5.10. Double Sequences and Double Series

This chapter will extend the notion of sequences: So far we have dealt with one-dimensional lists of real numbers, a point of view that is also supported by our notation. It would be an interesting option to extend our elaboration on two-dimensional lists as well. Although we won't get really new results (see e.g. [5.10.9] - [5.10.12]) double sequences play an important role under technical aspects. They will provide basic criteria ([5.10.23] and [5.10.26]) for changing the order of limit formation.

Definition: We call each function

(a_{n m}) : ℕ^{*} \times ℕ^{*} \to ℝ

[5.10.1]

a double sequence (in $ℝ$ ). The function

(\sum_{i = 0}^{n} \sum_{j = 0}^{m} a_{i j}) : ℕ \times ℕ \to ℝ

[5.10.2]

is called the double series with respect to ${(a_{n m})}_{\begin{array}{l} n \geq 0 \\ m \geq 0 \end{array}}$ .

Consider:

As with usual sequences we will call functions from $ℤ^{\geq k} \times ℤ^{\geq l} \to ℝ$ double sequences as well. We will note them (as already done in [5.10.2]) appropriately like this ${(a_{n m})}_{\begin{array}{l} n \geq k \\ m \geq l \end{array}}$ .
Value tables to double sequences are of course to be displayed in a plane. In doing this we regard n as the row index running down and m as the column index running to the right. As an example we have:

$(n + {(- 1)}^{n} m) = (\begin{matrix} 0 & - 1 & - 2 & - 3 & \dots \\ 3 & 4 & 5 & 6 & \dots \\ 2 & 1 & 0 & - 1 & \dots \\ 5 & 6 & 7 & 8 & \dots \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋱ \end{matrix})$

Now we transfer the standard terms from the common sequences.

Definition: A double sequence $(a_{n m})$ is called

1. bounded, if there is an $n_{0} \in ℕ^{*}$ and a $c \in ℝ$ such that $\| a_{n m} \| \leq c for all n, m \geq n_{0}$	[5.10.3]
2. increasing, if $n_{1} \leq n_{2} \land m_{1} \leq m_{2} \Rightarrow a_{n_{1} m_{1}} \leq a_{n_{2} m_{2}}$ holds for each $(n_{1}, m_{1}), (n_{2}, m_{2}) \in ℕ^{} \times ℕ^{}$	[5.10.4]
3. decreasing, if $n_{1} \leq n_{2} \land m_{1} \leq m_{2} \Rightarrow a_{n_{1} m_{1}} \geq a_{n_{2} m_{2}}$ holds for each $(n_{1}, m_{1}), (n_{2}, m_{2}) \in ℕ^{} \times ℕ^{}$	[5.10.5]
4. convergent to $g \in ℝ$ (symbolically: $a_{n m} \to g$ ), if there is an $n_{0} \in ℕ^{*}$ for every $ε > 0$ such that $\| a_{n m} - g \| < ε for all n, m \geq n_{0}$	[5.10.6]
5. Cauchy sequence, if there is an $n_{0} \in ℕ^{*}$ for each $ε > 0$ such that $\| a_{n_{1} m_{1}} - a_{n_{2} m_{2}} \| < ε for all n_{1} > n_{2} \geq n_{0} \land m_{1} > m_{2} \geq n_{0}$	[5.10.7]

Consider:

The formulation of boundedness (and of the Cauchy criterion) has been adapted to the nature of double sequences and is not a pure copy of the original definition. Nevertheless they are a sound - to witness subsequently - generalization of [5.3.11] as we see from

$| a_{n} | \leq c for all n \geq n_{0} \Rightarrow | a_{n} | \leq \max {c, | a_{1} |, \dots, | a_{n_{0} - 1} |} for all n$ .

Example:

${(- 1)}^{n + m} (\frac{1}{n} + \frac{1}{m}) \to 0$

[5.10.8]

For a given $ε > 0$ choose $n_{0} > \frac{2}{ε}$ , then every $n, m \geq n_{0}$ holds:

| {(- 1)}^{n + m} (\frac{1}{n} + \frac{1}{m}) | = \frac{1}{n} + \frac{1}{m} < \frac{ε}{2} + \frac{ε}{2} = ε

With the new terms being quite close to the old ones we expect similar properties. For proving them the following proposition is helpful. Before let us call a function

ϕ = (ϕ_{1}, ϕ_{2}) : ℕ^{*} \to ℕ^{*} \times ℕ^{*}

strictly increasing if both coordinate functions $ϕ_{1} und ϕ_{2}$ are strictly increasing.

