$\lim_{n\to\infty}\sum_{k=0}^{\lfloor\alpha n \rfloor}C_n^k(-1)^k(1-\frac{k}{\alpha n})$, where $C_n^k$ is the  binomial coefficient  and $0 <\alpha <1$.


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The above question has been solved by Losif Pinelis. A variation is $\lim_{n\to \infty}\sum_{k=0}^{\lfloor\alpha n \rfloor}C_n^k(-1)^k(1-\frac{k}{\alpha n})^n$. 

How can we handle with this sum?