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

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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All i could do was find an upper bound for the summation as $(\beta - 1)^{N-1}$ –  Nischal Jan 17 '12 at 7:14
    
Have you tried using Gosper's algorithm or (since it has compact support) Zeilberger's algorithm? –  Peter Taylor Jan 17 '12 at 11:55
    
@PeterTaylor Had not heard of the algorithms that you have mentioned. Will read up on them and give them a try.Thanks –  Nischal Jan 17 '12 at 15:02
    
WolframAlpha's answer in hypergeometric form. !result –  Raymond Manzoni Mar 17 '12 at 10:50
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1 Answer

$\sum_{k_1=0}^{N-1}\frac{\binom{N-1}{k_1}(\beta -2)^{N-1-k_1}}{1+k_1+\alpha(N-1-k_1)}< \frac{1}{N}\sum_{k_1=0}^{N-1} \binom{N-1}{k_1}(\beta -2)^{N-1-k_1}$ 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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Thanks for getting a tighter bound –  Nischal Jan 17 '12 at 15:02
    
Likewise, this sum is $\gt(\beta-1)^{N-1}/(1+\alpha(N-1))$. –  Did Feb 16 '12 at 9:31
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