<s>[INST] How to show $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$ I am new to the subject of rings of formal power series. The problem is to show that $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$, where $\mathbb{Z_6}[[x]]$ is the ring of formal power series with coefficients in $\mathbb{Z}_6$, also known as $ \mathbb{Z }/6\mathbb{Z}$. 
My work is as follows:
$ \begin{align} (1+3x)^{-1} &= \frac{1}{1+3x}
\\&= \frac{1}{1-(-3x)}
\\ &= 1 +(-3x) + (-3x)^2 + (-3x)^3 ... 
\\ &= 1 - 3x + 3^2x^2 -3^3x^3 ... 
\\ &=_6 1 + 3x +3x^2 + 3x^3 + ... \end{align}$
This is supposed to show that the inverse of $1 + 3x$ is $ 1 + 3x + 3x^2  + ...$ in the ring $\mod 6$. I am confused what this actually means. When I plug in $x =3$ into $1+3x$, this produces $4$ in mod 6 and there is no modular inverse of $4$. So why is $1+3x$ called the inverse?
I notice that we can check the calculation manually.
$\begin{align}( 1 + 3x) ( 1 + 3x + 3x^2 + 3x^3 + ...) &= 1 + 3x + 3x + 3^2x^2 + 3x^2 +...
\\ &=_6 1 + 3x + 3x + 3x^2 + 3x^2  +...
\\ &=_6 1 + 6x + 6x^2 + ...
\\ &=_6 1 + 0 + 0 + ...
\\ &=_6 1
\end{align}$
What bothers me is that some values you cannot plug into x, the modulo inverse is not defined. 
 [/INST] You need to show that 
$$(1 + 3x + 3x^2 + 3x^3 + \cdots)(1 + 3x) = 1$$
using the addition and multiplication rules in $\Bbb Z_6[[x]]$. The element $1 + 3x + 3x^2 + 3x^3 + \cdots$ is $\sum\limits_{j = 0}^\infty 3x^j$, so over $\Bbb Z_6[[x]]$ we write
$$\left(\sum_{j = 0}^\infty 3^jx^j\right)(1 + 3x) = \sum_{j = 0}^\infty (3x^j + 3x^{j+1}) = 1 + \sum_{j = 1}^\infty 3x^j + \sum_{j = 0}^\infty 3x^{j+1} = 1 + 2\sum_{j = 1}^\infty 3x^j$$
Now since $3^j \equiv 3\pmod 6$ for all $j \ge 1$, 
$$2\sum_{j = 1}^\infty 3^jx^j = 2\sum_{j = 1}^\infty 3x^j = \sum_{j = 1}^\infty 6x^j = 0$$
and so $$\left(\sum_{j = 0}^\infty 3^jx^j\right)(1 + 3x) = 1$$
</s>