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

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?

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

The above question has been solved by Iosif Pinelis. A variation is $$ \lim_{n\to \infty}\sum_{k=0}^{\lfloor\alpha n \rfloor}C_n^k(-1)^k\left(1-\frac{k}{\alpha n}\right)^n. $$ How can we handle this sum?

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

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

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?