We are asked to evaluate the infinite sum $\sum_{n=1}^{\infty} n (\frac{1}{3})^n$.

1. Problem Description

We are asked to evaluate the infinite sum $\sum_{n=1}^{\infty} n (\frac{1}{3})^n$.

2. Solution Steps

We can use the formula for the sum of the series $\sum_{n=1}^{\infty} nx^{n}$, where $|x|<1$.

Consider the geometric series:

$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x| < 1$.

Differentiating both sides with respect to $x$, we get:

$\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$ for $|x| < 1$.

Multiplying both sides by $x$, we get:

$\sum_{n=1}^{\infty} nx^{n} = \frac{x}{(1-x)^2}$ for $|x| < 1$.

Now, we can substitute $x = \frac{1}{3}$ into the formula.

$\sum_{n=1}^{\infty} n(\frac{1}{3})^n = \frac{\frac{1}{3}}{(1-\frac{1}{3})^2} = \frac{\frac{1}{3}}{(\frac{2}{3})^2} = \frac{\frac{1}{3}}{\frac{4}{9}} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$.

3. Final Answer

$\frac{3}{4}$