Prove the following sum: $\sum_{r=0}^{n} \frac{(-1)^{r}}{m+r+1} \binom{n}{r} = \frac{m! n!}{(m+n+1)!}$

Question

$$\sum_{r=0}^{n} \frac{(-1)^{r}}{m+r+1} \binom{n}{r} = \frac{m! n!}{(m+n+1)!}$$

I expanded the series and I can see that the denominator of the RHS = LHS but I don't really know how to prove it properly.

Answer 1

Hint: 2 approaches

• Consider the $n$-th difference of function $\frac{1}{m+1+x}$ using the formula https://en.wikipedia.org/wiki/Finite_difference#n-th_difference. (
  This also means you can prove the formula by induction on $n$ using minus two formulas)
• Consider the generating function $f(x)=(1-x)^n\ln(1-x) $, whose coefficient of $x^{n+m+1}$ is $f_{n,m}=(-1)^{n+1}\mathrm{LHS}$. Since $f'=-n(1-x)^{n-1}\ln(1-x)+-(1-x)^{n-1}$, we obtain $(n+m+1)f_{n,m}=-nf_{n-1,m}$.

Answer 2

Consider the sum

$$ \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{m+r} = x^m \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} x^{r} = x^m(1-x)^n.$$

Integrating both sides of the above equation with respect to $x$ from $0$ to $1$, we have

$$ \sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1} x^{m+r+1}\Big|_{0}^{1}=\int_{0}^{1}x^m(1-x)^n dx .$$

The last integral is nothing but the beta function,then we get

$$ \sum_{r=0}^{n} (-1)^{r} \binom{n}{r}\frac{1}{m+r+1}= \beta(m+1,n+1)=\frac{m!n!}{(m+n+1)!}. $$