# Iterated sums--something like a differsum

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

• This is the discrete convolution, en.wikipedia.org/wiki/Convolution#Discrete_convolution In the particular case of sequences $u:\mathbb{Z}\to\mathbb{C}$ supported on $\mathbb{Z}_+$, it is also called "one-side convolution", and corresponds to the Cauchy product of power series. – Pietro Majer Nov 25 '16 at 19:56
• @PietroMajer I'm well aware of that, I guess I didn't make myself clear enough. I was more interested in the properties of the iterated sum, and then the complex iterated sum. If this construction exists in literature somewhere. – user78249 Nov 25 '16 at 19:58
• This fails when $n=0,$ right? – Pat Devlin Nov 25 '16 at 20:45
• And this even fails for positive integers $s$ and $q$. Right? – Pat Devlin Nov 25 '16 at 20:50
• Should it be $1\times\ldots\text{($k$times)}\ldots\times 1 = {n+k-1\choose k-1}$? – Julian Rosen Nov 25 '16 at 23:18

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 $$G_s (x) = \sum_{n=0}^{\infty} {n+s-1 \choose n} x^n = \frac{1}{(1-x)^{s}}.$$
• I don't see how the OP's question is asking what you say it is really asking (which is a form of the well-known Chu-Vandermonde identity). But perhaps the OP should clarify what is meant by the expression $\binom{n}{s}$ (for complex $s$): is it $\frac{\Pi(n)}{\Pi(s)\Pi(n-s)}$ where $\Pi(s) := \int_0^\infty x^s e^{-x} dx$? – Todd Trimble Nov 26 '16 at 1:19
• What he asked is false and not analogous with his convolution formula. His convolution formula is supposed to be that of my post when $s$ is a positive integer. In his notation, $1\times 1$ is supposed to have generating function $1/(1-x)^2$, but his didn't. – Pat Devlin Nov 26 '16 at 1:22
• Even after fixing the typo, I still don't understand how this would answer the (fixed-up) question. Roughly speaking: how is an expression of type $\binom{\text{integer}}{s}$, by my reading a transcendental function in the complex variable $s$, supposed to relate to expressions of type $\binom{z}{\text{integer}}$ which are polynomials in $z$? – Todd Trimble Nov 26 '16 at 1:38
• It's supposed to be ${n+s-1 \choose s-1} = {n+s-1 \choose n}$ – Pat Devlin Nov 26 '16 at 1:40