Evaluate the sum...
\[\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k} + \cdots \]
∑ 3^(-n) n = 1/4 * 3^(-k) (-2 k + 3^(k + 1) - 3)
