Suppose there's a discrete hazard function $\displaystyle h(t) = \frac{f(t)}{S(t)}$ that is defined on a discrete domain $\mathcal{T} = \{t_1 < t_2 < ... < t_n\}$, where $f(t)$ is the PMF $P(T_i = t)$ and $S(t)$ is the complementary CDF $P(T_i \geq t)$.
It's easy to derive that if $n<\infty$, $\displaystyle \frac{f(t_n)}{S(t_n)} = 1$. But what if $n \rightarrow \infty$, is $\displaystyle \frac{f(t_n)}{S(t_n)}$ still 1?
For insight, consider a continuous distribution supported on the positive numbers with a nonzero differentiable survival function (complementary CDF) $S$ so that $$f(t) = -S^\prime (t)$$ for all $t\gt 0.$ In this case the condition
$$1 = \lim_{t\to \infty} \frac{f(t)}{S(t)} = -\lim_{t\to\infty} \frac{\mathrm{d}}{\mathrm{d}t} \log(S(t))$$
indicates that asymptotically (for large $t$) the survival function behaves exponentially.
Consequently, when the survival function decays at a different rate, we should expect different limiting behavior. Changing the rate can readily be achieved by scaling $t,$ suggesting we consider $S(t)$ proportional to $\exp(-\lambda t)$ for $\lambda \gt 0.$ Here,
$$f(t) \ \propto\ -\frac{\mathrm{d}}{\mathrm{d}t} \exp(-\lambda t)\ \propto\ \exp(-\lambda t).$$
Let us therefore examine its discrete analog given by
$$f(n) = C\, \exp(-\lambda n) = C\, a^n$$
for $n = 1, 2, 3, \ldots$ where $a = \exp(-\lambda) \lt 1$ and $C$ is a normalizing constant (which we won't have to compute since it will disappear in the ratio $f/S;$ all we need to check is that it's not infinite).
The survival function is easily computed as the sum of a geometric series
$$S(n) = \sum_{i=n}^\infty f(i) = C\sum_{i=n}^\infty a^i = \frac{Ca^n}{1-a},$$
whence
$$\frac{f(n)}{S(n)} = 1-a \lt 1.$$
Finally, since it is necessary that $1=S(1)$ we find $C = 1-a,$ verifying $S$ really defines a genuine probability distribution.
This exhibits explicit discrete distributions with constant hazard functions $f/S$ for any possible positive constant. In particular, their limiting values are not $1.$
Further reasoning along these lines will help you generate sequences of probabilities $f(n),$ $n=1,2,3,\ldots,$ for which the ratios $f(n)/S(n)$ have any specified limit on the interval $[0,1]$ or do not reach a limit at all. In particular,
$f(n) = 1/n-1/(n+1)$ gives $S(n)=1/n,$ whence $$\lim_{n\to\infty} \frac{f(n)}{S(n)} = \lim_{n\to\infty} 1 - \frac{n}{n+1} = 0.$$
$f(n) = 1/(e\,(n-1)!)$ (a shifted Poisson distribution) defines a distribution with countable support having a limiting value of $1$ for its hazard function.
Create the sequence $f(n)$ by concatenating constant sequences of length $2^i-1$ and values $2^{-\binom{i}{2}}$ for $i=1,2,3,\ldots.$ This sequence begins $$f(n) = \left(\frac{1}{2}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{64}, \ldots, \frac{1}{64}, \frac{1}{1024}, \ldots\right)$$ where the constant subsequences have lengths $1,$ $3,$ $7,$ and so on. It sums to $1,$ making it a probability mass function. You may compute that $f(2^i-i-1)/S(2^i-i-1) = 1/2$ for $i=2, 3, 4, \ldots$ while $f(2^i-i)/S(2^i-i) = 2^{-i}$ grows arbitrarily small. Consequently $f(n)/S(n)$ has no limit.
