Questions regarding p-adic expansion and numbers 
As opposed to real number expansions which extend to the right as sums
  of ever smaller, increasingly negative powers of the base $p$,
  $p$-adic numbers may expand to the left forever, a property that can
  often be true for the $p$-adic integers. For example, consider the
  $p$-adic expansion of 1/3 in base 5. It can be shown to be
  $…1313132_5$, i.e., the limit of a sequence $2_5$, $32_5$, $132_5$,
  $3132_5$, $13132_5$, $313132_5$, $1313132_5$, … :
$\dfrac{5^2-1}{3}=\dfrac{44_5}{3} = 13_5; \, \dfrac{5^4-1}{3}=\dfrac{4444_5}{3} = 1313_5$
  $\Rightarrow-\dfrac{1}{3}=\dots 1313_5$
  $\Rightarrow-\dfrac{2}{3}=\dots 1313_5 \times 2 = \dots 3131_5$
  $\Rightarrow\dfrac{1}{3} = -\dfrac{2}{3}+1 = \dots 3132_5.$ (Wikipedia, p-adic number)

I am unable to comprehend this. How does $\dfrac{5^2-1}{3}=\dfrac{44_5}{3} = 13_5; \, \dfrac{5^4-1}{3}=\dfrac{4444_5}{3} = 1313_5$ result in $-\dfrac{1}{3}=\dots 1313_5$?
 A: Because is says that if $x=\cdots1313_5$ then $3x+1\equiv 0 \text{ mod } 5^n$ for all $n$, which is precisely what it means to be $0$ in $\mathbb{Q}_5$. Thus, you see that $3x+1=0$ so that $\displaystyle x=\frac{-1}{3}$. 
EDIT:
Now that I have more time, let me be less glib about this response.
Whenever possible, we want to turn problems about $\def\Qp{\mathbb{Q}_5}$ $\def\Zp{\mathbb{Z}_5}$ $\Qp$ into problems about $\Zp$ since they are easy to deal with. So, how can we interpret $\displaystyle \frac{-1}{3}\in\Qp$, well since $\Qp$ is $\text{Frac}(\Zp)$ the only clear interpretation is that it is the element $x$ of $\Zp$ which satisfies $3x+1=0$. So, instead let us try to look for a solution $3x+1=0$.
To begin, let us recall how we can think about $\Zp$. Intuitively, $\Zp$ is the set $\{z\}$ of solutions to systems of equations as follows:
$$\begin{cases}z &\equiv a_1 \mod 5\\ z &\equiv a_2 \mod 5^2\\ z &\equiv a_3 \mod 5^3\\ &\vdots\end{cases}$$
where the equations are "consistent" (i.e. $a_i\equiv a_j\mod p^i$ for $i\leqslant j$). So, now if $x$ satisfies $3x+1=0$ then this should translate to mean
$$\begin{cases}3x+1 &\equiv 3a_1+1 \equiv 0 \mod 5\\ 3x+1 &\equiv 3a_2+1 \equiv 0 \mod 5^2\\ 3x+1 &\equiv 3a_3+1\equiv 0 \mod 5^3\\ &\vdots\end{cases}$$
So, we can solve each of these equations piecewise and find that 
$$(a_1,a_2,a_3,\ldots)=(3,8,83,\ldots)$$
Ok, so, this tells us that $x=(3,8,83,\ldots)$...this doesn't look right? How can we go from this to the desired $x=\ldots1313_5$? The key is that we have the same element of $\Qp$ expressed in different forms. Indeed, the notation $x=1313_5$ means that 
$$x=3+1\cdot 5+3\cdot 5^2 +\cdots$$
To reconcile this ostensible difference, let us write $3+1\cdot 5+3\cdot 5^2+\cdots$ in the same notation that we already have $x$ in. Recall that the correspondence between $\mathbb{Z}$ and these sequences is 
$$m\mapsto (m\mod 5,m\mod 5^2,m\mod 5^3,\ldots)$$
Thus, we see that 
$$\begin{aligned} 3 & \mapsto (3,3,3,\cdots)\\ 5 &\mapsto (0,5,5,\ldots)\\ 5^2 & \mapsto (0,0,5^2,\ldots)\end{aligned}$$
Thus, we see that 
$$ \begin{aligned}3+5+3\cdot 5^2 &=(3,3,3,\ldots)+(0,5,5,\ldots)+(0,0,75,\ldots)\\ &= (3,8,83,\ldots)\end{aligned}$$
and voilà!
