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](https://en.wikipedia.org/wiki/Kronecker_delta).



Generating function of a sequence involving reciprocals of binomial coefficients
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](http://eqworld.ipmnet.ru/en/solutions/ode/ode0103.pdf), 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
$$G(x) = \frac{k(1-xz)^{k-1}}{x^k(1-x)} \int \frac{x^{k-1}}{(1-xz)^k}\,dx$$
$$= \frac{(1-xz)^{k-1}}{1-x} \sum\_{t\geq 0} \frac{k}{k+t} \binom{-k}{t} (-z)^t x^t$$
$$ = \frac{(1-xz)^{k-1}}{1-x}\ {}\_2F\_1(k,k+2;k+1;-xz).$$




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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 $k\geq 1$.


