Find the jump of force of mortality under Balducci assumption

Question

I need to find the jump of $\mu_x$ (force of mortality) under Balducci assumption at the point n $\in \Bbb N$. Under the assumption of Baoducci force of mortality function has the form: $$\mu_{x}= \frac{q_n} {(p_n + (x-n)q_n)}, x \in (n, n+1)$$ where $q_n$ is the probability of death between the ages of $x$ and age $x + 1$ and $p_n$ is the probability that a life age $x$ will survive to age $x + 1$. It is also known that: $$p_n = \frac{s(n+1)} {s(n)} \\ q_n = 1- p_n$$ I found the left and right limits: $$\lim\limits_{x \to n+0} \mu_x = \frac{q_n}{p_n}$$ and $$\lim\limits_{x \to n-0} \mu_x = \frac{q_{n-1}}{p_{n-1}+q_{n-1}}$$ So jump will be $$\Delta = \frac{q_n}{p_n}- \frac{q_{n-1}}{p_{n-1}+q_{n-1}}=\frac{1-p_n}{p_n}- q_{n-1}= \frac{1}{p_n}- q_{n-1}$$$$$$And then I don't know if I can leave the answer in this form or whether it is still necessary to simplify. I tried to simplify the expression but it turned out only more complicated. And I doubt if I did everything right. I would be grateful for any advice.

Answer 1

Your answer looks fine until right at the end, where I would say: $\Delta = \frac{q_n}{p_n}- \frac{q_{n-1}}{p_{n-1}+q_{n-1}}=\frac{q_n}{p_n}- q_{n-1}$