Upper bound on $ \binom{a}{m+1}\sum ^m_{j=0} \binom{a-m-1}{j}/\binom{b}{j+m+1}$

Question

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.

Answer 1

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 $$ \begin{array}{rcl}S&=&\binom{a}{m+1}\sum ^m_{j=0} \frac{\binom{a-m-1}{j}}{\binom{b}{j+m+1}} \\ &\leq & \binom{a}{m+1} (m+1)(a-m-1)!\\ &=& \frac{a!(m+1)(a-m-1)!}{(m+1)!(a-m-1)!}\\ &=& \frac{a!}{m!}\end{array}.$$

Answer 2

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