real analysis – How does $frac29 + frac{2}{9^2}+frac{2}{9^3}+cdots=frac14$ imply that the base 3 expansion of $1/4$ is $0.020202…$?

Converting between base $b$ and base $b^k$ for $kge2$ is extremely simple: one base-$b^k$ digit corresponds one-for-one with $k$ base-$b$ digits.

The given result shows that the base-$9$ expansion of $frac14$ is $0.overline2$. Each digit $2$ becomes $02$ in base $3$, yielding the desired result $frac14=0.overline{02}$.