Alternating summation identity for $m!$ in terms of two variables

Question

As part of a larger problem, I'm trying to show that

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

I've tried to see if any of the factorials can be simplified with one another, but even if I did I don't understand how the sum would simplify.

I have tested some values to check that there is no simple telescoping or anything.

Any help would be much appreciated.

Answer 1

We have that

$$\frac{(n+m+1)!}{(i+m+1)(n-i)!i!}=\binom{n}{i}\frac{(n+m+1)!}{(i+m+1)n!}$$

and then

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

which can be proved as shown in the related

• Summing a series with binomial coefficients without calculus.
• Combinatorial interpretation of identity
• Prove the following sum: $\sum_{r=0}^{n} \frac{(-1)^{r}}{m+r+1} \binom{n}{r} = \frac{m! n!}{(m+n+1)!}$