How does interchange of counter variable in this double summations work? I'm reading a book on linear algebra. $F$ is a field and $S$ is the set of nonnegative integers. It say the the set of all functions from $S$ into $F$ is a vector space over $F$. We denot it $F^{\infty}$ and also define a product in $F^\infty$ by associating with each pair of vectors $f$ and $g$ in $F^\infty$ the vector $fg$ which is given by
$$(fg)_n = \sum\limits_{i=0}^n f_ig_{n-i}, \hspace{10 mm} n=0,1,2,...$$
When the book want to prove associativity of the product it goes like this:
$$[(fg)h] = \sum\limits_{i=0}^n (fg)_ih_{n-i} = \sum\limits_{i=0}^n \left( \sum\limits_{j=0}^if_jg_{i-j}\right)h_{n-i}=\sum\limits_{i=0}^n\sum\limits_{j=0}^if_ig_{i-j}h_{n-i}=\sum\limits_{j=0}^nf_j\sum\limits_{i=0}^{n-j}g_ih_{n-i-j}=\sum\limits_{j=0}^nf_j(gh)_{n-j}=[f(gh)]_n$$
1.I think in the third sum it shoud be $f_j$ instead of $f_i$. Tell me if I am wring.
2.I don't understand the fourth equality when the order of the double sum is reversed. How can we prove $\sum\limits_{i=0}^n\sum\limits_{j=0}^if_ig_{i-j}h_{n-i}=\sum\limits_{j=0}^nf_j\sum\limits_{i=0}^{n-j}g_ih_{n-i-j}$
 A: I think you're right that there is a typo. In the second sum (the first double sum), we have $f_jg_{i-j}$, and in the next expression, the $j$ subscript on $f$ inexplicably changes into an $i$. In the next expression, it changes back to a $j$. It should have been $j$ all along.
Once that is corrected, we can address your second question. It is valid to pull the $f_j$ out of the inner summation, because it has no dependence on $i$. As for the other terms, something a bit more subtle happens.
Consider the double summation $\sum\limits_{i=0}^n\sum\limits_{j=0}^i$. If you imagine an $ij$-plane, the terms in this sum occupy a triangle whose edges are: the $i$-axis from $0$ to $n$, the vertical line from $(0,n)$ to $(n,n)$, and the diagonal line $i=j$ from $(0,0)$ to $(n,n)$. If you want to switch the order of the sum and run through the points in that same triangle, you're going to write
$$\sum\limits_{i=0}^n\sum\limits_{j=0}^i X = \sum\limits_{j=0}^n\sum\limits_{i=j}^n X$$
where $X$ represents whatever expression we have in the summation.
Once that's clear, we can address how the inner sum is re-indexed. We have to see why:
$$\sum\limits_{i=j}^n g_{i-j}h_{n-i} = \sum\limits_{i=0}^{n-j} g_{i}h_{n-i-j}$$
This is the usual re-indexing trick where we want to start at zero instead of somewhere else, so we shift our index. Basically, replace every instance of $i$ on the left-hand side with the expression $i+j$. When doing this, consider that the upper limit ($n$) is really shorthand for $i=n$. After simplifying, you should end up with the right-hand side.
Does this help?
A: Here is a derivation which also could be helpful.

We obtain
  \begin{align*}
\color{blue}{[(fg)h]}&=\sum_{i=0}^n(fg)_ih_{n-i}\\
&=\sum_{i=0}^n\left(\sum_{j=0}^i f_jg_{i-j}\right)h_{n-i}
=\sum_{\color{blue}{0\leq j\leq i\leq n}}f_jg_{i-j}h_{n-i}\tag{1}\\
&=\sum_{j=0}^n\sum_{i=j}^nf_jg_{i-j}h_{n-i}\\
&=\sum_{j=0}^nf_j\sum_{i=0}^{n-j}g_ih_{n-i-j}\tag{2}\\
&=\sum_{j=0}^nf_{n-j}\left(\sum_{i=0}^jg_ih_{j-i}\right)\tag{3}\\
&=\sum_{j=0}^nf_{n-j}(gh)_j\left(=\sum_{j=0}^nf_j(gh)_{n-j}\right)\\
&\color{blue}{=[f(gh)]}
\end{align*}

Comment:


*

*In (1) the representation with indices $0\leq j\leq i\leq n$ helps to identify upper and lower limit of $i$ and $j$.

*In (2) we factor out $f_j$. We shift the index $i$ to start from $i=0$ and substitute $i\rightarrow i+j$ in the summation terms accordingly.

*In (3) we change the order of summation of the outer sum by $j\rightarrow n-j$.
Note: You are right with 1.) and 2.) is shown above.
