\(2,\frac{2}{3},\frac{2}{9},\frac{2}{27},\cdots=\left\{2\left(\frac{1}{3}\right)^{n-1}\right\}\)
\(1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots=\left\{\left(\frac{1}{2}\right)^{n-1}\right\}\)
\(3,-1,\frac{1}{3},-\frac{1}{9},\cdots=\left\{3\left(-\frac{1}{3}\right)^{n-1}\right\}\)