An infinite cascade of T-networks

The equivalent resistance is a sequence function of the number $n$ of the cascaded T-networks, $R_{eq} = R_{eq}(n)$

We have,

$R_{eq}(1) = 2 R$

And iteratively, we can write,

$R_{eq}(n+1) = R + R // (R + R_{eq}(n) )$

$= R + \dfrac{R (R + R_{eq}(n) )} { (2 R + R_{eq}(n) )}$

the fixed point of this recursive equation is obtained by setting $R_{eq}(n+1) = R_{eq}(n) = R^*$ , hence

$R^* = R +\dfrac{ (R^2 + R R^*)}{ (2 R + R^*) }$

Multiplying both sides by $(2R + R^* )$ , we get,

$2 R R^* + R^{*2} = R( 2 R + R^*) + R^2 + R R^*$

re-arranging and simplifying, we obtain,

$R^{*2} = 3 R^2$

so that the limiting $R_{eq}$ is $R^* = \sqrt{3} R$ , therefore, the answer is $\boxed{\sqrt{3} }$

Let the equivalent resistance of infinite cascade of $T$ -network be $R_\infty$ . From the figure above, we have:

$\begin{aligned} R_\infty & = R + R || (R+R_\infty) \\ R_\infty & = R + \frac {R(R+R_\infty)}{2R + R_\infty} & \small \blue{\text{Multiply both sides by }2R+R_\infty} \\ 2RR_\infty + R_\infty^2 & = 2R^2 + RR_\infty + R^2 + RR_\infty \\ R_\infty^2 & = 3R^2 \\ \frac {R_\infty^2}{R^2} & = 3 \\ \implies \frac {R_\infty}R & = \sqrt 3 \approx \boxed{1.732} \end{aligned}$