Sum of $\prod 1/n_i$ where $n_1,\ldots,n_k$ are divisions of $m$ into $k$ parts. Fix $m$ and $k$ natural numbers. Let $A_{m,k}$ be the set of all  partitions  divisions of $m$ into $k$ parts. That is:
$$A_{m,k} = \left\{ (n_1,\ldots,n_k) : n_i >0, \sum_{i=1}^k n_i = m \right\} $$
We are interested in the following sum $s_{m,k}$:
$$s_{m,k} = \sum_{ (n_1,\ldots,n_k) \in A_{m,k} } \prod_{i=1}^k \frac{1}{n_i} $$
Can you find $s_{m,k}$ explicitly, or perhaps its generating function or exponential generating function?
EDIT: Since order matters in the $(n_1,\ldots,n_k)$ this is not exactly a partition. 

I am not sure I understand the problem correctly, but
$$ g(z) = 
\left( \frac{z}{1} + \frac{z^2}{2} +\frac{z^3}{3} + \ldots + \frac{z^q}{q} + \ldots \right)^k =
\left( \log \frac{1}{1-z} \right)^k $$
looks like a good candidate to me, so that
$$ s_{m,k} = [z^m] \left( \log \frac{1}{1-z} \right)^k.$$
This is the exponential generating function for a sequence of $k$ cycles containing a total of $m$ nodes,
$$\mathfrak{C}(\mathcal{Z}) \times \mathfrak{C(\mathcal{Z})} \times \mathfrak{C(\mathcal{Z})} \times \cdots \times \mathfrak{C(\mathcal{Z})} =
\mathfrak{C}^k(\mathcal{Z}) ,$$
so that $m! [z^m] g(z)$ gives the number of such sequences.
Since the components are at most $m$ we could truncate the inner logarithmic term at $z^m/m$, but I suspect the logarithmic form is more useful for asymptotics. 
