# 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.

• What does "I need for some how maybe taking $N\to\infty$?" mean? Mar 3, 2013 at 18:58
• Is $P$ the same as $p$, and if so, why does it appear outside of the summation?
– robjohn
Mar 3, 2013 at 18:58
• You can simplify by the second factors, and the denominators of the first fractions simplify easily. The remaining sums look nasty, but it seems likely they only differ in the last term or so. That should be enough to get asymptotic behaviour, if that is what you need. Mar 3, 2013 at 19:06

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}