Geometric progression sum

$A=1+\dfrac 1 2+\dfrac 1 4+\dfrac 1 8+\ldots$

$2A=2+1+\dfrac 1 2+\dfrac 1 4+\dfrac 1 8+\ldots$

or $2A=2+A$

or $A=\boxed2$

Alternate method

$A=\dfrac 1 {\frac 1 2}=\boxed2$ (sum of a GP with infinite terms)

$A = \displaystyle \sum_{r=0}^{\infty} \dfrac{1}{2^r}\\ =1+ \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \ldots$

Notice that this is a sum of a geometric progression to infinity, where $a=1$ and $r = \dfrac{1}{2}$ , where $0 < r < 1$

Therefore, we can apply the formula:

$S_{\infty} = \dfrac{a}{1-r}\\ =\dfrac{1}{1-\frac{1}{2}}\\ =\dfrac{1}{\left(\frac{1}{2}\right)}\\ =\boxed{2}$