Regular sum of hypercube regions' volumes

Question

I need to calculate, or at least to find a good estimate, for the following sum $$ \sum_{n_1+n_2 + \dots + n_k = N}\frac{1}{(n_1+1)(n_2+1)\dots (n_k+1)},\quad (1)$$ where $n_i \ge 1$. These numbers represent volumes of particular hyperpyramids in a hypercube, therefore the title of the question.

Update: The motivation section below contains an arithmetic error, but the required sum seems to appear nonetheless in an updated formula, and I also guess that this kind of sum may appear quite naturally in all sorts of tasks.

Motivation: I have independent equidistributed $\mathbb{R}$-valued random variables $\xi_1,\dots,\xi_{N+1}$ with $P(\xi_i = \xi_j) = 0$. Denote by $\Diamond_i$ either $<$ or $>$. Then, provided $\sharp\{\Diamond_i \text{ is} >\} = k$ the probability of the event $$P\left(\xi_1\Diamond_1\xi_2, \xi_3\Diamond_2\max(\xi_1,\xi_2),\dots, \xi_{N+1}\Diamond_N\max(\xi_1,\dots,\xi_{N})\right) = \frac{1}{(n_1+1)(n_2+1)\dots (n_k+1)}, \quad(2)$$ where $n_1 + \dots + n_k = N$ and $n_i \ge 1$ and correspond to the places where $\Diamond_i$ is a $>$. By design, all events of the form $\sharp\{\Diamond_i \text{ is} >\} = k$ are mutually exclusive, so $P(\sharp\{\Diamond_i \text{ is} >\} = k)$ is the sum of all possible events of the from $(2)$, which gives $(1)$.

Extended task: What I am actually about to calculate is $P(\sharp\{\Diamond_i \text{ is} >\} \le k)$, which thus gives a formula $$\sum_{l=1}^k \sum_{n_1+n_2 + \dots + n_l = N}\frac{1}{(n_1+1)(n_2+1)\dots (n_l+1)}.\quad (3)$$ This formula, though more complex, may have some nice cancellations in it, perhaps.

Answer 1

Note that $$ \left( {\ln \left( {\frac{1}{{1 - x}}} \right)} \right)^{\,m} = \sum\limits_{0\, \le \,k} {\frac{m!}{k!}{k\brack m}\,x^{\,k} } $$ where the square brackets indicate the (unsigned) Stirling N. of 1st kind.

From that one obtains \begin{align*} {n\brack m} &=\frac{n!}{m!}\sum_{\substack{1\,\leq\,k_j\\k_1\,+\,k_2\,+\,\cdots\,+\,k_m\,=\,n}}\frac{1}{k_1\,k_2\,\cdots\,k_m} \\&=\frac{n!}{m!}\sum_{\substack{0\,\leq\,k_j\\k_1\,+\,k_2\,+\,\cdots\,+\,k_m\,=\,n-m}}\frac{1}{(1+k_1)(1+k_2)\cdots(1+k_m)} \end{align*} which is an alternative definition for such numbers.

In the referenced link you can also find the asymptotic formulation.

Answer 2

The sum in $(1)$ is equal to the coefficient of $x^N$ in $$\Big(\sum_{n=1}^{\infty}\frac{x^n}{n+1}\Big)^k=\Big(\frac{-x-\ln(1-x)}{x}\Big)^k.$$ This alone can already be used for computations. A closer look at $$\Big(\frac{-x-\ln(1-x)}{x^2}\Big)^k=\sum_{n=0}^{\infty}a_{n,k}x^n$$ (the sum in $(1)$ is thus $a_{N-k,k}$) reveals a better-to-use recurrence $$a_{n,k}=\frac{k}{n+2k}\sum_{m=0}^{n}a_{m,k-1}.\qquad(k>0)$$ This can also be used for estimates and asymptotic analysis (if needed).