Note that for any double sequence $(a_{n m})$ the function $(a_{n m}) \circ ϕ = (a_{ϕ (n)})$ is a common sequence. Furthermore, by an easy induction argument, we have:

ϕ_{i} (n) \geq n

for all n

Proposition: For any double sequence $(a_{n m})$ we have:

1. $(a_{n m})$ bounded $\Leftrightarrow (a_{ϕ (n)})$ bounded for all strictly increasing $ϕ$	[5.10.9]
2. $(a_{n m})$ monotone $\Leftrightarrow (a_{ϕ (n)})$ monotone for all strictly increasing $ϕ$	[5.10.10]
3. $a_{n m} \to g \Leftrightarrow a_{ϕ (n)} \to g$ for all strictly increasing $ϕ$	[5.10.11]
4. $(a_{n m})$ Cauchy sequence $\Leftrightarrow (a_{ϕ (n)})$ Cauchy sequence for all strictly increasing $ϕ$	[5.10.12]

Proof:

1. ►
" $\Rightarrow$ "  Let $| a_{n m} | \leq c for all n, m \geq n_{0}$ .

For $n \geq n_{0}$ we have $ϕ_{i} (n) \geq ϕ_{i} (n_{0}) \geq n_{0}$ and thus: $| a_{ϕ (n)} | \leq c for all n \geq n_{0}$ . Consequently:

$| a_{ϕ (n)} | \leq \max {c, | a_{ϕ (1)} |, \dots, | a_{ϕ (n_{0} - 1)} |} for all n$ .

" $\Leftarrow$ "  is shown indirectly: If $(a_{n m})$ is unbounded, the assertion

$| a_{n m} | \leq k for all n, m \geq k + 1$

can't be true for any k. So there are numbers $n_{k}, m_{k} \in ℕ^{> k}$ for every $k \in ℕ^{*}$ such that

$| a_{n_{k} m_{k}} | > k$ .

Now we set by recursion

$\begin{array}{l} ϕ (1) ≔ (n_{1}, m_{1}) \\ ϕ (i + 1) ≔ (n_{p}, m_{p}) \end{array}$
with $p ≔ \max {ϕ_{1} (1), ϕ_{2} (1), \dots, ϕ_{1} (i), ϕ_{2} (i), i + 1}$ .

$ϕ$ is strictly increasing as we have for all $i, j \in ℕ^{*}, i \geq j$ :

$\begin{array}{l} ϕ_{1} (i + 1) = n_{p} > p \geq ϕ_{1} (j) \\ ϕ_{2} (i + 1) = m_{p} > p \geq ϕ_{2} (j) \end{array}$

The estimate $| a_{ϕ (i + 1)} | = | a_{n_{p} m_{p}} | > p \geq i + 1$ now proves $(a_{ϕ (n)})$ to be an unbounded sequence.   Contradiction!

2. ►
For simplicity we only consider increasing sequences.

" $\Rightarrow$ ":  For $j \leq i$ we have $ϕ_{1} (j) \leq ϕ_{1} (i) \land ϕ_{2} (j) \leq ϕ_{2} (i)$ and thus: $a_{ϕ (j)} \leq a_{ϕ (i)}$ .

" $\Leftarrow$ ":  Again we proceed indirectly. Suppose there are two pairs of numbers $(n_{1}, m_{1}), (n_{2}, m_{2})$ with $n_{1} < n_{2} \land m_{1} < m_{2}$ such that $a_{n_{1} m_{1}} > a_{n_{2} m_{2}}$ . We set

$ϕ (1) ≔ (n_{1}, m_{1}), ϕ (2) ≔ (n_{2}, m_{2}) and ϕ (i + 2) ≔ (n_{2} + i, m_{2} + i)$ .

