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$\begingroup$

$ \sum_{k_1=0}^{N-1}\frac{\binom{N-1}{k_1}(\beta -2)^{N-1-k_1}}{1+k_1+\alpha(N-1-k_1)} $

where $\alpha \in \mathbb{R}$ and $\alpha >1$ (1.3 say), $\beta \in \mathbb{R}$ and $\beta >2 $, and N is a finite natural number

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    WolframAlpha's [answer](http://www.wolframalpha.com/input/?i=sum%28%28binomial%28n-1,k%29*%28b-2%29%5E%28n-1-k%29%29/%281%2bk%2ba*%28n-1-k%29%29,k=0..n-1%29) in hypergeometric form. ![result](http://i.stack.imgur.com/fEO3B.gif)2012-03-17

1 Answers 1

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$\sum_{k_1=0}^{N-1}\frac{\binom{N-1}{k_1}(\beta -2)^{N-1-k_1}}{1+k_1+\alpha(N-1-k_1)}< \frac{1}{N}\sum_{k_1=0}^{N-1} \binom{N-1}{k_1}(\beta -2)^{N-1-k_1}$ since $\alpha>1$.

Now, $\sum_{k_1=0}^{N-1} \binom{N-1}{k_1}(\beta -2)^{N-1-k_1}=(\beta-1)^{N-1}$ by binomial theorem. Hence, we have

$\displaystyle\sum_{k_1=0}^{N-1}\frac{\binom{N-1}{k_1}(\beta -2)^{N-1-k_1}}{1+k_1+\alpha(N-1-k_1)}< \frac{(\beta-1)^{N-1}}{N}.$

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    Likewise, this sum is $\gt(\beta-1)^{N-1}/(1+\alpha(N-1))$.2012-02-16