Answer
$a_{n}=(-\frac{1}{3})^{n-1}$, n goes from 1 to $\infty$.
Work Step by Step
{$1,~-\frac{1}{3},~\frac{1}{9},~-\frac{1}{27},~\frac{1}{81}$}
- The numerator: + or -1. It starts positive. Then, alternates between positive and negative: $(-1)^{n-1}$
- The denominator: $1,~3,~9,~27,~81,~...,~3^{n-1},~...$
$a_{n}=\frac{(-1)^{n-1}}{3^{n-1}}=(-\frac{1}{3})^{n-1}$
