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