Let $0 < \ell \leq b \leq a$. Let $a > b+\ell$. I am trying to upper bound the following expression such that the upper bound evaluates to $\leq 1$ (clearly 1 is an upper bound but i want something nontrivial such that the upper bound is $<1$ whenever possible):
\begin{equation} \sum_{i=1}^{\ell} {\ell \choose i} \frac{{b \choose i}}{{a \choose i}} (-1)^{i+1} \end{equation}
In this process of trying it : the best thing i could do is do the following approximation: \begin{equation} \frac{{b \choose i}}{{a \choose i}} \approx (\frac{b}{a})^i \frac{e^{-1/b} e^{-2/b}....e^{-(i-1)/b}}{e^{-1/a} e^{-2/a}....e^{-(i-1)/a}} \end{equation} This leads to a summation of the form: \begin{equation} \sum_{i=1}^{\ell} {\ell \choose i} e^{-c_1{i^2}} c_2^i (-1)^{i+1} \end{equation} for some $c_1,c_2$
But i do not know whether this approximation will give an upper bound or not. I also do not know how to evaluate the approximate summation given above.
I am going to use my actual summation by multiplying with a number tending to infinity. So i need a very good approximation and it needs to be $<1$ whenever possible.
[Update]: I was able to reduce the summation to the following expression : Coefficient of $z^{\ell}$ in the following expression: \begin{equation} (1+z)^b \int_{0}^{1} (z-(z+1)x)^{\ell} (1-x)^{a-\ell} dx \end{equation} Any help with above integral is also appreciated...
Also the summation can be converted to: \begin{equation} \sum_{i=1}^{\ell} {\ell \choose i} {b \choose i} \frac{1}{(i-1)!} \sum_{j=0}^{i-1} \frac{{(i-1) \choose j}}{a-j} (-1)^{i-1-j}(-1)^{i+1} \end{equation}
The sum has a closed form so Stirling's asymptotic formula for the ratio of factorials should do the trick: $$\sum_{k=1}^n (-1)^{k+1} \binom{n}{k} \frac{ \binom{b}{k}}{\binom{a}{k}}= 1-\frac{\Gamma(-a)\Gamma(n+b-a)}{\Gamma(b-a)\Gamma(n-a)} $$ The identity is from a Gauss hypergeometric 2F1 evaluated at 1. Although it was not stated that $a,b$ could be integer, in those cases you would need to use reflection formulas to avoid an indeterminate expression.