Understanding the Summation of Infinite Series: Is it True for i and j?

[yungman]

$$\hbox {Is }\;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}\;\hbox{?}$$

$$\hbox {Is }\;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;\hbox{?}$$

I think it is because even though the right side has two summation of $i$ , but both increment at the same time. So is $j$. therefore the result is the same.

 

[DrClaude]

$$\hbox {Is }\;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}\;\hbox{?}$$

Is $a_1 b_1 + a_2 b_2 = (a_1 +a_2) (b_1 + b_2)$?

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}$$

This is not correct. You cannot have two summations over the same index.

 

[yungman]

Thanks for the reply, but if you keep j=1, i=1,2,3...

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+...$$

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+...$$
If you start increment j, the series just repeat with j=2,3,4...

Because $i$ increment all at the same time. There should be no difference. This is not like trying to make

$a_1 b_1 + a_2 b_2 = (a_1 +a_2) (b_1 + b_2)$

 

[DrClaude]

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}$$

The way I read this is
$$
\left( \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \right) \left( \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j} \right)
$$
which means that addition and multiplication have been inverted, which is not correct.

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+...$$
If you start increment j, the series just repeat with j=2,3,4...

Because $i$ increment all at the same time.

Again, you cannot have two summations with the same index. It doens't make sense.

 

[Ray Vickson]

Thanks for the reply, but if you keep j=1, i=1,2,3...

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j} \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}B_{i,j}=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+...$$

$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \;\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}A_{i,j}B_{i,j}=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} A_{i,j}B_{i,j}\;=A_{1,1}B_{1,1}+A_{2,1}B_{2,1}+A_{3,1}B_{3,1}+...$$
If you start increment j, the series just repeat with j=2,3,4...

Because $i$ increment all at the same time. There should be no difference. This is not like trying to make

$a_1 b_1 + a_2 b_2 = (a_1 +a_2) (b_1 + b_2)$

No, you are wrong: it is exactly like that. In fact, if you take $A_{11} = a_1, A_{12}= a_2, B_{11} = b_1, B_{12} = b_2$ and all other $A_{ij}, B_{ij} = 0,$ then you are claiming that $\sum_i\sum_j A_{ij}B_{ij} = a_1 b_1 + a_2 b_2$ equals $\sum_i \sum_j A_{ij} \sum_l \sum_m B_{lm} = (a_1+a_2)(b_1+b_2),$ and that is false.

 

[yungman]

No, you are wrong: it is exactly like that. In fact, if you take $A_{11} = a_1, A_{12}= a_2, B_{11} = b_1, B_{12} = b_2$ and all other $A_{ij}, B_{ij} = 0,$ then you are claiming that $\sum_i\sum_j A_{ij}B_{ij} = a_1 b_1 + a_2 b_2$ equals $\sum_i \sum_j A_{ij} \sum_l \sum_m B_{lm} = (a_1+a_2)(b_1+b_2),$ and that is false.

I really don't get this, you are using $\sum_i \sum_j A_{ij}$ on the first and $\sum_l \sum_m$ on the second one. I am using $\sum_i \sum_j$ for both. The two sum cannot be independently incremented. When $i$ incremented by one, both has to be incremented by 1. Using what you say that $i=1$ and $j$=1,2 only all other are zeros.


$$\hbox{For }i=1,j=1,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{1}B_{1}$$

$$\hbox{For }i=1,j=2,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{2}B_{2}$$

So the sum will be $a_1b_1+a_2b_2$

 

[verty]

I really don't get this, you are using $\sum_i \sum_j A_{ij}$ on the first and $\sum_l \sum_m$ on the second one. I am using $\sum_i \sum_j$ for both. The two sum cannot be independently incremented. When $i$ incremented by one, both has to be incremented by 1. Using what you say that $i=1$ and $j$=1,2 only all other are zeros.$$\hbox{For }i=1,j=1,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{1}B_{1}$$

$$\hbox{For }i=1,j=2,\;\sum_i \sum_j A_{ij} \sum_i \sum_j B_{ij}=\left(\sum_i \sum_j A_{ij}\right) \left(\sum_i \sum_j B_{ij}\right)=A_{2}B_{2}$$

So the sum will be $a_1b_1+a_2b_2$

It doesn't work that way, you could even have this:

$\sum\limits_{i=0}^\infty \sum\limits_{i=0}^i A_i$

which would mean, A_0 + (A_0 + A_1) + (A_0 + A_1 + A_2) + ...

My point, each sigma has its own variable that it creates just for that summation.

 

[yungman]

thanks everybody. So I cannot count on $i$ and $j$ all increment at the same time. I have to treat is as if they are independent between the two summation?

Thanks

 

[yungman]

This is the link of the real question about summation. I just don't know how the book can move the summation inside:

https://www.physicsforums.com/showthread.php?t=711644

Since that is a different problem from this, I posted it as new thread. But where I got stuck is the same.

 

[DrClaude]

I've had a look at the other thread and I don't see anywhere the same indices used at the same time.