To infinity and Beyond

This is a sum of a geometric progression where $a=1$ , $r=\dfrac{1}{3}$ and $n=\infty$

Using the formula:

$1+ \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \ldots\\ = S_{\infty}\\ = \dfrac{1}{1-\dfrac{1}{3}}\\ = \dfrac{3}{2} = \boxed{1.5}$

It is a geometric series where r=1/3 and we need the infinite sum = 1/ (1-r) = 1.5

It's simple. Using your knowledge about adding fraction...

1 + $\frac{1}{3}$ = 1 $\frac{1}{3}$

1 $\frac{1}{3}$ + $\frac{1}{9}$ = 1 $\frac{4}{9}$

1 $\frac{4}{9}$ + $\frac{1}{27}$ = 1 $\frac{13}{27}$

1 $\frac{13}{27}$ + $\frac{1}{81}$ = 1 $\frac{40}{81}$

The sum will eventually be equal to 1 $\frac{1}{2}$ as you continue adding the succeeding fractions.