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.)
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 $i+j+k=n$.
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:
$$(C\star (A\star B)(t))(n)=(A\star(B\star C)(t))(n)$$
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 $j=0$.
