Trying to simplify a summation... Looking to simplify the following summation, tried to shift terms etc but can't see how the conditions are satisfied.
$$\large \frac{\sum\limits_{p=1}^{N-1} \frac{p(2(N-1)-1-p)!}{(N-1)!(N-1-p)!}\frac{(\beta)^{-p-1}-(\alpha)^{-p-1}}{(\beta)^{-1}-(\alpha)^{-1}}}{\sum\limits_{p=1}^N\frac{p(2N-1-p)!}{N!(N-p)!}\frac{(\beta)^{-p-1}-(\alpha)^{-p-1}}{(\beta)^{-1}-(\alpha)^{-1}}}$$
I need some how maybe taking N$\rightarrow \infty$? satisfy the following for the above summation; 
equal to 0.25 if $\alpha \geq 0.5$ and $\beta \geq 0.5$, then $\alpha(1-\alpha)$ if, $\alpha < 0.5$ and $\beta > \alpha $ and lastly $\beta(1-\beta)$ if $\beta < 0.5$ and $\alpha>\beta$.
Many thanks in advance any help would be most appreciated.
 A: Note that 
$$
\begin{align}
\frac{p(2N-1-p)!}{N!(N-p)!}
&=\frac{p}{N}\binom{2N-1-p}{N-p}\\
&=(-1)^{N-p}\frac{-p}{-N}\binom{-N}{N-p}\\
&=(-1)^{N-p}\left(1+\frac{N-p}{-N}\right)\binom{-N}{N-p}\\
&=(-1)^{N-p}\left(\binom{-N}{N-p}+\binom{-1-N}{N-1-p}\right)\tag{1}
\end{align}
$$
Then
$$
\begin{align}
&\sum_{p=1}^N\frac{p(2N-1-p)!}{N!(N-p)!}x^{-p-1}\\
&=\sum_{p=1}^N(-1)^{N-p}\left(\binom{-N}{N-p}+\binom{-1-N}{N-1-p}\right)x^{-p-1}\\
&=x^{-1-N}\sum_{p=1}^N(-1)^{N-p}\binom{-N}{N-p}x^{N-p}\\
&-x^{-N}\sum_{p=1}^N(-1)^{N-1-p}\binom{-1-N}{N-1-p}x^{N-1-p}\\
&=x^{-1-N}\sum_{p=0}^{N-1}(-1)^p\binom{-N}{p}x^p\\
&-x^{-N}\sum_{p=0}^{N-2}(-1)^p\binom{-1-N}{p}x^p\\
&\sim x^{-1-N}(1-x)^{-N}-x^{-N}(1-x)^{-1-N}\\
&=\frac{1-2x}{x^{N+1}(1-x)^{N+1}}\tag{2}
\end{align}
$$
Therefore, for $0\le\alpha\ne\beta\le1$,
$$
\begin{align}
&\frac{\displaystyle\sum_{p=1}^{N-1}\frac{p(2(N-1)-1-p)!}{(N-1)!(N-1-p)!}\frac{\beta^{-p-1}-\alpha^{-p-1}}{\beta^{-1}-\alpha^{-1}}}
{\displaystyle\sum_{p=1}^N\frac{p(2N-1-p)!}{N!(N-p)!}\frac{\beta^{-p-1}-\alpha^{-p-1}}{\beta^{-1}-\alpha^{-1}}}\\[9pt]
&\sim\frac{\displaystyle\frac{1-2\beta}{\beta^N(1-\beta)^N}-\frac{1-2\alpha}{\alpha^N(1-\alpha)^N}}
{\displaystyle\frac{1-2\beta}{\beta^{N+1}(1-\beta)^{N+1}}-\frac{1-2\alpha}{\alpha^{N+1}(1-\alpha)^{N+1}}}\\[9pt]
&\stackrel{N\to\infty}\longrightarrow\min(\alpha(1-\alpha),\beta(1-\beta))\tag{3}
\end{align}
$$
