A Trig-y Substitution

$\begin{aligned} a_n & = \frac {2a_{n-1}}{1-a_{n-1}^2} & \small \color{#3D99F6} \text{Let }a_n = \tan \theta_n \\ \tan \theta_n & = \frac {2\tan \theta_{n-1}}{1-\tan^2 \theta_{n-1}} = \tan (2\theta_{n-1}) \end{aligned}$

Since $a_1 = \dfrac 1{\sqrt 3}$ , $\implies \theta_1 = \dfrac \pi 6$ . Then we have:

$\begin{aligned} a_2 & = \tan (2\theta_1) = \tan \frac 13 \pi = \sqrt 3 \\ a_3 & = \tan \frac 23 \pi = - \sqrt 3 \\ a_4 & = \tan \frac 43 \pi = \tan \frac 13 \pi = \sqrt 3 \\ a_5 & = \tan \frac 23 \pi = - \sqrt 3 \\ \cdots & = \cdots \\ \implies a_n & = \begin{cases} \sqrt 3 & \text{if }n \text{ is even.} \\ - \sqrt 3 & \text{if }n \text{ is odd.} \end{cases} & \small \color{#3D99F6} \text{for }n \ge 2 \\ \implies a_{2018} & = \sqrt 3 \approx \boxed{1.732} \end{aligned}$

Note that $a_{1}=tan(\frac{\pi}{6})$ . Letting $a_{n-1}=tan(x)$ gives $a_{n}=tan(2x)$ . Because each successive term is tangent of two times the previous term, the $nth$ term can be written as $a_{n}=tan(\frac{2^{n-1}\pi}{6})$ or $a_{n}=tan(\frac{2^{n-2}\pi}{3})$ . The 2018th term in the sequence would then be the value of $tan(\frac{2^{2016}\pi}{3})$ . The value of $tan(\frac{\pi}{3})$ cycles mod 6, so we just need to find the value of $2^{2016}$ (mod 6). Using Chinese Remainder Theorem, we see that $2^{2016}\equiv{4}$ (mod 6), so $a_{2018}=tan(\frac{4\pi}{3})=\sqrt{3})$ .