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!


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 along a closed path $\gamma$ about $z=0$.

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


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