Let $Z\_N = \displaystyle{\sum\_{k+j\leq N}} \frac{N!N^{k+j}}{N^{N+1}}\frac{u^kv^j}{k!j!}\binom{N-j}{N-j-k}$ where $u$ and $v$ are two unknowns. 


My question is: Is there a closed-form for $Z\_N$ or is $Z\_N$ the multivariate taylor series expansion of some function $f(u,v)$ such that $f(u,v)$ has a closed form? Thanks very much!



A closed form of a summation or the taylor series expansion of some function with a closed form?
Let's focus on $P\_N:=\sum\_{k+j\leq N}\frac{N^{k+j}}{k!j!}\binom{N-j}ku^kv^j$. To get $Z\_N$, just multiply out by $\frac{N!}{N^{N+1}}$. 


The notation $[z^m]F(z)$ means the coefficient $a\_m$ of $z^m$ in the series expansion $F(z)=\sum\_na\_nz^n$.


Start by changing variables $m=k+j$ so that $j=m-k$ and hence
\begin{align} P\_N&=\sum\_{m=0}^N\sum\_{k=0}^m\frac{N^m}{k!(m-k)!}\binom{N-m+k}ku^kv^{m-k} \\
&=\sum\_{m=0}^N\frac{N^m}{m!}\sum\_{k=0}^m\binom{m}{m-k}v^{m-k}\binom{N-m+k}ku^k \\
&=\sum\_{m=0}^N\frac{N^m}{m!}\cdot[z^m]\left(\frac{(1+vz)^m}{(1-uz)^{N-m+1}}\right) \\
&=\frac1{2\pi i}\sum\_{m=0}^N\frac{N^m}{m!}\int\_{\gamma}\frac{(1+vz)^m}{(1-uz)^{N-m+1}}\frac{dz}{z^{m+1}};
\end{align}
where in the last line we invoked [Cauchy's Integral Formula](https://en.wikipedia.org/wiki/Cauchy's_integral_formula) along a closed path $\gamma$ about $z=0$.


I don't think we can expect a closed formula.


