# Generating function of a sequence involving reciprocals of binomial coefficients

Question: Is there a closed-form expression for the following sum

$$F(z,k,r)=\sum_{n=0}^{r} \frac{z^n}{{n+k} \choose {k}}\label{sum}\tag{1}$$

where $z\in\mathbb{C}$, and $r$, $k$ are non-negative integers.

Remark: Obviously, the above expression is a polynomial of degree $r$. However, I am interested in "alternative" expressions for it as later on I will need to compute residues of the product of above functions (see below).

Motivation: I am trying to compute a much nastier expression involving sum over multisets and ratios of multinomial coefficients:

$$\sum_{(n_1,\ldots,n_d)\vdash N-k}\dfrac{{{k} \choose {k_1,k_2,\ldots,k_d}} {{N-k} \choose {n_1,n_2,\ldots,n_d}} }{ {{N} \choose {n_1+k_1,n_2+k_2,\ldots,n_d+k_d}}}\label{ssum}\tag{2}$$

where $N,k$ are fixed non-negative integers and $(k_1,\ldots,k_d)$ is fixed and satisfies $(k_1,\ldots,k_d)\vdash k$.

I got expression $\eqref{sum}$ by trying to compute $\eqref{ssum}$ explicitly via the integral representation of Kronecker delta.

• Writing $\frac{F(z,k,r)}{k!} = \sum_{n=0}^r \frac{\Gamma(n+1)}{\Gamma(n+k+1)} z^n = \sum_{n=0}^r\frac{z^n}{(n)_k}$ looks close to certain hypergeometricish functions I have seen but I am unsure if I know the name of them. – Daniel Parry Sep 21 '15 at 12:11
• In case it is of any help, the generating function $G(x)=\sum_r F(z,k,r)x^r$ appears to satisfy $x(x-1)(xz-1)G'(x) + \big(x(2x-1)z -(k+1)x+k\big)G(x) - k = 0$. – Martin Rubey Sep 21 '15 at 13:56
• (The generating function according to $k$ does not appear to be as nice.) – Martin Rubey Sep 21 '15 at 14:04
• Mathematica writes it out as a difference of two ${}_2F_1$ hypergeometrics, but that seems to be perhaps not the "closed-form" you are after! – Suvrit Sep 21 '15 at 15:11
• @Suvrit Note that the first of the hypergeometrics does not involve $r$ – Igor Rivin Sep 21 '15 at 15:29

Let's solve Martin Rubey's differential equation, which I write as $$g(x)G'(x) = f_1(x)G(x) + f_0(x),$$ where $g(x) = x(x-1)(xz-1)$, $f_1(x) = -x(2x-1)z+(k+1)x-k$, and $f_0(x)=k$.
Then by the general formula, we compute $$F(x) = \int \frac{f_1(x)}{g(x)}\,dx = (k - 1) \ln(x z - 1) - k\ln(x) - \ln(x - 1),$$ and thus $$e^{F(x)} = \frac{(xz-1)^{k-1}}{x^k(x-1)}.$$ Then since $G(0)=1$, we get $$G(x) = \frac{k(1-xz)^{k-1}}{x^k(1-x)} \int \frac{x^{k-1}}{(1-xz)^k}\,dx$$ $$= \frac{(1-xz)^{k-1}}{1-x} \sum_{t\geq 0} \frac{k}{k+t} \binom{-k}{t} (-z)^t x^t$$ $$= \frac{(1-xz)^{k-1}}{1-x}\ {}_2F_1(k,k+2;k+1;-xz).$$
Extracting the coefficient of $x^r$: $$[x^r]\ G(x) = \sum_{n\leq r}(-z)^n\sum_{t\geq 0} \frac{k}{k+t} \binom{-k}{t} \binom{k-1}{n-t},$$ we arrive at a remarkable identity $$\sum_{t\geq 0} \frac{k}{k+t} \binom{-k}{t} \binom{k-1}{n-t} = \frac{(-1)^n}{\binom{n+k}k},$$ which holds for all $n\geq 0$ and $k\geq 1$.
• The identity is a special case of Saalschütz's theorem. It can be generalized to $$\sum_{t\geq 0} \frac{a}{a+t} \binom{-k}{t} \binom{k-1}{n-t} = (-1)^n\frac{\binom{a}{k}}{\binom{n+a}k}.$$ – Ira Gessel Apr 20 '18 at 22:51