Show equality between these two series I have a simple (?) index substitution (?) problem, where I want to show that
$$
\sum_{n=0}^\infty \sum_{m=0}^n a_n b_m c_{n-m} = \sum_{j=0}^\infty \sum_{k=0}^\infty a_{j+k} b_k c_j
$$
but whatever I try, I end up with $j$ also in the limits of the second sum, instead of it going to $\infty$ .
The way I got to this equivalence was by writing out the first few terms of the left-hand sum and sorting them by  $c_j$ :





m = n
m = n-1
m = n - 2
m = n-3
...




n = 0
$a_0b_0c_0$






n = 1
$a_1b_1c_0$
$a_1b_0c_1$





n = 2
$a_2b_2c_0$
$a_2b_1c_1$
$a_2b_0c_2$




n = 3
$a_3b_3c_0$
$a_3b_2c_1$
$a_3b_1c_2$
$a_3b_0c_3$



...
...
...
...
...
...




from there it looks fairly obvious that I can write it as the right-hand sum.
Or have I made a mistake there already?
I hope it's understandable what I want to achieve and I am not missing any crucial information. If so, let me know.
Thank you very much for your help!
 A: There was a small mistake in my previous answer, which I've deleted.
I'll explain this in the simplest possible language without leaving any logical gaps.
On the one hand, you have all 2-tuples of the form $(n,m)\in\mathbb{Z}^2$ where $0\leq m\leq n$.
On the other hand, you have all 2-tuples of the form $(k,j)\in\mathbb{Z}^2$ where $0\leq j$ and $0\leq k$.
$\\$
For simplicity, let me write $f(x,y,z)$ instead of $a_x b_y c_z$.
We can also write $f(n,m,n-m)$ as $g(n,m)$
and we can write $f(j+k,k,j)$ as $h(k,j)$.
$\\$
One way to prove that $\displaystyle\sum_{0\leq m\leq n}g(n,m)=\sum_{0\leq j\text{ and }0\leq k}h(k,j)$
is to show that you can pair up each 2-tuple $(n,m)$ with a 2-tuple $(k,j)$
such that each 2-tuple has a unique partner
and no 2-tuple is left out of the pairing
(in formal mathematical jargon, this is called a bijection)
and such that each specific pairing of $(n',m')$ with $(k',j')$ obeys $g(n',m')=h(k',j')$.
$\\$
You can achieve this by declaring the following:
$(n',m')$ shall be paired up with $(k',j')$, where $k'=m'$ and $j'=n'-m'$.
Prove that this pairing satisfies the criteria listed in the previous paragraph.
