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2. And also, if $\tau: \mathbb N \to \mathbb N \times \mathbb N$ is another bijection, arranging the sequence of indices into an "infinite matrix" and vice versa, and $b: \mathbb N \times \mathbb N \to X$ is such a matrix. Then if

1. First consider the case of a positive series, and establish that the sum of original (on $k$) series is $\leq$ the sum of the transposed series (on $\sigma(k)$). Since the former series may also be viewed as transposition of the latter, we also have the opposite inequality -- i.e. the sums are equal.

   The extrapolation onto the general case of arbitrary such series is done by virtue of grouping the elements of the same sign, i.e. representation of the series as double sum: first by the elements having the same sign, then by the resulting values applied to the corresponding signs. (e.g. in case of $X = \mathbb R$, it just two signs, $+$ and $-$)

2. In one direction, assuming $\bar{B} := \sum_{k=0}^{\infty} |b_{\tau(k)}|$ is defined.

$$ \left| B - \sum_{k = 0}^{k = \phi(\epsilon)} b_{\tau(k)} \right| = \left| \sum_{k = \phi(\epsilon) + 1}^{\infty} b_{\tau(k)} \right| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| \lt \epsilon $$

where $\phi: \mathbb R_{\gt 0} \to \mathbb N$ is the "estimate map", from the definition of the limit, for the absolute convergence of $b_{\tau(k)}$.

Then we need to twist the estimation a little bit:
Let $\psi (f, k) := \max_{0 \le x \le k} f(x)$, and $(m, n) := \psi (\tau, \phi (\epsilon))$ defines the "index rectangle", based at $(0, 0)$ and containing all the values $\tau(0), \dots, \tau(\phi(\epsilon))$. So that

$$ \sum_{k = 0}^{k = \phi(\epsilon)} |b_{\tau(k)}| \leq \sum_{i = 0}^{i = m} \sum_{j = 0}^{j = n} |b_{i,j}| \leq \sum_{i = 0}^{i = m} \sum_{j = 0}^{\infty} |b_{i,j}| $$

subtracting this inequality from $\bar{B}$

$$ \sum_{i = m + 1}^{\infty} \sum_{j = 0}^{\infty} |b_{i,j}| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| $$

and, as above, obtain

$$ \left| B - \sum_{i = 0}^{i = m} \sum_{j = 0}^{\infty} b_{i,j} \right| = \left| \sum_{i = m + 1}^{\infty} \sum_{j = 0}^{\infty} b_{i,j} \right| \leq \sum_{i = m + 1}^{\infty} \sum_{j = 0}^{\infty} |b_{i,j}| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| \lt \epsilon $$

So finally we've constructed the desired estimate map, taking $\epsilon$ to $m$, such that

$$ \left| B - \sum_{i = 0}^{i = m} B'_i \right| \lt \epsilon $$

It shows that

In the other direction, assuming $\bar{B} := \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} |b_{i, j}|$ is defined. Then

2. And also, if $\tau: \mathbb N \to \mathbb N \times \mathbb N$ is another bijection, arranging the sequence of indices into an "infinite matrix" and vice versa, and $b: \mathbb N \times \mathbb N \to X$ is such a matrix. Then

1. First consider the case of a positive series, and establish that the sum of original (on $k$) series is $\leq$ the sum of the transposed series (on $\sigma(k)$). Since the former series may also be viewed as transposition of the latter, we also have the opposite inequality -- i.e. the sums are equal.

   The extrapolation onto the general case of arbitrary such series is done by virtue of grouping the elements of the same sign, i.e. representation of the series as double sum: first by the elements having the same sign, then by the resulting values applied to the corresponding signs. (e.g. in case of $X = \mathbb R$, it just two signs, $+$ and $-$)

2. In one direction, assuming $\bar{B} := \sum_{k=0}^{\infty} |b_{\tau(k)}|$ is defined.

$$ \left| B - \sum_{k = 0}^{k = \phi(\epsilon)} b_{\tau(k)} \right| = \left| \sum_{k = \phi(\epsilon) + 1}^{\infty} b_{\tau(k)} \right| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| \lt \epsilon $$

where $\phi: \mathbb R_{\gt 0} \to \mathbb N$ is the "estimate map", from the definition of the limit, for the absolute convergence for $b_{\tau(k)}$.

And if $S$ is an arbitrary subset of indices, such that $[0, \phi(\epsilon)] \subseteq S \subseteq \mathbb N$, then also

$$ \left| B - \sum_{k \in S} b_{\tau(k)} \right| = \left| \sum_{k \in \mathbb N \setminus S} b_{\tau(k)} \right| \leq \sum_{k \in \mathbb N \setminus S} |b_{\tau(k)}| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| \lt \epsilon $$

generally, if $S$ and $S'$ are two such sets, and $S \subseteq S'$, then $\left| B - \sum_{k \in S'} b_{\tau(k)} \right| \leq \left| B - \sum_{k \in S } b_{\tau(k)} \right|$.

