Why is $\frac{1}{3}=0.2\overline{31}$ in 5-adic expansion ?
I get:
$\frac{1}{3}=\frac15+3\sum\limits_{k=1}^{\infty}\left(\frac15\right)^{2k}+\sum\limits_{k=1}^{\infty}\left(\frac15\right)^{2k+1}=0.13131313\dots=0.\overline{13}$
Something msut be wrong ?
On
$\frac{1}{3}$ is in base $5$ $$\frac{1}{3}= 0.131313\ldots_{(5)}= 0.\overline{13}_{(5)}$$ since $$0.131313\ldots_{(5)}= \frac{0.131313\ldots_{(5)}}{0.444444\ldots_{(5)}}= \frac{13_{(5)}}{44_{(5)}}= \frac{1}{3}$$
Therefore, $-\frac{1}{3}$ in $5$-adic is $$-\frac{1}{3}= \ldots 131313. $$
(use $\sum_{n \in \mathbb{Z}} x^n= 0$ ) so now substract ( that's why @Lubin: 's point of view is so useful) $-\frac{1}{3}$ from $0.$ and get in $5$-adics $$\frac{1}{3} = \ldots1313132.= \overline{13}2. $$
What you've written is the base $5$ representation of the real number $\frac{1}{3}$.
The value $\frac{1}{3}$ in $5$-adic numbers is not just a different representation, but in some sense, a different number.
The standard $5$-adic representation of $\frac{1}{3}$ is actually $$\frac{1}{3}=2.\overline{31}.$$
Note the different placement of the decimal point from the expression in your question. $0.2\overline{31}$ is actually $\frac{5}{3}$ in $5$-adic numbers.