$Problem$
Suppose that a and b belong to a commutative ring $R$ with unity. If $a$ is the unit of $R$ and $b^2 = 0$; Show that $a+b$ is a unit of $R$.
$ Attempt$
$(a+b)(a-b) a^{-2}= (a^2-b^2)a^{-2}= 1$.
How to show that $a+b$ is a unit if we replace $b^2$ with $b^n$
I have just started ring theory after finishing Group theory. Any hint or suggestion will be appreciated.
In general, even if you just have $b^n=0$, let $c=-a^{-1}$ and write $a+b$ as $a(1-cb)$ and $1=(1-cb)(1+cb+c^2b^2+\cdot\cdot\cdot)$, where the sum is finite because eventually the terms all vanish, so $1-cb$ as an inverse and thus so does $a+b=a(1-cb)$.
In general, when you have nilpotent elements floating around, remembering power series representations like $\frac{1}{1-x}=1+x+x^2+\cdot\cdot\cdot$ is very useful.