The Infinite Sum

First, let sum $S$ be $1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \ldots$ .

$S = 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \ldots$

Multiply it by 3

$3S = 3 + 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \ldots$

You can find the $" 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \ldots "$ inside the "Multiplied by 3", so substitute it inside and become:

$3S = 3 + S$

$2S = 3$

$S = 1.5$

So, the answer is 1.5

As this GP sum is infinite with initial term $a=1$ and a common ratio $r=\frac{1}{3}$ , the sum of them are: $\frac{a}{1-r}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}}=\Large \boxed{\frac{3}{2}}$