Lecture 17 - Double Sums

This idea stems from one of the cautionary tales from before.

Let $(a_{i j})$ be a doubly-indexed array of numbers:

What is $\sum a_{i j}$ ? Let's look at a few natural examples (candidates, if you will). Do we add up the rows then columns? Or the columns then the rows? We could try the first where we add the rows then the columns:

\sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} a_{i j} : rows first

Another natural candidate would be columns first:

\sum_{j = 1}^{\infty} \sum_{i = 1}^{\infty} a_{i j} : columns first

Maybe we want something that is more balanced. Recall that this rows/columns first mentality lead to some issues. We could try diagonals first:

\sum_{n = 1}^{\infty} \sum_{i + j = n + 1} a_{i j} : diagonals first

We could also sum over the $n \times n$ box of all the top-left $n$ -box elements, and let $n \to \infty$ :

lim_{n \to \infty} \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j} : box method

You can think of the inside as a partial sum $s_{n} = \sum^{n} \sum^{n} a_{i j}$ .

Recall from yesterday was that rearrangement for a conditional convergent series had the limit of partial sums approach any value of our choosing. The idea here is similar in that rearrangements of these double sums shouldn't change anything.

Double Sum Construction

If ${\sum_{(i, j) \in X} | a_{i j} | | X \subset N \times N, | X | is finite}$ is bounded, then all these values converge, and to the same value.

This is just saying that any finite sum of these $\sum a_{i j}$ should be bounded, and if that's the case then we can 'let' our $X$ approach the whole set of $N \times N$ .

If we let $T_{n} = \sum_{i = 1}^{n} \sum_{j = 1}^{n} | a_{i j} |$ , then this is equivalent to our double-sum construction condition, since then $X$ is just the $n \times n$ box of the upper-left terms. Clearly $T_{n}$ is an increasing sequence, so then if it's bounded by the construction, then it's equivalent to saying that $T_{n}$ is convergent (and thus Cauchy) since it's now monotone.

Proof
Let $S_{n} = \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j}$ (this is just like our $T_{n}$ , except without the absolute values). Note that if $n > m$ and for notation say $X_{n}, X_{m}$ are the $n \times n$ and $m \times m$ boxed respectively:

| S_{n} - S_{m} | = | \sum_{(i, j) \in X_{n}, \notin X_{m} X} a_{i j} |

Adds up the L-shaped region in the figure below:

Then:

\begin{aligned} \leq \sum_{^{″}} | a_{i j} | \\ = T_{n} - T_{m} \\ = | T_{n} - T_{m} | \end{aligned}

We showed that $T_{n}$ is cauchy, so then $| T_{n} - T_{m} | < ε$ for any $ε$ . Thus then if we let $ε > 0$ then $| S_{n} - S_{m} | < ε$ so then it must be Cauchy and thus converge.
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As a result we can call:

S = lim_{n \to \infty} S_{n} = lim_{n \to \infty} \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j}

Adding the Diagonals

Let's show that adding the diagonals works here:

Proof
Let $d_{n} = \sum_{i + j = n + 1}^{n} a_{i j}$ add up the $n$ entries

The claim is that $\sum_{n = 1}^{\infty} d_{n} = S$ . The idea is that $D_{m} = d_{1} + \dots + d_{m}$ is the $m$ -th partial sum of the $d_{n}$ 's. In that regard it add the top-left triangle:

So let $\varepsilon > 0$. Choose $N \in \mathbb{N}$ where: - (top-left box with only orange): $m \geq N$ implies - (the biggest box, as well as the green/red triangles summed up): $n > m \geq N$ implies

Suppose $m \geq 2 N \geq N$ (the idea is to get the biggest box). We NTS that $| D_{m} - S | < ε$ .

\begin{aligned} D_{m} & = \sum_{n = 1}^{m} d_{n} \\ = \sum_{n = 1}^{m} \sum_{j + j = n + 1} a_{i j} \\ = \underset{S_{N}}{\underset{⏟}{\sum_{(i, j) \in X_{n}} a_{i j}}} + \sum_{(i, j) \notin X_{n}, \in X_{2 m}} a_{i j} \\ | D_{m} - S | & \leq | S_{N} - S | + \sum_{(i, j) \notin X_{n}, \in X_{2 m}} | a_{i j} | & Triangle Inequality \\ \leq | S_{n} - S | + | T_{2 m} - T_{N} | \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}

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Adding the Rows

We can't go into the proof, but the idea is that if we have:

R_{i} = \sum_{j = 1}^{\infty} a_{i j}

we first need to show it converges to some $R_{i}$ . We then need to show that $\sum_{i = 1}^{\infty} R_{i} = S$ .