Regular sum of hypercube regions' volumes 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.
 A: 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).
A: 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.
