Fibbonaci geometric progression

$\begin{aligned} S & = \frac 12 + \frac 14 + \frac 28 + \frac 3{16} + \frac 5{32} + \frac 8{64} + \frac {13}{128} + \cdots & \small \color{#3D99F6} ...(1) \\ 2S & = 1 + \frac 12 + \frac 24 + \frac 38 + \frac 5{16} + \frac 8{32} + \frac {13}{64} + \cdots & \small \color{#3D99F6} ...(2) \\ S & = 1 + 0 + \frac 14 + \frac 18 + \frac 2{16} + \frac 3{32} + \frac 5{64} + \frac 8{128} + \cdots & \small \color{#3D99F6} ...(2)-(1) \\ S & = 1+ \frac 12 \left(\frac 12 + \frac 14 + \frac 28 + \frac 3{16} + \frac 5{32} + \frac 8{64} + \cdots\right) \\ S & = 1 + \frac 12 S \\ \frac 12 S & = 1 \\ \implies S & = \boxed 2 \end{aligned}$

It's been a long time and I haven't found the answer to the question, I shook it and it turned out right (what a luck) ☺☺☺