Formal Power Series and Inverse Limit Let $G$ be a free group with generators $g_1,g_2,\dots$. Consider the group ring $RG$, where $R$ is a commutative ring.
Let $\epsilon: RG\to G$ be the augmentation mapping and let $I=\ker\epsilon$ be the augmentation ideal of $RG$.
To be precise, $\epsilon$ is the map $$\epsilon(\sum_{g\in G}a_gg)=\sum_{g\in G}a_g$$
Let $A=\lim_n RG/I^n$, the inverse limit. How do we describe $A$ as a formal power series?
I was told that $A$ is an algebra of formal power series on non-commuting variables ($g_1+1,g_2+1,\dots$?), but I am not sure how to prove it.
Thanks.

I read the question in Exercise from Rotman: formal power series ring as inverse limit, possibly the first step is to construct an inverse system, which is:
$\psi_{n,m}: RG/I^m\to RG/I^n$
$\alpha+I^m\mapsto \alpha+I^n$ for all $m\geq n$.
 A: Noncommutative power series is by definition
$$\mathbb C\langle\langle y_1,\ldots, y_n\rangle\rangle = \lim_N\ \mathbb C\langle y_1,\ldots, y_n\rangle/(y_1,\ldots, y_n)^N$$
There is a linear change-of-variables isomorphism $\mathbb C\langle y_1,\ldots, y_n\rangle \cong \mathbb C\langle x_1,\ldots, x_n\rangle$ sending $y_i = 1 - x_i$, so noncommutative power series is also equal to
$$"\ = \lim_N\ \mathbb C\langle x_1,\ldots,x_n\rangle/(1-x_1,\ldots,1-x_n)^N$$
You could also flip the sign on the $x_i$ if you wanted.

The map
$$\mathbb C \langle x_1,\ldots,x_n, x_1^{-1},\ldots,x_n^{-1}\rangle \ \ \to\ \ \lim_N\ \mathbb C\langle x_1,\ldots,x_n\rangle/(1-x_1,\ldots,1-x_n)^N$$
sending $x_i \mapsto x_i$ is well-defined because $x_i^{-1} = \frac1{1-(1-x_i)} = \sum (1-x_i)^k$ is convergent with respect to the limit.
The augmentation ideal is the preimage of $J = (1-x_1,\ldots, 1-x_n)$ of the formal power series ring, because $1 - x_i^{-1} = -x_i^{-1}(1-x_i)$. So the homomorphism extends to the completion,
$$\lim_N\, \mathbb C\langle x_1,\ldots, x_n,x_1^{-1},\ldots,x_n^{-1}\rangle/I^N \ \ \to\ \ \lim_N\ \mathbb C\langle x_1,\ldots,x_n\rangle/(1-x_1,\ldots,1-x_n)^N$$
To prove it's an isomorphism you just check the inverse is well-defined, but this is the easier direction.
