Why is $\frac{1}{x+2} = \sum_k^\infty (-1)^k(1+x)^k \in \mathbb{Z} [[x]]$ not the inverse of $x+2$ in $\mathbb{Z}$?

Question

As also stated in here, a formal power series is a unit in $R[[x]]$ iff it is constant coefficient is a unit in the ring $R$. However, for example, we can find the inverse of $1+x$ by observing that $\frac{1}{1+x} = \sum_k^\infty (-x)^k$, so to find the inverse of $x+2$, I simply observed that $$\frac{1}{x+2} = \frac{1}{1- (-1-x) } = \sum_k^\infty (-1)^k(1+x)^k \in \mathbb{Z} [[x]].$$

However, this contradicts with the stated theorem that since $2$ is not unit in $\mathbb{Z}$, $x+2$ is a non unit, so what is wrong in my argument ?

Answer 1

Because $$\sum_k(-1)^k(1+x)^k=\sum_{k\geq 0}(-1)^k + \sum_{k \geq 0}\sum_{1\leq r\leq k}{k\choose r}x^r$$ is not convergent in $\mathbb{Z}[[x]]$.

Answer 2

Not that $ \sum_k^\infty (-1)^k(1+x)^k$ is not an element of $ \mathbb{Z} [[x]]$. For example the constant term of $\sum_k^\infty (-1)^k(1+x)^k $ is not an integer numbr, indeed, $ \sum_k^\infty (-1)^k(1+0)^k $ is divergence.