Does tail decay rate of R.V. implies corresponding p.m.f. decay? Suppose that $X$ is a random variable on the integers $n\in\mathbb Z$ that obeys the tail decay
$$\Pr[X>n]\sim c\cdot n^{-a}$$
as $n\to\infty$ for some $a,c>0$. Can we always infer from this that
$$\Pr[X=n]\sim ac\cdot n^{-a-1}$$
as $n\to\infty$?
The reason I suspect this to be the case is that I can easily prove an analogue statement for continuous random variables (see next paragraph below), though I can't generalize the method I've used there.

If $Y$ is continuous with density $f$ and CDF $F$, and
$$H(t):=\Pr[Y>t]\sim c\cdot t^{-a},$$
then a naive application of limits yields the indeterminate form
$$c=\lim_{t\to\infty}\frac{H(t)}{t^{-a}}=\frac{0}{0},$$
so by the Hôpital rule,
$$c=\lim_{t\to\infty}\frac{H'(t)}{-at^{-a-1}}.$$
Given that $H'(t)=-F'(t)=-f(t),$ it follows that
$$f(t)\sim ac\cdot t^{-a-1},$$
as desired.

The answer is no. For example, if we let $X$ be a random variable with CDF, $P(X\geq n) = n^{-a}$,
then $2X$ satisfies $P(2X \geq n) \sim c\cdot n^{-a}$, yet $P(2X = n) \nsim ac \cdot n^{-a-1}$ since $P(2X = n)=0$ for odd $n$.
In fact, this can be adapted to give a counterexample in the continuous case as well. If we let $U$ be a uniform random variable on $[0,1]$, say, independent of $X$, then $P(2X+U\geq t) \sim c\cdot t^{-a}$, yet the density $f(t)$ of $2X+U$ is zero on intervals $2n +1 < x < 2n$, so that $f(t) \nsim ac \cdot t^{-a-1}$. The proof you gave is incorrect: L'Hopital's rule determines the limit of an indeterminate form $f/g$ if the limit of $f'/g'$ is known to exist, but it does not work the other way around: the existence of the limit of $f/g$ does not imply that of $f'/g'$ (e.g., a simple counterexample is given by $f(x)=x+\sin(x)$ and $g(x)=x$, as $x\to\infty$).
