A machine works for $Y_0$ time then fails and takes $X_1$ time to repair. Then again works for $Y_1$ time and then fails and takes $X_2$ time to repair and so on. All the $X_n$'s and $Y_n$'s are independent non negative rvs. $X_n$'s are iid with pmf $f$ and $Y_n$'s are iid with pmf $g$. Let $r_n$ be the probability that the machine is working at time $n$. Find $\lim r_n$.
My work:
Let $W_i=X_i+Y_{i-1}$; $U_k=\sum_{i=1}^{k} W_i$. Let $R_n$ be the event that machine is working at time $n$.
$$R_0=(Y_0\ge 1)\cup (U_1=0, Y_1\ge 1)\cup (U_2=0,Y_2\ge 1) \cup \cdots$$ Therefore $$\begin{aligned}r_0 & =P(Y_0\ge 1)+P(U_1=0)P(Y_1\ge 1)+P(U_2=0)P(Y_2\ge 1)+\cdots \\ & = (1-g(0))(1+P(W_1=0)+P(W_1=0)^2+\cdots) \\ & = \frac{1-g(0)}{1-g(0)f(0)}\end{aligned}$$
Note that $$R_{n+1}=(Y_0\ge n+2)\cup (U_1\le n+1, U_1+Y_1\ge n+2)\cup (U_2\le n+1,U_2+Y_2\ge n+2) \cup \cdots$$ Note that $P(Y_0\ge n+2)=P(Y_0\ge n+1)-P(Y_0=n+1)$
$\begin{aligned}T_{n+1} & := P(U_k\le n+1, U_k+Y_k\ge n+2) \\ & =P(U_k\le n,U_k+Y_k\ge n+2)+P(U_k=n+1,U_k+Y_k\ge n+2) \\ & = P(U_k \le n, U_k+Y_k \ge n+1)-P(U_k\le n,U_k+Y_k=n+1)\\ & +P(U_k=n+1)P(Y_k\ge 1) \\ & = T_n -[P(U_k+Y_k=n+1)-P(U_k=n+1,U_k+Y_k=n+1)]\\ & +P(U_k=n+1)P(Y_k\ge 1)\\ & = T_n -P(U_k+Y_k=n+1)+P(U_k=n+1)\end{aligned}$
Therefore
$$r_{n+1}=r_n-P(Y_0=n+1)+\sum_{k=1}^{\infty} P(U_k=n+1) - \sum_{k=1}^{\infty} P(U_k+Y_k=n+1)$$ Please can someone verify this relation. Therefore $$\begin{aligned}\lim r_n -r_0 & = \sum_{n=0}^{\infty} (r_{n+1}-r_n)\\ & = \sum_{k=1}^{\infty} P(U_k\ge 1) -\sum_{k=1}^{\infty} P(U_k+Y_k\ge 1)-P(Y_0\ge 1) \\ & = \sum_{k=1}^{\infty} [1-P(U_k=0)-1+P(U_k=0)P(Y_k=0)]+(g(0)-1)\\ & = (g(0)-1)\sum_{k=1}^{\infty}P(U_k=0)+(g(0)-1)\\ & = \frac{g(0)-1}{1-g(0)f(0)}=-r_0\end{aligned}$$ $$\implies \lim r_n =0$$ However I firmly think that the limit is non zero. Can someone verify whether the interchanging of sum is justified or not? Is the limit really zero?