Let $\psi (f, k) := \max_{0 \le x \le k} f(x)$, and $(m, n) := \psi (\tau, \phi (\epsilon))$ defines the "index rectangle", based at $(0, 0)$ and containing all the values $\tau(0), \dots, \tau(\phi(\epsilon))$. We define $S$ to be $\tau^{-1}([0, m], [0, n])$, i.e. a set of such an indices, that are mapped by $\tau$ into the rectangle (large enough to satisfy the requirements on $S$). Then

$$ \left| B - \sum_{k \in S} b_{\tau(k)} \right| = \left| B - \sum_{i = 0}^{i = m} \sum_{j = 0}^{j = n} b_{i,j} \right| $$

and, since all such expressions are bound by $\epsilon$, and may only get smaller with $m$ and $n$, we also have

$$ \left| B - \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} b_{i,j} \right| < \epsilon $$

Since $\epsilon$ may be arbitrarily small, necessary

In the other direction, assuming $\bar{B} := \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} |b_{i, j}|$ is defined. Then

In the other direction, assuming absolute convergence for $b_{i,j}$:

First, we ensure absolute convergence of the series $b_{\tau(k)}$, by finding some "index rectangle" for each set $\tau(0), \dots, \tau(k)$, and comparing the sums corresponding to these index sets. The equality of the limits follows from the previous result.

In the other direction, assuming $\bar{B} := \sum_{i=0}^{\infty} \sum_{j=0}^{\infty} |b_{i, j}|$ is defined. Then

$$ \sum_{i=0}^{i=m} \sum_{j=0}^{j=n} |b_{i, j}| \leq \bar{B} $$

and

$$ \sum_{k=0}^{k=l} |b_{\tau(k)}| \leq \sum_{i=0}^{i=m} \sum_{j=0}^{j=n} |b_{i, j}| $$

where as before $(m, n) := \psi (\tau, l)$.

So $l \mapsto \sum_{k=0}^{k=l} |b_{\tau(k)}|$, the sequence of positive partial sums, is increasing and bounded, therefore, converging.

The equality of the limits follows from the previous result.

1. First consider the case of a positive series, and establish that the sum of original (on $k$) series is $\leq$ the sum of the transposed series (on $\sigma(k)$). Since the former series may also be viewed as transposition of the latter, we also have the opposite inequality -- i.e. the sums are equal.

   The extrapolation onto the general case of arbitrary such series is done by virtue of grouping the elements of the same sign, i.e. representation of the series as double sum: first by the elements having the same sign, then by the resulting values applied to the corresponding signs. (e.g. in case of $X = \mathbb R$, it just two signs, $+$ and $-$)

2. In one direction, assuming absolute convergence for $b_{\tau(k)}$:

The sum of absolute values

$$ \bar{B} := \sum_{k = 0}^{\infty} |b_{\tau(k)}| $$

is an upper bound for the increasing sequences of partial sums

$$ \bar{B'}_i(m) := \sum_{j = 0}^{j = m} |b_{i,j}| $$

so these are converging:

$$ B'_i := \sum_{j = 0}^{\infty} b_{i,j} $$

and we want yet to check that

$$ \sum_{i = 0}^{\infty} B'_i = B $$

where

$$ B := \sum_{k = 0}^{\infty} b_{\tau(k)} $$

Then we need to twist the estimation a little bit:
Let $\psi (f, k) := \max_{0 \le x \le k} f(x)$, and $(m, n) := \psi (\tau, \phi (\epsilon))$ defines the "index rectangle", based at $(0, 0)$ and containing all the values $\tau(0), \dots, \tau(\phi(\epsilon))$.
  So we can introduce yet another estimation:

and, as above (but harder to write), obtain

$$ \left| B - \sum_{i = 0}^{i = m} \sum_{j = 0}^{j = n} b_{i,j} \right| \lt \epsilon $$

and

$$ \left| B - \sum_{i = 0}^{i = m} \sum_{j = 0}^{\infty} b_{i,j} \right| \lt \epsilon $$

1. First consider the case of a positive series, and establish that the sum of original (on $k$) series is $\leq$ the sum of the transposed series (on $\sigma(k)$). Since the former series may also be viewed as transposition of the latter, we also have the opposite inequality -- i.e. the sums are equal.

   The extrapolation onto the general case of arbitrary such series is done by virtue of grouping the elements of the same sign, i.e. representation of the series as double sum: first by the elements having the same sign, then by the resulting values applied to the corresponding signs. (e.g. in case of $X = \mathbb R$, it just two signs, $+$ and $-$)

2. In one direction, assuming $\bar{B} := \sum_{k=0}^{\infty} |b_{\tau(k)}|$ is defined.

Then $\bar{B}$ is an upper bound for the increasing sequences of partial sums:

$$ \bar{B}'_i(m) := \sum_{j = 0}^{j = m} |b_{i,j}| \leq \bar{B} $$

so these sums are converging:

$$ \sum_{j = 0}^{\infty} |b_{i,j}| \text{ and } \sum_{j = 0}^{\infty} b_{i,j} =: B'_i $$

and we want yet to check that

$$ \sum_{i = 0}^{\infty} B'_i = B := \sum_{k = 0}^{\infty} b_{\tau(k)} $$

Then we need to twist the estimation a little bit:
Let $\psi (f, k) := \max_{0 \le x \le k} f(x)$, and $(m, n) := \psi (\tau, \phi (\epsilon))$ defines the "index rectangle", based at $(0, 0)$ and containing all the values $\tau(0), \dots, \tau(\phi(\epsilon))$. So that

subtracting this inequality from $\bar{B}$

$$ \sum_{i = m + 1}^{\infty} \sum_{j = 0}^{\infty} |b_{i,j}| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| $$

and, as above, obtain

$$ \left| B - \sum_{i = 0}^{i = m} \sum_{j = 0}^{\infty} b_{i,j} \right| = \left| \sum_{i = m + 1}^{\infty} \sum_{j = 0}^{\infty} b_{i,j} \right| \leq \sum_{i = m + 1}^{\infty} \sum_{j = 0}^{\infty} |b_{i,j}| \leq \sum_{k = \phi(\epsilon) + 1}^{\infty} |b_{\tau(k)}| \lt \epsilon $$