## Precalculus (6th Edition) Blitzer

The given equation is ${{a}_{n}}=\frac{{{\left( -1 \right)}^{2n}}}{3n}$ Now we have \begin{align} & {{\left( -1 \right)}^{2n}}={{\left\{ {{\left( -1 \right)}^{2}} \right\}}^{n}} \\ & ={{1}^{n}} \\ & =1 \end{align} Thus, the sequence will not be an alternative sequence. Therefore, the statement is illogical. Therefore, the statement does not make sense.