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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

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  • $\begingroup$ Could you please clarify what you mean in the first double sum? You have summed over $k$ twice on the right hand side, which doesn't make consistent since, and some of the answerers have taken it to mean different things! $\endgroup$ Mar 26, 2013 at 20:19
  • $\begingroup$ You are right, so one of the k's could just be replaced with an l, in fact I will edit it. Pardon me $\endgroup$
    – TestGuest
    Mar 26, 2013 at 20:42
  • $\begingroup$ Actually I see now that on wikipedia too it is summed twice over k as it was done previously in my post: de.wikipedia.org/wiki/Summe $\endgroup$
    – TestGuest
    Mar 26, 2013 at 23:21
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    $\begingroup$ Just for clarity, the wikipedia page only sums over k once on the left hand side in the article, (the right hand side of your original post) where you were summing twice (and over different ranges). This is what was inconsistent. Using $k$ twice on the right hand side is okay (but arguably poor notation) since the $k$'s do not interact. This inequality comes from the difference between adding over the whole grid, and just adding up the diagonals, if you think about the sum being over a grid, as in my answer. Brian Scott provided an example, but almost any values will show this. $\endgroup$ Mar 26, 2013 at 23:29

4 Answers 4

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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.

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  • $\begingroup$ The pdf link is no longer avaiable :( $\endgroup$
    – drewdles
    Sep 30, 2015 at 16:11
  • $\begingroup$ @Anant: I found the new address and fixed the link. $\endgroup$ Sep 30, 2015 at 16:38
  • $\begingroup$ @BrianM.Scott it has been removed from that link as well $\endgroup$
    – stateless
    Apr 7, 2021 at 16:35
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    $\begingroup$ @stateless: I’ve replaced that link with one that works. I’m not positive that it’s the same PDF, but I think that it is, and it does at least appear to be a workable substitute. $\endgroup$ Apr 7, 2021 at 17:38
  • $\begingroup$ @BrianM.Scott thanks! $\endgroup$
    – stateless
    Apr 7, 2021 at 19:46
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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.

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  • $\begingroup$ Thank you very much, I have another question actually too: Is $\sum_{k,l}^m a_kb_l$ equal to $\sum_{k}^m \sum_{l}^m a_kb_l$? $\endgroup$
    – TestGuest
    Mar 26, 2013 at 22:46
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    $\begingroup$ @TestGuest Yes, it's just another way of writing the same thing (although if you don't know that $k$ and $l$ both go up to $m$ the left hand version might be a bit vague. Happy to help! $\endgroup$ Mar 26, 2013 at 22:51
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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

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  • $\begingroup$ Hi, welcome to the website! This isn't really an answer I'm afraid, things like this are better off in the comments section underneath the original question. $\endgroup$ Mar 26, 2013 at 19:54
  • $\begingroup$ Hi, yeah, I totally agree with you, but I can't find such possibility to put a comment under original question $\endgroup$ Mar 26, 2013 at 20:01
  • $\begingroup$ @Tom: Users with less than 50 reputation points cannot comment on posts they don't own. $\endgroup$ Mar 26, 2013 at 21:52
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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 $n=k+1$:

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

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