Obviously $ϕ$ is strictly increasing as is easily seen from the value tables:

$\begin{array}{l} (ϕ_{1}) = (n_{1}, n_{2}, n_{2} + 1, n_{2} + 2, \dots) \\ (ϕ_{2}) = (m_{1}, m_{2}, m_{2} + 1, m_{2} + 2, \dots) \end{array}$

but $(a_{ϕ (n)})$ is no increasing sequence.   Contradiction!

3. ►
" $\Rightarrow$ ":  For $ε > 0$ there is an $n_{0} \in ℕ^{*}$ such that

$| a_{n m} - g | < ε for all n, m \geq n_{0}$ .

As $ϕ_{i} (n) \geq n$ we have more than ever for all $n \geq n_{0}$ :

$| a_{ϕ (n)} - g | < ε$

" $\Leftarrow$ ":  As usual we argue indirectly. Suppose $(a_{n m})$ does not converge to g. Then there is an $ε > 0$ such that there are numbers $n_{k}, m_{k} \in ℕ^{\geq k + 1}$ for every $k \in ℕ^{*}$ with

$| a_{n_{k} m_{k}} - g | \geq ε$ .

Similar to 1. the recursion

$\begin{array}{l} ϕ (1) ≔ (n_{1}, m_{1}) \\ ϕ (i + 1) ≔ (n_{p}, m_{p}), \end{array}$

with $p ≔ \max {ϕ_{1} (1), ϕ_{2} (1), \dots, ϕ_{1} (i), ϕ_{2} (i)}$ provides a strictly increasing $ϕ$ such that

$| a_{ϕ (n)} - g | \geq ε for all n$ .

$(a_{ϕ (n)})$ thus is divergent.   Contradiction!

4. ►
" $\Rightarrow$ ":  If for an arbitrary $ε > 0$

$| a_{n_{1} m_{1}} - a_{n_{2} m_{2}} | < ε holds for all n_{1} > n_{2} \geq n_{0} \land m_{1} > m_{2} \geq n_{0}$ ,

we especially have (note: $ϕ_{i} (n) > ϕ_{i} (m) \geq m \geq n_{0}$ )

$| a_{ϕ (n)} - a_{ϕ (m)} | < ε for all n > m \geq n_{0}$ .

This estimate stays valid if $n \leq m$ (interchanging the addends) and thus is valid for all $n, m \geq n_{0}$ .

" $\Leftarrow$ ":  If $(a_{n m})$ is no Cauchy sequence there is an $ε > 0$ such that there are numbers $n_{k} > r_{k} > k$ and $m_{k} > s_{k} > k$ for every $k \in ℕ^{*}$ so that

$| a_{n_{k} m_{k}} - a_{r_{k} s_{k}} | \geq ε$ .

From the two-stage recursion

$\begin{array}{l} ϕ (1) ≔ (r_{1}, s_{1}) \land ϕ (2) ≔ (n_{1}, m_{1}) \\ ϕ (2 i + 1) ≔ (r_{p}, s_{p}) \land ϕ (2 i + 2) ≔ (n_{p}, m_{p}) \end{array}$

with $p ≔ \max {ϕ_{1} (1), ϕ_{2} (1), \dots, ϕ_{1} (2 i), ϕ_{2} (2 i)}$ we get a strictly increasing $ϕ$ such that

$| a_{ϕ (2 n)} - a_{ϕ (2 n - 1)} | \geq ε for all n$ .

From that we see that $(a_{ϕ (n)})$ is no Cauchy sequence.   Contradiction!

With [5.10.11] the uniquness of the limit is guaranteed for double sequences as well: If there were two different limits for the double sequence $(a_{n m})$ the common sequence $(a_{n n})$ e.g. will also allow two different limits. Thus we may note the convergence $a_{n m} \to g$ in the more familiar way:

\lim_{n, m \to \infty} a_{n m} = g

Many properties of common sequences are now easily transferred using [5.10.9] to [5.10.12]. We start with the basic limit theorems:

Proposition:

1. $a_{n m} \to a \land b_{n m} \to b \Rightarrow a_{n m} + b_{n m} \to a + b$	[5.10.13]
2. $a_{n m} \to a \land b_{n m} \to b \Rightarrow a_{n m} - b_{n m} \to a - b$	[5.10.14]
3. $a_{n m} \to a \land b_{n m} \to b \Rightarrow a_{n m} \cdot b_{n m} \to a \cdot b$	[5.10.15]
4. $a_{n m} \to a \land b_{n m} \to b \Rightarrow \frac{a_{n m}}{b_{n m}} \to \frac{a}{b}$ , if $b, b_{n m} \neq 0$	[5.10.16]

