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

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

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.

• Wow, this is a very beautiful solution, thank you! @Max
– luw
Apr 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(n-m, k)/(n-k), k = 0 .. n-m))


$$-{\frac {{\it JacobiP} \left( n-m,-n,-n+m-1,3 \right) \left( -n-1 \right) !\, \left( n-m \right) !}{ \left( -m \right) !}}$$

• Those factorials of negative integers in the Maple answer are a problem. Doing it slightly differently gives $(-1)^{n+m} \operatorname{JacobiP}(n-m,-n,-n+m-1,3)(n-m)!\,(m-1)!/n!$. Apr 14 '19 at 2:21
• @Brendan McKay: Thank you for your valuable comment. Apr 14 '19 at 4:02

If you like, there is a 2nd-order recurrence for it.

Let $$a_n=\sum_{k=0}^{n-m}\frac1{n-k}\binom{n-m}k$$. Then, $$(n+2)a_{n+2}+(m-3n-4)a_{n+1}+2(n+1-m)a_n=0.$$