Lecture 18 - More on Double Sums

Recall what we talked about last time here. Namely we had to add up the sums in a table:

We specifically talked about either adding up via the $n \times n$ box:

S_{n} = \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j}

and taking that limit, or doing adding those absolute values:

T_{n} = \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j}

Lemma

$(T_{n})$ is bounded iff:

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

converges (or alternatively if

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

converges)

Proof
$(\to)$ : Let $T$ be an upper bound for $(T_{n})$ . First the inner sum. For any $i$ -th row the $m$ -th partial sum of it is:

| a_{i 1} | + | a_{i 2} | + \dots + | a_{i m} | \leq max (T_{i}, T_{m}) \leq T

We've just showed that the partial sums of non-negative numbers is bounded. But these sequences are increasing (all components are positive), so using Monotone Convergence Theorem gives that $T_{m}$ converges. Since it's the partial sum, that is equivalent to saying that $\forall i$ we have $\sum_{j = 1}^{\infty} | a_{i j} | = b_{i}$ . Remember, $b_{i}$ is the sum of terms in the $i$ -th row.

Now for the outer sum. We'll let $ε > 0$ and we'll show that $\forall m$ that the partial sum $T_{m} = b_{1} + \dots + b_{m} < T + ε$ , which would imply that the partial sum itself is bounded by $T$ . To do this, let $m \in N$ be arbitrary. Let $n \in N$ be large enough where for each row $i, \dots, m$ we have:

b_{i} - \sum_{j = 1}^{n} | a_{i j} | < \frac{ε}{m}

Then using some algebra:

\begin{aligned} b_{1} + \dots + b_{m} & = \underset{< \frac{ε}{m}}{\underset{⏟}{(b_{1} - \sum_{j = 1}^{n} | a_{1 j} |)}} + \dots + \underset{< \frac{ε}{m}}{\underset{⏟}{(b_{m} - \sum_{j = 1}^{n} | a_{m j} |)}} \\ + \underset{< T}{\underset{⏟}{\sum_{i = 1}^{m} \sum_{j = 1}^{n} | a_{i j} |}} \\ < m (\frac{ε}{m}) + T \\ = T + ε \end{aligned}

Which completes the proof.

Why is the

< T

apply? Whichever

m, n

is larger, the sum is smaller than just adding the

X_{m}, X_{n}

box terms, which itself is bounded by

T

$(\leftarrow)$ : Suppose $\sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} | a_{i j} |$ converges. Then by definition $T_{n}$ must converge to the same limit of the partial sums of our series. As a result since all convergent sequences are bounded then $T_{n}$ must be bounded.

#todo/school : Finish this proof 📅 2024-10-25 ✅ 2024-10-29

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Absolute convergence for Double Sums

If $\sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} | a_{i j} |$ converges, then $\sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} a_{i j}$ converges. Furthermore, it converges to the value $S := lim_{n \to \infty} s_{n} = lim_{n \to \infty} \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i j}$ .

Proof

Using the ACT, since our first series converges then we get that the second series converges. It converges to the limit described above by construction.

#todo/school : And this proof 📅 2024-10-27 ✅ 2024-10-29

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Products

Let:

\sum a_{i} = a_{1} + a_{2} + a_{3} + \dots

\sum b_{j} = b_{1} + b_{2} + b_{3} + \dots

What should be meant by $(\sum a_{i}) \cdot (\sum b_{j})$ ? Formally if we did:

(a_{1} + a_{2} + \dots) \cdot (b_{1} + b_{2} + \dots) = ?

should have each term $a_{i} b_{j} | i, j \in N$ . We can imagine having some big matrix of these terms:

Cauchy Product

Given $\sum a_{i}$ and $\sum b_{j}$ , define the Cauchy Product as:

\sum_{n = 1}^{\infty} c_{n} : c_{n} = \sum_{i + j = n + 1} a_{i} b_{j}

where $c_{n} = a_{n} b_{1} + a_{n - 1} b_{2} + \dots + a_{1} b_{n}$ .

The idea is that each $c_{n}$ is the sum of all the backwards diagonal terms in the image above (see the circles).

Theorem

Suppose $\sum_{i = 1}^{\infty} a_{i}$ converges absolutely to $A$ , and $\sum_{j = 1}^{\infty} b_{j}$ absolutely to $B$ . Then the Cauchy Product will converge to $A B$ .

Proof
Recall that since our sums converge absolutely then let $A^{'} = \sum_{i = 1}^{\infty} | a_{i} |$ and $B^{'} = \sum_{j = 1}^{\infty} | b_{j} |$ which exist. Similarly we can define the partial sums:

A_{n} = \sum_{i = 1}^{n} a_{i}, B_{n} = \sum_{j = 1}^{n} b_{j}

So then $(A_{n}) \to A$ and $(B_{n}) \to B$ by construction.

Using the ALT, we get:

\begin{aligned} \sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} | a_{i} b_{j} | & = \sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} | a_{i} | | b_{j} | \\ = \sum_{i = 1}^{\infty} | a_{i} | \sum_{j = 1}^{\infty} | b_{j} | & ALT, constant \\ = A^{'} B^{'} \end{aligned}

So then the Cauchy Product sum is absolutely convergent. Now to show that it converges specifically to $A B$ . We see that from idea that doing the sums in any order doesn't matter that this diagonal summation is equivalent to adding the $n \times n$ box, so then:

\begin{aligned} \sum_{n = 1}^{\infty} c_{n} & = lim_{n \to \infty} \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i} b_{j} \\ = lim_{n \to \infty} \sum_{i = 1}^{n} a_{i} \sum_{j = 1}^{n} b_{j} \\ = lim_{n \to \infty} (A_{n} B_{n}) \\ = lim A_{n} \cdot lim B_{n} & ALT \\ = A B \end{aligned}

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