Proving that $\sum_{i=0}^{n-p} \frac{i!}{(p+i)!} = \frac{1}{p-1}[\frac{1}{(p-1)!}-\frac{(n-p+1)!}{n!}]$

Question

I'm trying to prove that $$\sum_{i=0}^{n-p} \frac{i!}{(p+i)!} = \frac{1}{p-1}\left[\frac{1}{(p-1)!}-\frac{(n-p+1)!}{n!}\right]$$

for $p,n \geq 2$, $p, q \in \mathbb{N}$.

I'm trying to use induction but these two variables confuses me.

Can someone give me a hint?

Thanks.

Answer 1

Induction works: $$\frac1{p-1}\left(\frac1{(p-1)!}-\frac{(n-p+1)!}{n!}\right)-\frac1{p-1}\left(\frac1{(p-1)!}-\frac{(n-p)!}{(n-1)!}\right)\\ =\frac{(n-p)!}{p-1}\left(\frac1{(n-1)!}-\frac{n-p+1}{n!}\right)=\frac{(n-p)!}{p-1}\frac{p-1}{n!}=\frac{(n-p)!}{n!}$$ So the difference of two successive right hand sides gives us the last term in the sum on the left. The base case is trivial when you start at $n=p-1$.