Let $T = \{ t \in \mathbb{Q}: 0 \leq t \leq 1 \} $. With the parameterization of the unit circle given by $x(t) = \frac{1-t^2}{1+t^2}$ and $y(t) = \frac{2t}{1+t^2}$.
With an indexing of the set $T$ by a natural number $n$. What is an approach to show if the following sequence has a limit of $\frac{\pi}{2}$?
$x_0 = 0$
$x_n = x_{n-1} + |x(t_{n-1})-x(t_{n})| + |y(t_{n-1})-y(t_{n})|$ where $t_n \in T$
Each entry of the sequence is the summation of previous entries and the taxicab distance between current rational points on the circle. The intent is to create a convergence to the arc length of the first quadrant of a unit circle.
The answer is no, for two reasons. Given any subsdivion $0=s_0 \le \ldots \le s_n= 1$, the taxicab distance of the polygone line $(x(s_k),y(s_k))_{0 \le k \le n}$ is 2. Moreover, if the sequence $(t_n)_{n \ge 0}$ is an enumeration of $T=\mathbb{Q} \cap [0,1]$, it cannot be monotonic, the set of all limit points is the whole interval $[0,1]$ so the polygonal line given by the points $(x(t_n),y(t_n))$ makes infinitely many steps forward and backwards and the total length is infinite.