Let $F(n,p)=\sum_{m=p}^{\infty} \frac{(n+m)!}{n!\ m!\ 2^m}$. What is the upper bound for $F(n,p)$, where $p\geq n+1$ and $n\geq 0$? Does there exist a tight bound where the value of $p$ could be easily estimated under a constraint $F(n,p)\leq 2^{-k}$, where $k>1$?
Note that $F(n,p)=\frac{(n+p)!}{n!\ p!}\left(\frac{1}{2^{p-1}}+2^{n+1}n\ B(\frac{1}{2},p+1,n)\right)$, where $B(x,y,z)=\int_0^x t^{y-1}(1-t)^{z-1}dt$ is the incomplete Beta function.
By analyzing the growth rate of $\frac{(n+m)!}{n!\ m!\ 2^m}$, I am able to get that in order to have $F(n,p)\leq 2^{-k}$, it is sufficient to take $p\geq 4n+\frac{12}{7}k+\frac{19}{7}$, but it is not tight for large values of $n$.