Power law tail constant Suppose I know that the CDF $F(x)$ of a positive integer-valued random variable satisfies:
$$ 1-F(x) \sim \frac{C}{x^\alpha \Gamma(1-\alpha)} $$
where $\alpha \in (0,1)$. Then is $C>1$ necessarily? Doing an integral on the continuous density $f(x) \propto 1/x^{1.5}$ suggests that this is true, but I'm not sure in general whether it makes sense at all.
 A: The notation $1-F(x) \sim \frac{C}{x^\alpha \Gamma(1-\alpha)}$ here means 
$$\lim_{x\to \infty} \frac{1-F(x)}{\frac{C}{x^\alpha \Gamma(1-\alpha)}} = 1$$ 
Since $0\leq 1-F(x) \leq 1$, and  $x>0,\;\Gamma(1-\alpha)>0$, we can rearrange to get:
$$\lim_{x\to \infty} (1-F(x))x^\alpha \Gamma(1-\alpha) = C$$
However, $C$ can be any positive value, so it is not necessary that $C>1$

Here is a specific counterexample:
$$1-F(x) = \frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{x}\right), x\in \mathbb{N}$$
This corresponds to the monotonically decreasing pmf
$$ f(x)=\frac{0.25\sqrt{x}+0.5}{x^2}, x\in \mathbb{N}$$
Calculating our limit with $\alpha=0.5$ we get:
$$ \lim_{x\to \infty} \frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{x}\right)\sqrt{x}\Gamma(0.5) = 0.886227 < 1 $$
Therefore, with $C=0.886227$ we get our answer:
$$\lim_{x\to \infty} \frac{\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{x}\right)}{\frac{0.886227}{\sqrt{x}\Gamma(0.5)}}= 1 \implies \frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{x}\right) \sim \frac{0.886227}{x^{0.5} \Gamma(1-0.5)}$$
