Solution
= Solution
In the <P-adic numbers>, $\sum_{j\ge0}p^{2j}$ converges to $1/(1-p^2)$. Consequently
$$
\frac{1+2p}{p-p^3}=(p^{-1}+2)\sum_{j\ge0}p^{2j}
=\boxed{p^{-1}+2+p+2p^2+p^3+2p^4+\cdots.}
$$
Since $p\ne2$, the digits one and two are both between zero and $p-1$, so no carrying is required. These are its <p-adic digits>, with the unique negative-power digit at exponent minus one.