Is there a quick way to see these sums are equal? I can show that these two sums are equal by writing out all terms, but is there a quick way to see the equality:
$$\sum_{i=0}^n\sum_{j=0}^i a_jb_{i-j}c_{n-i} = \sum_{i=0}^n\sum_{j=0}^{n-i} a_ib_{j}c_{n-i-j} $$
(this question popped up while checking that the formal power series over a commutative ring is a commutative ring, in particular, when checking that the multiplicative associativity is satisfied.)
 A: Both sides are $\sum a_i b_j c_k$ where the sum is over all triples
$(i,j,k)$ of non-negative integers with $i+j+k=n$.
A: This should also follow from the fact that both sides are convolutions of the sequences $A(i)=a_i,B(i)=b_i,C(i)=c(i)$, and convolutions are well known to be commutative and associative: 
$$(C\star (A\star B)(t))(n)=(A\star(B\star C)(t))(n)$$
A: We can transform the left-hand side in a few steps in order to get the right-hand side.

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^n\sum_{j=0}^ia_jb_{i-j}c_{n-i}}&=\sum_{j=0}^n\sum_{i=0}^ja_ib_{j-i}c_{n-j}\tag{1}\\
&=\sum_{0\leq i\leq j\leq n}a_ib_{j-i}c_{n-j}\tag{2}\\
&=\sum_{i=0}^n\sum_{j=i}^na_ib_{j-i}c_{n-j}\tag{3}\\
&\color{blue}{=\sum_{i=0}^n\sum_{j=0}^{n-i}a_ib_jc_{n-i-j}}\tag{4}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we observe that in the left-hand side of OPs identity we see $a_i$ while on the right-hand side of OPs identity we see $a_j$. So in the first step we consequently exchange each occurrence of $i$ with $j$ and vice versa.

*In (2) we just use another representation of (1) to better see the index range.

*In (3) we exchange the order of summation guided by the index range representation in (2).

*In (4) we shift the index of the inner sum to start with $j=0$.
