How to calculate:
$$\sum \_{k=0}^{n-m} \frac{1}{n-k} {n-m \choose k}.$$



How to calculate: $\sum\limits_{k=0}^{n-m} \frac{1}{n-k} {n-m \choose k}$
Notice that $\frac{1}{n-k} = \int\_0^1 x^{n-k-1} dx$. Hence,
$$\sum\_{k=0}^{n-m} \frac{1}{n-k}\binom{n-m}k = \int\_0^1 dx \sum\_{k=0}^{n-m} x^{n-k-1}\binom{n-m}k = \int\_0^1 x^{m-1}(1+x)^{n-m}dx = (-1)^m B\_{-1}(m,n-m+1),$$
where $B\_{\cdot}(\cdot,\cdot)$ is [incomplete beta function](https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function).


