Simplify and upper bound of sum equal zero

Question

Let $\beta_1$ , $\beta_2$ and $S$ real numbers, and $N$ and $M$ natural numbers. Can we simplify this formula. $$ \sum_{n=0}^{N-1}\sum_{m=0}^{NM-1}(-1)^{n+m}\binom{N-1}{n}\binom{NM-1}{m} \beta_{1}\beta_{2}\begin{pmatrix} \frac{\beta_{1}(n+1)+\beta_{2}(m+1)}{\beta_{1}(n+1)\beta_{2}(m+1)S} \end{pmatrix} .$$ After simplification if there is any upper bound for let Because this sum it equal zero. Any solution please.

Answer 1

Since $\frac{a+b}{ab} = \frac 1a + \frac 1b$, your sum simplifies to $$\frac 1S\sum_{n=0}^{N-1}\sum_{m=0}^{NM-1}(-1)^{n+m}\binom{N-1}{n}\binom{NM-1}{m}\left(\frac{\beta_2}{n+1} + \frac{\beta_1}{m+1}\right)$$

Which further simplifies to $$\frac 1S\left[\beta_2\left(\sum_{n=0}^{N-1}\binom{N-1}n(-1)^n\frac1{n+1}\right)\left(\sum_{m=0}^{NM-1}\binom{NM-1}m(-1)^m\right) \\+ \beta_1\left(\sum_{m=0}^{NM-1}\binom{NM-1}m(-1)^m\frac1{m+1}\right)\left(\sum_{n=0}^{N-1}\binom{N-1}n(-1)^n\right)\right]$$

Values for each of the expressions in $(\,)$ can be found with a little searching (though actually, you really only need to find the value of two of them).