\( \sum _ { k = 2 } ^ { \infty } ( \frac { 3 ^ { - k } } { 3 ^ { n } } ) \)

We start with the sum \[ S = \sum_{k=2}^{\infty} \left(\frac{3^{-k}}{3^n}\right). \] Since \(3^n\) does not depend on the index \(k\), we can factor it out of the summation: \[ S = \frac{1}{3^n} \sum_{k=2}^{\infty} 3^{-k}. \] Now, observe that the series \[ \sum_{k=2}^{\infty} 3^{-k} \] is a geometric series. A geometric series has the form \[ \sum_{k=0}^{\infty} ar^k, \] where \(a\) is the first term and \(r\) is the common ratio. In our case, we can write the series starting at \(k=2\) with the first term \[ a = 3^{-2} = \frac{1}{9}, \] and the common ratio is \[ r = \frac{1}{3}. \] For a geometric series with \(|r| < 1\), the sum is given by \[ \sum_{k=0}^{\infty} ar^k = \frac{a}{1 - r}. \] Because our series begins at \(k = 2\), we directly use \(a = 3^{-2}\) and \(r = \frac{1}{3}\): \[ \sum_{k=2}^{\infty} 3^{-k} = \frac{3^{-2}}{1 - \frac{1}{3}}. \] Compute the denominator: \[ 1 - \frac{1}{3} = \frac{2}{3}. \] Thus, the sum becomes \[ \sum_{k=2}^{\infty} 3^{-k} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6}. \] Returning to the original expression, we now have \[ S = \frac{1}{3^n} \cdot \frac{1}{6} = \frac{1}{6 \cdot 3^n}. \]