Infinite sum of ratio of factorials Mathematica tells me that for $r, p$ positive integers with $\mathcal{r}  < p$ we have
$$\sum_{i=p+1}^\infty \frac{(i-r)!}{(i+r)!} = \frac{(p+1-r)!}{(2r-1)(p+r)!}. $$
Can anyone point me in the direction of a proof of this?
For $r=1$, this is reduces to 
$$ \sum_{i=p+1}^\infty \frac{1}{i(i+1)} = \frac{1}{1+p}$$ which can be shown be a partial fraction expansion and telescoping sum argument.
Thanks.
 A: Let $$f(p) = \dfrac{(p+1-r)!}{(2r-1)(p+r)!}$$
and note that $$f(p) - f(p-1) = - \dfrac{(p-r)!}{(p+r)!}$$
which tells you that your sum is $f(p)+c$ for some constant $c$.
Now just look at the limit as $p \to \infty$.
A: You need to now the the evaluation of the hypergeometric Gauss function $_2F_1(a,b;c;1)$ which is given by:
$$_2F_1(a,b;c;1)=\sum_{i=0}^{\infty} \frac{\Gamma(a+i)\cdot \Gamma(b+i)\cdot \Gamma(c)}{\Gamma(a)\cdot \Gamma(b)\cdot \Gamma(c+i)\cdot \Gamma(i+1)}=\frac{\Gamma(c)\cdot \Gamma(c-a-b)}{\Gamma(c-a) \cdot \Gamma(c-b)}$$
With $\mathfrak R(c-a-b)\gt 0$ and where $\Gamma(\cdot)$ denotes the Euler Gammafunction (so $\Gamma(n+1)=n!).$
If you transform your sum in an appropriate way, you get:
$$\sum_{i=p+1}^{\infty}\frac{(i-r)!}{(i+r)!}=\sum_{i=0}^{\infty}\frac{(i+p+1-r)!}{(i+p+1+r)!}=\sum_{i=0}^{\infty}\frac{\Gamma(p+2-r+i)}{\Gamma(p+2+r+i)}=\frac{\Gamma(p+2-r)}{\Gamma(p+2+r)}\cdot \sum_{i=0}^{\infty}\frac{\Gamma(1+i)\cdot \Gamma(p+2-r+i)\cdot \Gamma(p+2+r)}{\Gamma(1)\cdot \Gamma(p+2-r)\cdot \Gamma(p+2+r+i)\cdot \Gamma(i+1)}=\frac{\Gamma(p+2-r)}{\Gamma(p+2+r)}\cdot _2F_1(1,(p+2-r);(p+2+r);1)=\frac{\Gamma(p+2-r)}{\Gamma(p+2+r)}\cdot \frac{\Gamma(p+2+r)\cdot \Gamma(p+2+r-1-(p+2-r))}{\Gamma(p+2+r-1) \cdot \Gamma(p+2+r-(p+2-r))}=\frac{\Gamma(p+2-r)\cdot \Gamma(2r-1)}{\Gamma(p+1+r) \cdot \Gamma(2r)}=\frac{\Gamma(p+2-r)\cdot \Gamma(2r-1)}{\Gamma(p+1+r) \cdot (2r-1)\cdot \Gamma(2r-1)}=\frac{(p+1-r)!}{(p+r)! \cdot (2r-1)}$$
