I want to find the limit of the following equations:
$\lim_{n\to +\infty} \sum_{k=0}^{n-2} \alpha_{\!_{(k)}} {{n-1}\choose{k}} [1-p_{\!_{(n)}}]^{k} [p_{\!_{(n)}}]^{n-1-k} \quad (1)$
where the first term $\alpha_{\!_{(k)}}$ can be viewed as sequence indexed by $k$, strictly positive and upper bounded by a constant, say $v$;
the second term ${{n-1}\choose{k}} [1-p_{\!_{(n)}}]^{k} [p_{\!_{(n)}}]^{n-1-k}$ is standard binomial probabilities;
$p_{\!_{(n)}}$ is probabilities where $p_{\!_{(n)}} \in (0,1)$ and $\lim_{n\to +\infty} p_{\!_{(n)}} = 1 $ and $\lim_{n\to +\infty} p_{\!_{(n)}} ^n= 0 $.
To find the limit of Equation (1). I first find the maximum of second term (the binomial). When $k=(n-1+1)p_{\!_{(n)}} = n p_{\!_{(n)}}$, the max is $(1 - p)^{n p} p^{-1 + n - n p} {{n-1}\choose{np}} =: \bar{b} $.
For example if Equation (1) is less than
$\lim_{n\to +\infty} \sum_{k=0}^{n-2} v \bar{b} = (n-1) v \bar{b} \quad (2)$
How can I prove that $\bar{b}$ is less than something, for example $1/n^2$, then Eq(2) is less than $(n-1) v \frac{1}{n^2}$ which is zero at the limit. So I show that Eq(1) is zero when $n\to +\infty$. Thank you.
Let $S_{n}:=\sum_{k=0}^{n-2}a_{k}\left(\begin{array}{c} n-1\\ k \end{array}\right)\left(1-p\left(n\right)\right)^{k}\left(p\left(n\right)\right)^{n-1-k}$ and $a_{k}\leq v$ for all $k$. We have $v\cdot\sum_{k=0}^{n-2}\left(\begin{array}{c} n-1\\ k \end{array}\right)\left(1-p\left(n\right)\right)^{k}\left(p\left(n\right)\right)^{n-1-k}=v\cdot\frac{p\left(n\right)-1+\left(1-p\left(n\right)\right)^{n}}{p\left(n\right)-1}$. Observe that $\lim_{n}\frac{p\left(n\right)-1+\left(1-p\left(n\right)\right)^{n}}{p\left(n\right)-1}=1-\lim_{n}\left(1-p\left(n\right)\right)^{n}=0$. By the sandwich theorem we have that $\lim_{n}S_{n}=0$.