How to calculate: $$\sum _{k=0}^{nm} \frac{1}{nk} {nm \choose k}.$$
Notice that $\frac{1}{nk} = \int_0^1 x^{nk1} dx$. Hence, $$\sum_{k=0}^{nm} \frac{1}{nk}\binom{nm}k = \int_0^1 dx \sum_{k=0}^{nm} x^{nk1}\binom{nm}k = \int_0^1 x^{m1}(1+x)^{nm}dx = (1)^m B_{1}(m,nm+1),$$ where $B_{\cdot}(\cdot,\cdot)$ is incomplete beta function.

$\begingroup$ Wow, this is a very beautiful solution, thank you! @Max $\endgroup$– luwApr 13 '19 at 17:20
Just to demonstrate abilities of CASes. Mathematica produces
Sum[1/(n  k)*Binomial[n  m, k], {k, 0, n  m}] // FullSimplify
$(1)^{n+1} B_{1}(n,m+n+1) $
Maple says
simplify(sum(binomial(nm, k)/(nk), k = 0 .. nm))
$${\frac {{\it JacobiP} \left( nm,n,n+m1,3 \right) \left( n1 \right) !\, \left( nm \right) !}{ \left( m \right) !}} $$

1$\begingroup$ Those factorials of negative integers in the Maple answer are a problem. Doing it slightly differently gives $(1)^{n+m} \operatorname{JacobiP}(nm,n,n+m1,3)(nm)!\,(m1)!/n!$. $\endgroup$ Apr 14 '19 at 2:21

$\begingroup$ @Brendan McKay: Thank you for your valuable comment. $\endgroup$ Apr 14 '19 at 4:02
If you like, there is a 2ndorder recurrence for it.
Let $a_n=\sum_{k=0}^{nm}\frac1{nk}\binom{nm}k$. Then, $$(n+2)a_{n+2}+(m3n4)a_{n+1}+2(n+1m)a_n=0.$$