For my brilliant 43 followers

From the first few values of $z_n$ , we note that $z_n = \tan \left(\dfrac{\pi}{2^{n+1}} \right)$ . Let us proof it by induction.

1. For $n=1$ , $z_1 = \tan \left(\dfrac{\pi}{2^{2}} \right) = \tan \frac{\pi}{4} = 1$ . Therefore it is true for $n=1$ .

2. Assuming that it is true for all $n \ge 1$ , then:

$\begin{aligned} z_{n+1} & = \tan \left(\frac{\pi}{2^{n+2}} \right) = \tan \left(\frac{1}{2}\dot{}\frac{\pi}{2^{n+1}} \right) \\ \quad \quad \text{Since } \tan \left(\frac{\pi}{2^{n+1}} \right) & = \frac{2 \tan \left(\frac{1}{2}\dot{}\frac{\pi}{2^{n+1}} \right)}{1-\tan^2 \left(\frac{1}{2}\dot{}\frac{\pi}{2^{n+1}}\right)} \\ \Rightarrow z_n & = \frac{2z_{n+1}}{1-z^2_{n+1}} \\ \Rightarrow \frac{z_n}{1+\sqrt{1+z_n^2}} & = \frac{2z_{n+1}}{1-z_{n+1}^2 + \sqrt{1-2z_{n+1}^2+z_{n+1}^4+4z^2_{n+1}}} \\ & = \frac{2z_{n+1}}{1-z_{n+1}^2 + \sqrt{1+2z_{n+1}^2+z_{n+1}^4}} \\ & = \frac{2z_{n+1}}{1-z_{n+1}^2 + 1 + z_{n+1}^2} \\ & = z_{n+1} \end{aligned}$

$\quad$ It is true for $n+1$ too and therefore true for all $n \ge 1$ .

Therefore,

$\begin{aligned} \lim_{n \to \infty} 2^{n+1} z_n & = \lim_{n \to \infty} 2^{n+1} \tan \left(\frac{\pi}{2^{n+1}} \right) \\ & = 2^{n+1} \left(\frac{\pi}{2^{n+1}} \right) \\ & = \boxed{\pi} \end{aligned}$

Take $z_n$ as $tan\psi$