Infinite RC Circuit!

The equivalent series resistance is:

$R_{eq} = R + \frac{R}{2} + \frac{R}{4} + \ldots = R\left( {1 + \frac{1}{2} + \frac{1}{4} + \ldots } \right)$ $R_{eq} = 2R$

The equivalent series capacitance is:

$C_{eq} = \frac{1}{{\frac{1}{C} + \frac{1}{{2C}} + \frac{1}{{4C}} + \ldots }} = \frac{1}{{\frac{1}{C}\left( {1 + \frac{1}{2} + \frac{1}{4} + \ldots } \right)}}$ $C_{eq} = \frac{C}{2}$

The time constant of the equivalent RC circuit is:

$\tau = R_{eq} C_{eq} = RC$

The equivalent capacitor voltage in a RC circuit at a given time $t$ is $V_C = V\left( {1 - e^{ - t/\tau } } \right)$ and the equivalent resistor voltage is $V_R = Ve^{ - t/\tau }$ . Therefore, at a time $t = RC$ :

$V_C = V\left( {1 - e^{ - 1} } \right), \qquad V_R = Ve^{ - 1}$

The quotient of voltages is:

$\frac{{V_C }}{{V_R }} = \frac{{V\left( {1 - e^{ - 1} } \right)}}{{Ve^{ - 1} }} = e - 1 \approx \boxed{1.718}$

Req = 2R using infinite G.P series.
Similarly, Ceq = C/2.
Now use formula.
for capacitor Vc=V[1-e^(t/CR)]
for resistor which is Vr=V[e^(t/CR)] and as given t=CR

Now just take ratio and u'll get the answer