Proof: As an example we prove 1.: For each strictly increasing $ϕ$ we have:

a_{ϕ (n)} \to a \land b_{ϕ (n)} \to b

From the first limit theorem we thus get (cf. [5.6.1]): $a_{ϕ (n)} + b_{ϕ (n)} \to a + b$ , but that means:

a_{n m} + b_{n m} \to a + b

But also all the important convergence criteria are transferred to double sequences without difficulty.

Proposition:

1. $(a_{n m})$ convergent $\Rightarrow (a_{n m})$ bounded	[5.10.17]
2. $(a_{n m})$ monotone and bounded $\Rightarrow (a_{n m})$ convergent	[5.10.18]
3. $(a_{n m})$ convergent $\Leftrightarrow (a_{n m})$ Cauchy sequence	[5.10.19]

Proof: As an example again we only prove the first assertion:

	$(a_{n m})$ convergent
$\Rightarrow$	$(a_{ϕ (n)})$ convergent for all strictly increasing $ϕ$	[5.10.11]
$\Rightarrow$	$(a_{ϕ (n)})$ bounded for all strictly increasing $ϕ$	[5.5.1]
$\Rightarrow$	$(a_{n m})$ bounded	[5.10.9]

The two-dimensional nature of double sequences allows to create a further notion of convergence. This new concept will sometimes be beneficial for calculating the limit in [5.10.6].

Definition: A double sequence $(a_{n m})$ is called row-convergent if the common sequence $(a_{n m})$ is convergent for all fixed $n \in ℕ^{*}$ . The numbers

\lim_{m \to \infty} a_{n m}

[5.10.20]

are called row limits. In an analogous way we introduce the notion column-convergent and the term column limit.

Consider:

Both versions of covergence, the new and the old one are not compatible:

According to our eaxmple [5.10.8] the double sequence $({(- 1)}^{n + m} (\frac{1}{n} + \frac{1}{m}))$ e.g. tends to 0, but this sequence is neither row-convergent nor column-convergent. For (even) every fixed n the convergences

$\begin{array}{l} {(- 1)}^{n + 2 m} (\frac{1}{n} + \frac{1}{2 m}) = \frac{{(- 1)}^{n}}{n} + \frac{{(- 1)}^{n}}{2 m} \to \frac{{(- 1)}^{n}}{n} \\ {(- 1)}^{n + 2 m + 1} (\frac{1}{n} + \frac{1}{2 m + 1}) = - \frac{{(- 1)}^{n}}{n} - \frac{{(- 1)}^{n}}{2 m + 1} \to - \frac{{(- 1)}^{n}}{n} \end{array}$

show that the sequence $({(- 1)}^{n + m} (\frac{1}{n} + \frac{1}{m}))$ has two different accumulation points and thus is divergent. This proves row-divergence and similarly we get column-divergence.

The double sequence $({(1 - \frac{1}{n})}^{m})$ is row-convergent ( ${(1 - \frac{1}{n})}^{m} \to 0$ ) and column-convergent ( ${(1 - \frac{1}{n})}^{m} \to 1^{m} = 1$ ) but due to the next proposition it is not convergent.

Although - as just shown - both notions of convergence do not relate to each other, those convergent double sequences that are also row- and column-convergent play an important role. Actually they are the ones that allow to interchange the order of limit formation. Please note that from now on we no longer can support the difference between fixed and variable indices by using the italic and the normal format respectively.

Proposition: Let the double sequence $(a_{n m})$ converge to g. Then the following holds:

If $(a_{n m})$ is row-convergent the sequence of row limits will converge to g:

\lim_{n \to \infty} (\lim_{m \to \infty} a_{n m}) = g

[5.10.21]

If $(a_{n m})$ is column-convergent the sequence of column limits will converge to g:

\lim_{m \to \infty} (\lim_{n \to \infty} a_{n m}) = g

[5.10.22]

If $(a_{n m})$ is row-convergent and column-convergent the successive limits exist and are identical:

\lim_{n \to \infty} (\lim_{m \to \infty} a_{n m}) = \lim_{m \to \infty} (\lim_{n \to \infty} a_{n m})

[5.10.23]

Proof: We only show 1. because 2. is very similar and 3. is a direct result from 1. and 2.

