Upper bound on $ \binom{a}{m+1}\sum ^m_{j=0} \binom{a-m-1}{j}/\binom{b}{j+m+1}$ Given $a,b,m$ such that $0<2m<a<b$.
I would like to find out upper bound of $$S = \binom{a}{m+1}\sum ^m_{j=0} \frac{\binom{a-m-1}{j}}{\binom{b}{j+m+1}}$$
Anyone can help me please?
Thank you so much.
 A: It is crude but you may proceed as follow. For every $0 \leq j \leq m$ we have
$$\frac{\binom{a-m-1}{j}}{\binom{b}{j+m+1}} \leq \binom{a-m-1}{j} \leq (a-m-1)!$$
and so 
$$ \begin{array}{rcl}S&=&\binom{a}{m+1}\sum ^m_{j=0} \frac{\binom{a-m-1}{j}}{\binom{b}{j+m+1}} \\ &\leq & \binom{a}{m+1} (m+1)(a-m-1)!\\ &=& \frac{a!(m+1)(a-m-1)!}{(m+1)!(a-m-1)!}\\ &=& \frac{a!}{m!}\end{array}.$$
A: I begin by using this identity: $\int_0^1 t^{\alpha} (1-t)^{\beta}=\frac{!}{\alpha+\beta+1}\frac{1}{\binom{\alpha+\beta}{\alpha}}$, for some positive $\alpha$ and $\beta$. This can be alternatively expressed as follows: for some positive $n$ and $k$, we have $\frac{1}{\binom{n}{k}}= (n+1) \int_0^1 t^{k}(1-t)^{n-k} dt$. Applying this to the right hand side, we have: 
$$S=(b+1) \binom{a}{m+1} \sum_{j=0}^m \binom{a-m-1}{j} \int_0^1 t^{j+m+1}(1-t)^{b-j-m-1} dt\\
=  (b+1) \binom{a}{m+1} \int_0^1 (1-t)^{b-m-1} t^{m+1} \left[\sum_{j=0}^m \binom{a-m-1}{j} \left( \frac{t}{1-t}\right)^j \right] dt \\
\leq (b+1) \binom{a}{m+1} \int_0^1 (1-t)^{b-m-1} t^{m+1} \left[\sum_{j=0}^{a-m-1} \binom{a-m-1}{j} \left( \frac{t}{1-t}\right)^j \right] dt
\\
= (b+1) \binom{a}{m+1} \int_0^1 (1-t)^{b-m-1} t^{m+1}  \left[ \left(1+ \frac{t}{1-t}\right)^{a-m-1} \right] dt
\\
= (b+1) \binom{a}{m+1} \int_0^1  t^{m+1} (1-t)^{b-a} \\ 
=  \frac{b+1}{b-a+m+2}    \frac{ \binom{a}{m+1}}{\binom{b-a+m+1}{m+1}}
\\
=(b+1)\frac{a!(b-a)!}{(b-a+m+2)!(a-m-1)!}
$$ 
In case I have made arithmetic mistakes, I am sure you can pick it up and mend it, since I hope the main logistics are clear
