Solution: Let $\epsilon > 0$ be given. Since the sequences $\{ p_n \} $ and $\{ q_n \} $ are Cauchy, there exists $n_1, n_2 \in \mathbb{N} $ such that \begin{align} & d \left( p_m, p_n \right) < \frac{\epsilon }{2} \quad \forall\ m,n \geq n_1 \label{eq:16Jan2025-1} \\ & d \left( q_m, q_n \right) < \frac{\epsilon }{2} \quad \forall\ m,n \geq n_2 \label{eq:16Jan2025-2}. \end{align} Let $n_0 = \max \{ n_1, n_2 \} $. Then for any $m,n \geq n_0$, \begin{align*} d(p_m, p_n) < \frac{\epsilon }{2} \quad \text{ and } \quad d(q_m, q_n) < \frac{\epsilon }{2}. \end{align*}

Let $m,n \geq n_0$. Then using triangle's inequality and equations \eqref{eq:16Jan2025-1}, \eqref{eq:16Jan2025-2}, we have \begin{align*} d\left( p_n, q_n \right) & \leq d\left( p_n, p_m \right) + d\left( p_m, q_m \right) + d\left( q_m, q_n \right) \\ & < \frac{\epsilon }{2 } + d\left( p_m, q_m \right) + \frac{\epsilon }{2}. \end{align*} This implies \[ \left\vert d\left( p_n, q_n \right) - d\left( p_m, q_m \right) \right\vert < \epsilon . \] This proves that the sequence $\{ d\left( p_n, q_n \right) \} $ is Cauchy in $\mathbb{R} $ and hence convergent in $\mathbb{R} $.