1. ►

We abbreviate $g_{n} ≔ \lim_{m \to \infty} a_{n m}$ and show: $g_{n} \to g$ . To that end we take an arbitrary $ε > 0$ . As $a_{n m} \to g$ there is an $n_{0} \in ℕ^{*}$ such that all $n \geq n_{0}$ satisfy:

| a_{n m} - g | < \frac{ε}{2} for all m \geq n_{0}

but that means $g - \frac{ε}{2} < a_{n m} < g + \frac{ε}{2}$ for these m and consequently we get for the limit $g_{n}$ :

g - \frac{ε}{2} \leq g_{n} \leq g + \frac{ε}{2}

So we have $| g_{n} - g | \leq \frac{ε}{2} < ε$ for all $n \geq n_{0}$ .

All the results developed so far for double sequences are of course valid for double series as well. We specially focus on [5.10.23] which we will restate in connection with absolute convergent double series. We start studying double series with positive members.

Proposition: The double series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} a_{i j})$ with positive members is convergent if and only if it is bounded. In that case the following estimate holds:

\sum_{i = 0}^{n} \sum_{j = 0}^{m} a_{i j} \leq \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} a_{i j} für alle n, m \in ℕ

[5.10.24]

Proof: This direction " $\Rightarrow$ " is valid due to [5.10.17].

" $\Leftarrow$ ": As all the numbers $a_{i j}$ are positive the double series is an increasing sequence thus convergent according to [5.10.18].

[5.10.11] allows for each strictly increasing $ϕ$ to calculate the limit according to the proof of [5.7.1] as the supremum of the increasing ([5.10.10]) common sequence $(\sum_{i = 0}^{ϕ_{1} (n)} \sum_{j = 0}^{ϕ_{2} (n)} a_{i j})$ . For $ϕ : n \mapsto (n, n)$ e.g. we get for all n, m (taking $k = \max {n, m}$ ):

\sum_{i = 0}^{n} \sum_{j = 0}^{m} a_{i j} \leq \sum_{i = 0}^{k} \sum_{j = 0}^{k} a_{i j} \leq \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} a_{i j}

With this result we are now able to state and prove [5.9.13] for double series.

Proposition:

Every absolute convergent double series is convergent.

[5.10.25]

Proof: As $| a_{i j} | \pm a_{i j} \geq 0$ the double series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} \frac{1}{2} (| a_{i j} | + a_{i j}))$ and $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} \frac{1}{2} (| a_{i j} | - a_{i j}))$ have positive members so that [5.10.24] provides the following estimates:

\sum_{i = 0}^{n} \sum_{j = 0}^{m} \frac{1}{2} (| a_{i j} | \pm a_{i j}) \leq \sum_{i = 0}^{n} \sum_{j = 0}^{m} \frac{1}{2} (| a_{i j} | + | a_{i j} |) = \sum_{i = 0}^{n} \sum_{j = 0}^{m} | a_{i j} | \leq \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} | a_{i j} |

They guarantee (again using [5.10.24]) the convergence of both double series and due to [5.10.14] also the convergence of

(\sum_{i = 0}^{n} \sum_{j = 0}^{m} a_{i j}) = (\sum_{i = 0}^{n} \sum_{j = 0}^{m} \frac{1}{2} (| a_{i j} | + a_{i j})) - (\sum_{i = 0}^{n} \sum_{j = 0}^{m} \frac{1}{2} (| a_{i j} | - a_{i j}))

Proposition: If the series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} | a_{i j} |)$ is bounded the following equality holds:

\sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} a_{i j} = \sum_{j = 0}^{\infty} \sum_{i = 0}^{\infty} a_{i j}

[5.10.26]

Proof: Together with $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} | a_{i j} |)$ all of it's row series and all of it's column series are increasing and bounded, thus convergent ([5.10.18] and [5.7.1] respectively). According to [5.10.25] and [5.9.13] this is also true for the series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} a_{i j})$ so that the assertion follows from [5.10.23].

5.9. 5.11.