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. 
 A: 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. 
A: In fact the expansion coefficients of g(z) above provide the answer, and they have quite a simple closed-form expression.
Changing k to q (just to allow and use k and j as a summation indices) we have in fact:
$$
\eqalign{
  & \ln \left( {{1 \over {1 - x}}} \right)^{\,q} \quad \left| {\;0 \le {\rm integer}\,{\rm  }q} \right.\quad  = \left( {\sum\limits_{1\, \le \,k_{\,1} } {{{x^{\,k_{\,1} } } \over {k_{\,1} }}} } \right)\left( {\sum\limits_{1\, \le \,k_{\,2} } {{{x^{\,k_{\,2} } } \over {k_{\,2} }}} } \right) \cdots \left( {\sum\limits_{1\, \le \,k_{\,q} } {{{x^{\,k_{\,q} } } \over {k_{\,q} }}} } \right) =   \cr 
  &  = \sum\limits_{0\, \le \,j} {\left( {\sum\limits_{\scriptstyle 1\, \le \,k_{\,i}  \atop 
  \scriptstyle \,k_{\,1} \, + \,k_{\,2} \, + \, \cdots \, + \,k_{\,q} \, = \,j}  {{1 \over {k_{\,1} k_{\,2}  \cdots k_{\,q} }}} } \right)x^{\,j} }  =   \cr 
  &  = \sum\limits_{0\, \le \,j} {{{q!} \over {j!}}\left[ \matrix{
  j \cr 
  q \cr}  \right]\,x^{\,j} }  = x^{\,q} \sum\limits_{0\, \le \,j} {{{q!} \over {\left( {q + j} \right)!}}\left[ \matrix{
  q + j \cr 
  q \cr}  \right]\,x^{\,j} }  \cr} 
$$
where ${\left[ \matrix{  j \cr   q \cr}  \right]}$ indicates the (unsigned) Stirling Number of 1st kind. 
The above deriving from the double expansion :
$$
{1 \over {\left( {1 - x} \right)^{\,y} }} = \left\{ \matrix{
   = \sum\limits_{0\, \le \,j} {\,\left( { - 1} \right)^{\,j} \left( \matrix{
   - y \cr 
  j \cr}  \right)x^{\,j} }  = \sum\limits_{0\, \le \,j} {{{\left( { - 1} \right)^{\,j} } \over {j!}}\,x^{\,j} \left( { - y} \right)^{\,\underline {\,j} } }  = \sum\limits_{\scriptstyle 0\, \le \,k \atop 
  \scriptstyle 0\, \le \,j}  {{1 \over {j!}}\left[ \matrix{
  j \cr 
  k \cr}  \right]\,x^{\,j} y^{\,k} }  \hfill \cr 
   = \exp \left( {y\ln \left( {{1 \over {1 - x}}} \right)} \right) = \sum\limits_{0\, \le \,k} {{{y^{\,k} } \over {k!}}\left( {\ln \left( {{1 \over {1 - x}}} \right)} \right)^{\,k} }  \hfill \cr}  \right.
$$
