Why is $\hat{I}$ contained in the Jacobson radical $J(\hat{R})$? Suppose $I$ is an ideal of a commutative ring $R$, and $\hat{R}$ is the $I$-adic completion. I don't follow why $\hat{I}$ is in $J(\hat{R})$. 
I know $\hat{R}$ is complete wrt the $\hat{I}$-adic topology. Let $a\in\hat{I}$. Since $a^n+a^{n+1}+\cdots+a^m\in(\hat{I})^n$, by completeness the series $1+a+a^2+\cdots+a^n$ converges to some $b\in\hat{R}$. Supposedly, from the identity $(1-a)(1+a+\cdots+a^n)=1-a^{n+1}$, taking $n\to\infty$ implies $(1-a)b=1$, and then $1-a$ is a unit, so $1+ax$ is a unit for all $x\in\hat{R}$, and thus $a\in J(\hat{R})$, and we are done.
The part I don't follow is why does taking $n\to\infty$ imply that $1-a^{n+1}$ goes to $1$? 
I should add that this is from Robert Ash's Commutative Algebra notes Chapter 4.

That's what the $\hat I$-adic topology is. The basis neighbourhoods of a point $r\in \hat R$ are the sets of the form $r+\hat I^n$ for natural $n$. If $a\in \hat I$, then the sequence $r+a^n$ is eventually contained in any fixed open neighbourhood of $r$, which means that the sequence converges to $r$.
