Sum over the inverse cumulative product of integer compositions

Question

I've been trying to get an inroad on the following problem: \begin{align} S(N,k) = \sum_{a_1,\dots,a_k\geq 1; \sum a_i = N} \frac1{a_1(a_1+a_2)\cdots(a_1+\cdots+a_k)}, \end{align} where $N$ is a positive integer.

I have seen the question combinatorics: sum of product of integer compositions and from there found methods for counting sums of the form $\sum\prod f(a_i)$ but that does not seem to fit here?

Eventually, I want to then get to the sum $\sum_{k\geq 1}S(N,k)\alpha^k$, with some positive real $\alpha$. In fact an approximation for $N\gg1$ would already be useful.

This might have a known solution or be obvious to an expert in combinatorics, but I haven't been able to find something.

Answer 1

As $(a_1,\dots,a_{k-1},a_k)$ ranges over postive integers summing to $N$, then $$(a_1,a_1+a_2,a_1+a_2+a_3,\dots,a_1+\dots+a_k)$$ will range over subsets of $\{1,\dots,N\}$ of size $k$ which include $N$ (written in increasing order). Therefore, \begin{align} S(N,k) &=\sum_{1\le b_1<\dots<b_{k-1}<N}\frac1{b_1\cdots b_{k-1}N} \\&=\frac1N\sum_{1\le b_1<\dots<b_{n-1}\le N-1}\frac1{b_1\cdots b_{k-1}} \\&=\frac1N[x^{k-1}](1+x)(1+\tfrac{x}2)\cdots (1+\tfrac{x}{N-1}) \\&=\frac1{N!}[x^{k}]x(1+x)(2+x)\cdots ((N-1)+x) \\&=\frac1{N!}{N\brack k} \end{align} where ${N\brack k}$ is the $(N,k)$ Stirling number of the first kind, equal to the number of permutations of $\{1,\dots,N\}$ with $k$ cycles. That is, $S(N,k)$ is the probability a random such permutation has $k$ cycles.

A fact which we used above is that $$ \sum_{k=0}^N {N \brack k}x^k=x(1+x)(2+x)\cdots (N-1+x) $$ which implies that $$ \sum_{k=0}^N S(N,k)\alpha^k=\frac{\alpha(1+\alpha)(2+\alpha)\cdots(N-1+\alpha)}{N!}=\binom{\alpha+N-1}{N}=(-1)^N\binom{-\alpha}{N} $$ so you have a very nice "closed form" for your summation. The above equation represents the probability generating function for a random variable equal to the number of cycles of a random permutation of $N$ symbols.