If $a_0\in R$ is a unit, then  $\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$ Let $R$ a ring, and let $$\displaystyle R[[x]]=\left\{\sum_{k=0}^{\infty}a_k x^k\;\middle\vert\; a_k\in R\right\}$$ with addition and multiplication as defined for polynomials.  We have that $R[[x]]$ is a ring containing $R[x]$ as a subring.
How to prove that if $a_0\in R$ is a unit, then  $\displaystyle\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$?
 A: Hint $\rm\displaystyle\quad 1\: =\: (a-xf)(b-xg)\ \Rightarrow\ \color{#c00}{ab=1}$
$$\Rightarrow\ \ \displaystyle\rm\frac{1}{a-xf}\ = \dfrac{b}{\color{#c00}b(\color{#c00}a-xf)}\ =\ \frac{b}{1-bxf}\ =\ b\:(1+bxf+(bxf)^2+(bxf)^3+\:\cdots\:)$$
By the way, such rings are called rings of (formal) power series.
Corollary $ $ When $R$ is a field the fraction field of $R[[x]]$ has Laurent form, i.e. every fraction can be written with denominator $\rm\,x^n,\,$ by $\rm\,g/h = g/(x^n(a\!-\!xf)) = gh'/x^n\,$ for $\rm\, h'\! = (a\!-\!xf)^{-1}\!\in R[[x]]$
A: One trick is to get the geometric series involved. We can write
$$\left(\sum_{k=0}^\infty a_k x^k\right)^{-1}=\frac{1}{a_0}\frac{1}{1+\underbrace{\left(\sum\limits_{k=1}^\infty a_0^{-1}a_kx^k\right)}_{u(x)}}. \tag{$\circ$}$$
Now $x|u(x)$, so in the $l$-adic topology the sequence $u(x),u(x)^2,u(x)^3,\cdots$ converges to $0$, or equivalently the terms contributed to the coefficient of a $x^n$ from these powers is gauranteed to be finite for all $n$. Thus if you expand $(\circ)$ in a geometric series, the element of $R[[x]]$ it converges to will be a legitimate multiplicative inverse (by the usual algebra establishing the geo sum formula).
A: Let $a=\sum_{k=0}^\infty a_kx^k\in R[[x]]$, where $a_0$ is a unit. We want to construct some $b=\sum_{k=0}^\infty b_kx^k\in R[[x]]$ such that $ab=1$, or after expanding,
$$ab=a_0b_0+(a_1b_0+a_0b_1)x+\cdots=1+0x+0x^2+\cdots$$
We therefore need $b_0=a_0^{-1}$ (recall that $a_0$ is a unit). We want to have $a_1b_0+a_0b_1=0$, so our only choice for $b_1$ is $$b_1=\frac{-a_1b_0}{a_0}=-a_1a_0^{-2}.$$ We want $a_2b_0+a_1b_1+a_0b_2=0$, so we must have $$b_2=\frac{-a_2b_0-a_1b_1}{a_0}=-a_2a_0^{-2}+a_1^2a_0^{-3}.$$
Prove that, by continuing this process, you get a $b$ such that $ab=1$.
