So I've been fiddling around with the cauchy product of sequences lately, and am curious about a little identity I've found (which I'm sure is ubiquitous in finite differences, as I can't be the only one to have thought of it). I have been having trouble proving it, but the idea is rather straight forward. I think I might have seen it somewhere before but can't find any literature on the subject through searches and am exhausted for a proof (even though I know it should be simple).


Consider a ring of sequences $a:\mathbb{N} \to \mathbb{C}$ endowed with the usual addition $+$ and the multiplication operator $\times$ where


$$(a \times b)(n) = \sum\_{j=0}^n a\_j b\_{n-j}$$


The first identity, which at hand is very simple to prove, follows from associativity of the product. Firstly


$$(a \times 1)(n) = \sum\_{j=0}^n a\_j$$


and


**I am correcting the terrible mistake I made**


$$1 \times 1 \times ...(k\,times)...\times 1 = \dbinom{n+k-1}{k-1}$$ 


**end of terrible mistake**


giving the usual iterated sum formula


**edited to not be a terrible mistake**


$$\sum^k a(n) = (a\times \dbinom{n+k-1}{k-1})$$


$$\sum^k a(n) = \sum\_{j=0}^n a\_j\dbinom{n+k-1-j}{k-1}= \sum\_{n\_{k-1}=0}^n \sum\_{n\_{k-2}=0}^{n\_{k-1}}...k\,times...\sum\_{n\_0=0}^{n\_1} a(n\_0)$$


where $\sum^{k}\sum^{j} = \sum^{k+j}$. The question I am curious about is if this has ever been investigated for complex numbers--i.e; does anyone have any references to the operator


$$\sum^{s} a = \sum\_{j=0}^n a\_j \dbinom{n+s-j-1}{s-1}$$


and if perhaps it satisfies the identity


$$\sum^{s} \sum^{q} = \sum^{s+q}$$


so that it forms a sort of differsum (like a differintegral but with sums). By associativity of the Cauchy product the semigroup property is equivalent to showing


**edited to remove the typo**
$$\sum\_{j=0}^n \dbinom{j+s-1}{s-1}\dbinom{n+q-j-1}{q-1} = \dbinom{n+s+q-1}{s+q-1}$$


which follows for $s,q \in \mathbb{N}$ (by the above) but I am unsure for $\Re(s),\Re(q) > 0$.


This is something I imagine must be somewhere in the vastness of literature on calculus of variations or finite differences. A proof or a reference to a proof is what I'm really looking for. I'm particular to a reference though, as I'd like to read what else people have written on the subject.


**the typo was the whole problem to be honest, I just made a typo. /shameface**



Iterated sums--something like a differsum
Ok. Fixing the typo in the post, we can prove the claim without much trouble.


After fixing the typo, the question is really asking to prove the following for all nonnegative integers $n$ and all complex $s$ and $q$:
$$
\sum\_{k=0}^{n} {k+s-1 \choose k} {n-k + q-1 \choose n-k} = {n+q+s-1 \choose n},
$$
where ${z \choose n} = \frac{z(z-1)(z-2)\cdots (z-n+1)}{n!}$.


This is easy with generating functions. Define the formal power series $$G\_s (x) = \sum\_{n=0} ^{\infty} {n+s-1 \choose n} x^n.$$ Then the claim is the same as showing $G\_s(x) G\_q (x) = G\_{s+q} (x).$ But this follows immediately from the fact that 
$$
G\_s (x) = \sum\_{n=0}^{\infty} {n+s-1 \choose n} x^n = \frac{1}{(1-x)^{s}}.
$$


See [this link for more](https://en.m.wikipedia.org/wiki/Binomial_series).


