Chapter 1.10, Problem 19P

The following series are not power series, but you can transform each one into a power series by a change of variable and so find out where it converges.

${\displaystyle {\sum }_0^{\infty }{8}^{-n}{(x^2-1)}^n}$ Method: Let $y=\text{}{x}^2-1$.

The power series ${\displaystyle {\sum }_0^{\infty }{8}^{-n}{y}^n}$ converges for $\left|y\right|<8$, so the original series converges for $|x^2-1|<8$, which means $\left|x\right|<3$.