Let $a:=\alpha\in(0,1)$. By induction on $m=0,1,\dots$, 
$$\sum_{k=0}^m \binom nk(-1)^k\Big(1-\frac k{a n}\Big) \\ 
=(-1)^{m+1} (a+m-na)\frac{m+1}{an (n-1)}\,\binom n{m+1}.$$
So, letting $S_n$ denote the sum in question, we have 
$$S_n\sim(-1)^{\lfloor a n \rfloor+1}(a-\{na\}) \,\frac1n\,\binom n{m+1}
=(-1)^{\lfloor a n \rfloor+1}(a-\{na\})(2+o(1))^n$$ 
as $n\to\infty$, where $\{na\}$ is the fractional part of $na$. Depending on the arithmetical properties of $a$, the factor $(-1)^{\lfloor a n \rfloor+1}$ will alternate between $1$ and $-1$ and the factor $a-\{na\}$ will oscillate between $a-1<0$ and $a>0$, whereas the factor $(2+o(1))^n$ will of course go to $\infty$. So, the sum $S_n$ will not converge to any limit.