How to summate fractional binomial coefficients?

Question

I have fractional binomial coefficients like this

$$g_j^{(\beta)}=(-1)^j{\beta \choose j}$$

where $\beta \in (1,2)$ and $j = 0,1,2,\cdots$.

In a paper, the author says that $\sum_{j=0}^{\infty}g_j^{(\beta)}=0$. I don't know how to prove it.

Answer 1

Note that $ \binom{-1}{j} = (-1)^j $, so we can write the partial sum as $$ \sum_{j=0}^k (-1)^{k-j} \binom{\beta}{j} = (-1)^k \sum_{j=0}^k (-1)^{k-j} \binom{\beta}{j} = (-1)^k \sum_{j=0}^k \binom{-1}{k-j} \binom{\beta}{j} = (-1)^k \binom{\beta-1}{k} $$ by Vandermonde's identity. It remains to find out what happens to the latter as $k \to \infty$. The product can be written as $$ (-1)^k \binom{\beta-1}{k} = (-1)^k \frac{\beta-1}{1} \frac{\beta-2}{2} \dotsb \frac{\beta-k}{k} = \prod_{j=1}^k \left( 1 - \frac{\beta}{j} \right). $$ Now, we can bound this product using $1-x < e^{-x}$: if $a_i<1$, $ \prod_i (1-a_i) < e^{-\sum_i a_i} $, so if $0<\beta < n$, $$ 0 < \prod_{j=n}^k \left( 1 - \frac{\beta}{j} \right) < \exp{\left( - \beta \sum_{j=n}^k \frac{1}{j} \right)} $$ But the sum diverges to $\infty$ as $k \to \infty$, so the term on the right tends to $0$ as $k \to \infty$. Hence by the sandwich theorem, if $\beta>0$, $$ \sum_{j=0}^k (-1)^{k-j} \binom{\beta}{j} = (-1)^k \binom{\beta-1}{k} \to 0 $$ as $k \to \infty$, and so $$ \sum_{j=0}^{\infty} (-1)^{k-j} \binom{\beta}{j} = 0 $$