Forever

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...=S$ $is~~ geometric~~ progression~~ with~~ infinite~~ terms$ . $S=\frac{0.5}{1-\frac{1}{2}}=\color{#3D99F6}{\boxed{1}}=\color{#3D99F6}{\boxed{0.9999...}}$

summation of infinite series is = $\frac{a}{1-r}$

so, the summation of this infinite series is $\frac{0.5}{1-0.5}$ = $1$ ...........[where a= $\frac{1}{2}$ and r= $\frac {1}{2}$

FYI: 0.9999.... = 1

Given series is an geometric progression where $a = \dfrac{1}{2} , r = \dfrac{1}{2} , n = \infty \\ S = \dfrac{a}{1-r} \\ = \dfrac{\frac{1}{2}}{1 -\frac{1}{2}} \\ = \dfrac{\frac{1}{2}}{\frac{1}{2}} = \color{#20A900}{\boxed1}$