The sequence $(-1)^n\binom{\alpha-1}{n}$ converges. I need to show that for $n \in \mathbb N_0$ and $\alpha \ge 0$ the sequence $(-1)^n\binom{\alpha-1}{n}$ converges.
It can be shown that the sequence convereges to zero using a theorem claiming that $|\binom{\alpha}{n}| \sim \frac{c}{n^{1+\alpha}}$ for a $c \in \mathbb R$ and $\alpha \in \mathbb R \setminus \mathbb N$. But I am wondering whether there is a direct proof.
 A: A general approach is to consider the convergence of the related binomial series and infer that the general term converges to $0$. However, here we have the more troublesome case at the endpoint $x = -1$ of the interval of convergence.
Consider the general term
$$a_n = (-1)^n\binom{\alpha}{n}.$$
Using induction we can show that
$$\sum_{n=0}^ma_n = \prod_{n=1}^m\left(1 - \frac{\alpha}{n}\right).$$
For example,
$$1 - \binom{\alpha}{1} + \binom{\alpha}{2} = 1 - \alpha + \frac{\alpha(\alpha-1)}{2} \\ = 1 - \frac{3\alpha}{2} + \frac{\alpha^2}{2} \\ = \left(1 - \frac{\alpha}{1}\right)\left(1 - \frac{\alpha}{2}\right)$$
Since the harmonic series $\sum (1/n)$ diverges, we can show that both the product and series $\sum a_n$ converge.
With $\alpha > 0$ fixed, there exists $N \in \mathbb{N}$ such that $0 < \alpha /n < 1$ for $n > N$.  Hence,
$$1 - \frac{\alpha}{n} \leqslant \exp\left(-\frac{\alpha}{n} \right),$$
and
$$0 < \prod_{n=N+1}^m\left(1 - \frac{\alpha}{n}\right) \leqslant \exp \left(-\alpha\sum_{n=N+1}^m \frac{1}{n}\right).$$
In the limit as $m \to \infty$ we have
$$\prod_{n=N+1}^\infty\left(1 - \frac{\alpha}{n}\right) = 0$$ 
Hence, for $\alpha > 0,$
$$\sum_{n=0}^\infty a_n < \infty,$$
and 
$$\lim_{n \to \infty}(-1)^n\binom{\alpha}{n} = 0.$$
This argument also carries over to show that if $\alpha > 1$, then
$$\lim_{n \to \infty}(-1)^n\binom{\alpha-1}{n} = 0$$
