Summation Symbol: Changing the Order

Question

I have some questions regarding the order of the summation signs (I have tried things out and also read the wikipedia page, nevertheless some questions remained unanswered):

Original 1. wikipedia says that:

$$\sum_{k=1}^m a_k \sum_{\color{red}{k}=1}^n b_l = \sum_{k=1}^m \sum_{l=1}^n a_k b_l$$

does not necessarily hold. What would be a concrete example for that?

Edited 1. wikipedia says that:

$$\sum_{k=1}^m a_k \sum_{\color{red}{l}=1}^n b_l = \sum_{k=1}^m \sum_{l=1}^n a_k b_l$$

does not necessarily hold. What would be a concrete example for that?

2.As far as I see generally it holds that:

$$\sum_{j=1}^m \sum_{i=1}^n a_ib_j = \sum_{i=1}^n \sum_{j=1}^m a_ib_j $$

why is that? It is not due to the property, that multiplication is commutative, is it?

3.What about infinite series, when does: $$\sum_{k=1}^{\infty}\sum_{l=1}^{\infty} a_kb_l = \sum_{k=1}^{\infty}a_k \sum_{l=1}^{\infty}b_l$$ hold? And does here too $$\sum_{k=1}^{\infty}\sum_{l=1}^{\infty} a_kb_l = \sum_{l=1}^{\infty}\sum_{k=1}^{\infty} a_kb_l$$ hold?

Thanks

Answer 1

For the original first question where $l = k$, let $m=n=2$, $a_1=b_1=1$, and $a_2=b_2=2$; then

$$\sum_{k=1}^2a_k\sum_{k=1}^2b_k=\sum_{k=1}^2a_k(1+2)=1\cdot3+2\cdot3=9\;,$$

but $$\sum_{k=1}^2\sum_{k=1}^2a_kb_k=\sum_{k=1}^2(1^2+2^2)=5+5=10\;.$$

For the second question, imagine arranging the terms $a_ib_j$ in an $n\times m$ array:

$$\begin{array}{ccccc|c} a_1b_1&a_1b_2&a_1b_3&\dots&a_1b_m&\sum_{j=1}^ma_1b_j\\ a_2b_1&a_2b_2&a_2b_3&\dots&a_2b_m&\sum_{j=1}^ma_2b_j\\ a_3b_1&a_3b_2&a_3b_3&\dots&a_3b_m&\sum_{j=1}^ma_3b_j\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ a_nb_1&a_nb_2&a_nb_3&\dots&a_nb_m&\sum_{j=1}^ma_nb_j\\ \hline \sum_{i=1}^na_ib_1&\sum_{i=1}^na_ib_2&\sum_{i=1}^na_ib_3&\dots&\sum_{i=1}^na_ib_m \end{array}$$

For each $j=1,\dots,m$, $\sum_{i=1}^na_ib_j$ is the sum of the entries in column $j$, and for each $i=1,\dots,n$, $\sum_{j=1}^ma_ib_j$ is the sum of the entries in row $i$. Thus,

$$\begin{align*} \sum_{j=1}^m\sum_{i=1}^na_ib_j&=\sum_{j=1}^m\text{sum of column }j\\ &=\sum_{i=1}^n\text{sum of row }i\\ &=\sum_{i=1}^n\sum_{j=1}^ma_ib_j\;. \end{align*}$$

For infinite double series the situation is a bit more complicated, since an infinite series need not converge. However, it is at least true that if either of

$$\sum_{j=1}^m\sum_{i=1}^n|a_ib_j|\quad\text{and}\quad\sum_{i=1}^n\sum_{j=1}^m|a_ib_j|$$

converges, then the series without the absolute values converge and are equal. This PDF has much more information on double sequences and series.

Answer 2

First of all, by the distributivity of multiplication over addition, the following is true:

$$\bigg(\sum_{l=1}^m a_l\bigg)\bigg( \sum_{k=1}^n b_k\bigg) = \sum_{l=1}^m \bigg( a_l\sum_{k=1}^n b_k\bigg) = \sum_{l=1}^m \sum_{k=1}^n a_l b_k$$

This can be seen by writing out the sums explicitly.

This is also true:

$$\sum_{j=1}^m \sum_{i=1}^n a_ib_j = \sum_{i=1}^n \sum_{j=1}^m a_ib_j $$ Commutativity is not necessarily involved because each pair of numbers being multiplied together are also done so in the same order. One reason equality holds is because of the commutitivity of addition. Think of an $n\times m$ grid in the $xy$ plane. If the point with co-ordinate $(i,j)$ has the number $a_ib_j$ written on it, the sum of all the numbers on the grid is the same if we add along the rows first (the left hand sum) or if we add along the columns first (the right hand sum).

When it comes to infinite series, things get a lot more complicated. One thing that is true is that if

$$\sum_{k=1}^{\infty}\sum_{l=1}^{\infty} |a_kb_l|$$ converges, then:

$$\sum_{k=1}^{\infty}\sum_{l=1}^{\infty} a_kb_l = \sum_{l=1}^{\infty}\sum_{k=1}^{\infty} a_kb_l$$

You can follow this link:

http://www.math.ubc.ca/~feldman/m321/twosum.pdf

To see an example of where changing the order does matter.

Answer 3

shouldn't the first one be:

$$\sum_{k=1}^m a_k \sum_{k=1}^n b_k = \sum_{k=1}^m \sum_{l=1}^n a_k b_l$$ ?

anyway,

$$ \sum_{k=1}^m a_k = a_1 + ... + a_m\\ \sum_{k=1}^n b_k = b_1 + ... + b_n\\ \sum_{k=1}^m a_k \sum_{k=1}^n b_k = (a_1 + ... + a_m)(b_1 + ... + b_n) = \\ (1) =a_1b_1 + ... + a_1b_n + ... + a_mb_1 + ... a_mb_n \\ \sum_{k=1}^m \sum_{l=1}^n a_k b_l=\sum_{k=1}^m (a_kb_1 + ... + a_kb_n)=\\ (2) =a_1b_1 + ... + a_1b_n + ... + a_mb_1 + ... a_mb_n $$ (1) and (2) looks the same to me

Answer 4

Here is a proof by induction for (2).

base case: $n=1$ $$\sum_{j=1}^m \sum_{i=1}^1 a_i\times b_j =\sum_{j=1}^ma_1\times b_j = \sum_{i=1}^1\sum_{j=1}^m a_i\times b_j$$

Assume the property holds for $n=k$, and now prove if for $n=k+1$:

$$\sum_{j=1}^m \sum_{i=1}^{(k+1)} a_i\times b_j = \sum_{j=1}^m (\sum_{i=1}^k (a_i\times b_j) +a_{k+1}\times b_{j}) = \sum_{j=1}^m \sum_{i=1}^k a_i\times b_j+\sum_{j=1}^ma_{k+1}\times b_j \\= \sum_{i=1}^k\sum_{j=1}^m a_i\times b_j + \sum_{j=1}^ma_{k+1} b_j \\ =\sum_{i=1}^{k+1}\sum_{j=1}^m (a_i\times b_j) $$