A sum of absolute values of binomial coefficients Let $p>2$ be a real number  and consider the sum $J=\sum_{d=0}^{\infty}|\binom{p-2}{d}|$.
I want to know whether $J$ is a finite quantity or not?
Indeed, if we consider $I=\sum_{d=0}^{\infty}\binom{p-2}{d}$, then we have $I=2^{p-2}$ which is finite. But I am unable to verify whether $I$ (which involves modulus of each quantity in the summation) is finite or not?
Remark: $p$ need not be an integer, and I'm using the general definition of the  binomial coefficient as
$$
\binom{\alpha}{k} = \frac{\alpha(\alpha-1) \ldots(\alpha-k+1)}{1 \cdot 2 \cdots k}
$$
for real numbers $\alpha$ and positive integers $k$.
 A: It suffices to investigate the case that $x = p-2 > 0$ is not an integer, because the sum is finite otherwise.
Let $n \ge 0$ be an integer with $n < x < n+1$. For $d \ge n+2$ we have
$$
\left| \binom xd \right| = \frac{x(x-1)\cdots (x-n)}{(n+1)!} \times
\frac{(n+1-x)(n+2-x)\cdots (d-1-x)}{(n+2)(n+3) \cdots d} \, .
$$
The first factor does not depend on $d$, therefore it suffices to show that the series
$$
 \sum_{d=n+2}^\infty a_n \quad \text{with } a_n = \prod_{k=n+2}^d  \left( \frac{k-1-x}{k} \right) > 0
$$
is convergent. Using the well-known estimate $\ln (1+t) \le t$ we get
$$
\ln a_n = \sum_{k=n+2}^d \ln \left( \frac{k-1-x}{k} \right) \le -(x+1) \sum_{k=n+2}^d \frac 1k \\
\le -(x+1) \int_{n+2}^{d+1} \frac{dt}{t} 
= -(x+1) \ln \frac{d+1}{n+2} \, .
$$
It follows that
$$
 0 < a_n \le \frac{(n+2)^{x+1}}{(d+1)^{x+1}}
$$
and that implies the convergence of $\sum_{d=n+2}^\infty a_n$ because $x+1 > 1$.
A: If $x = p-2 > 0$ is not an integer then one can apply Raabe's test to $\sum_{d=0}^\infty b_d$ with $b_d = \left| \binom xd \right| $: For $d > x$ is
$$
d \left( \frac{b_d}{b_{d+1}} - 1\right) = d \left( \frac{d+1}{|x-d|} - 1\right) = (x+1)\frac{d}{d-x}
$$
and that converges to $x+1 > 1$ for $d \to \infty$, which implies the convergence of the series.
This works also for $\sum_{d=0}^\infty \left| \binom zd \right|$ with $z \in \Bbb C \setminus \Bbb N$ and $\operatorname{Re} z > 0$. In that case we get for $d > \operatorname{Re} z$
$$
d \left( \frac{b_d}{b_{d+1}} - 1\right) = d \left( \frac{d+1}{|z-d|} - 1\right) \le d \left( \frac{d+1}{d  - \operatorname{Re}(z)} - 1\right)
= (\operatorname{Re}(z)+1)\frac{d}{d-\operatorname{Re}(z)}
$$
which converges to $\operatorname{Re}(z)+1 > 1$.
