Given the model: \begin{aligned} Y_t &= \delta Y_{t-1}+u_t, \\ u_t &= \rho u_{t-1}+\epsilon_t, \end{aligned} where $\epsilon_t\sim i.i.d. (0,\sigma^2)$, $|\delta|,|\rho|<1$. Then how to find the prob. limit of the OLS estimator $\hat\delta$?

I have tried to use WLLN, since the $Y_t$ is not i.i.d., however the variance of $Y_t$ doesn't converge to 0. WLLN doesn't applies. So how to find the prob. limit of the OLS estimator $\hat\delta$?

Actually i calculated $\hat\delta$ directly using OLS estimator formula, and get \begin{equation} \hat\delta=\delta+\sum_{t=2}^T y_tu_{t-1}/\sum_{t=2}^Ty_{t-1}^2 \end{equation}. I wonder if i can expand $Y_t$ to infinite horizon to conclude $Y_t$ as covariance-stationary process or i can just expand $Y_t$ to finite horizon as $Y_t=\sum_{j=0}^{t-1}\delta^ju_{t-j}$. In order to use WLLN, I also used the above finite horizon expansion to calculate the $E(\sum_{t=2}^Ty_{t-1}^2/T-2)$ and find the probability limit is $\frac{(1+\rho \delta)\sigma^2}{(1-\rho \delta)(1-\delta^2)(1-\rho^2)}$. However, if i just consider the finite horizon expansion,then the fourth moment expection $E[(\sum_{t=2}^Ty_{t-1}^2/T-2)^2]$ is evidently doesn't converge to zero. So now i am quite confused. Is the OLS estimator $\hat\delta$ converge to some distribution in probability?

update: uploaded the full question Thanks for everyone, i'd appreciate it if you can add your comment to this question.