4em} (\mbox{row 1}) \\[1ex] \vdots & & \vdots & & & \hspace{3em} \vdots \\[1ex] a_1^{(k)} & + & a_2^{(k)} & + & \cdots \cdots \cdots \cdots \cdots \cdots & \hspace{4em} (\mbox{row k}) \\[1ex] \vdots & & \vdots & & & \hspace{3em} \vdots \\[1ex] = c^{(1)} & & = c^{(2)} & & \cdots \cdots \cdots \cdots \cdots \cdots \end{array} $$

with $\sum_{n = 1}^{\infty} d_n$ the term-by-term diagonal series as indicated in Fig. 1 above. Then $\sum_{n=1}^{\infty} c^{(n)}$ is convergent iff $\sum_{n = 1}^{\infty} d_n$ is convergent, and in this case these sums are equal.

Proof

Apply Theorem 3 to the transpose of the double sequence $(a_n^{(k)})$ and use the fact that the term-by-term diagonal series for this tranpose is a standard rearrangement of the diagonal series $\sum_{n=1}^{\infty} d_n$ of Theorem 3, so that the latter two have the same sum and convergence. QED

References

[1] Konrad Knopp (1954), Theory and Application of Infinite Series, 2nd English Edition translated from 4th German Edition, Blackie & Sons.

[2] Walter Rudin (1976), Principles of Mathematical Analysis, 3rd Edition, McGraw-Hill.

[3] Sheldon Axler (2020), Measure, Integration & Real Analysis, Springer Graduate Texts in Mathematics, https://measure.axler.net/.

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