An infinite cascade of $\Pi$ networks

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

$\begin{aligned} R_\infty & = R||(R+R||R_\infty) \\ & = R \bigg| \bigg| \left(R + \frac {RR_\infty}{R+R_\infty} \right) \\ & = R \bigg| \bigg| \frac {R^2+2RR_\infty}{R+R_\infty} \\ & = \frac {R\left(\frac {R^2+2RR_\infty}{R+R_\infty}\right)}{R+\frac {R^2+2RR_\infty}{R+R_\infty}} \\ & = \frac {\frac {R^2+2RR_\infty}{R+R_\infty}}{1+\frac {R+2R_\infty}{R+R_\infty}} \\ & = \frac {R^2+2RR_\infty}{2R+3R_\infty} & \small \blue{\text{Multiply both sides by }2R+3R_\infty} \\ 2RR_\infty + 3R_\infty^2 & = R^2+2RR_\infty \\ 3R_\infty^2 & = R^2 \\ \implies \frac {R_\infty}R & = \frac 1{\sqrt 3} \approx \boxed{0.577} \end{aligned}$

We have the following recursive relation between $R_{eq}(n+1)$ and $R_{eq}(n)$ ,

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

where $R_1 // R_2$ denotes the equivalent resistance of the parallel connection of $R_1$ and $R_2$ , so that $R_1 // R_2 = \dfrac{ R_1 R_2 }{R_1 + R_2}$

Using this formula in the above recursive equation, we can write,

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

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

Therefore,

$R_{eq}(n+1) = \dfrac{ R ( R^2 + 2 R x(n) ) / (R + x(n) ) }{ R + (R^2 + 2 R x(n) ) / (R + x(n) ) }$

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

$= \dfrac{ R^2 + 2 R R_{eq}(n) }{ 2 R+ 3 R_{eq}(n) }$

Now as $n \to \infty$ , both $R_{eq}(n)$ and $R_{eq}(n+1)$ approach $R_{\infty}$ , hence,

$R_{\infty} (2 R + 3 R_{\infty}) = (R^2 + 2 R R_{\infty})$

from which,

$3 R_{\infty}^2 = R^2$

Hence,

$\dfrac{R_{\infty}}{R} = \dfrac{1}{ \sqrt{3} } \approx \boxed{ 0.57735}$