How to prove that convolution of sequences is associative? Let {$a_n$} and {$b_n$} be finite real sequences with $n\ge0$. Convolution ($\ast$) of two sequences defined as
$$
\{a_n\}\ast\{b_n\}=\{\sum_{i=0}^{n} a_ib_{n-i}\}.
$$
The convolution of three sequences is:
$$
(\{a_n\}\ast\{b_n\})\ast\{c_n\}=(\{\sum_{i=0}^{n} a_ib_{n-i}\})\ast\{c_n\}=\{\sum_{j=0}^{n}\sum_{i=0}^{j} a_ib_{j-i}c_{n-j}\}
$$
$$
\{a_n\}\ast(\{b_n\}\ast\{c_n\})=\{a_n\}\ast(\{\sum_{k=0}^{n} b_kc_{n-k}\})=\{\sum_{l=0}^{n}\sum_{k=0}^{n-l} a_lb_kc_{n-l-k}\}.
$$
I'm not sure how to show both of the double sum is same. can anyone give me advice?
 A: We start with the second double sum and do some transformations to obtain the first double sum.

We obtain
\begin{align*}
\color{blue}{\sum_{l=0}^n\sum_{k=0}^{n-l}a_lb_kc_{n-l-k}}
&=\sum_{l=0}^n\sum_{k=0}^la_{n-l}b_{k}c_{l-k}\tag{1.1}\\
&=\sum_{l=0}^n\sum_{k=0}^la_{n-l}b_{l-k}c_k\tag{1.2}\\
&=\sum_{k=0}^n\sum_{l=k}^na_{n-l}b_{l-k}c_k\tag{1.3}\\
&=\sum_{k=0}^n\sum_{l=0}^{n-k} a_{n-l-k}b_lc_k\tag{1.4}\\
&=\sum_{k=0}^n\sum_{l=0}^k a_{k-l}b_lc_{n-k}\tag{1.5}\\
&\,\,\color{blue}{=\sum_{k=0}^n\sum_{l=0}^ka_lb_{k-l}c_{n-k}}\tag{1.6}
\end{align*}
and the claim follows.

Comment:

*

*In (1.1) we change the order of summation of the outer sum $l\to n-l$.


*In (1.2) we change the order of summation of the inner sum $k\to l-k$.


*In (1.3) we exchange inner and outer sum according to $\sum_{0\leq k\leq l\leq n}a_{n-l}b_{l-k}c_k$.


*In (1.4) we shift the index of the inner sum to start with $l=0$.


*In (1.5) we change the order of summation of the outer sum $k\to n-k$.


*In (1.6) we change the order of summation of the inner sum $l\to k-l$.
