 Solution

Suppose the inverse norms did not tend to infinity. Some subsequence would satisfy $\Vert a_n^{-1}\Vert \le C$. Since $a_n\to a$, for its large indices $\Vert a_n^{-1}(a-a_n)\Vert <1$. Thenis invertible in $A$ by the Neumann series, contradicting the hypothesis. Hence $\boxed{\Vert a_n^{-1}\Vert \to \infty }$, the inverse norm divergence at noninvertible boundary points property.

If $a$ were invertible in $B$, then continuity of inversion in a Banach algebra would give $a_n^{-1}\to a^{-1}$ in $B$. Explicitly, the same factorization around $a$ gives a Neumann series and the bound $\Vert a_n^{-1}-a^{-1}\Vert \le \Vert a_n^{-1}\Vert \Vert a-a_n\Vert \Vert a^{-1}\Vert$, with the inverse norms locally bounded in $B$. Since every $a_n^{-1}$ belongs to the closed subalgebra $A$, so would $a^{-1}$. That would make $a\in G(A)$, again a contradiction. Therefore $a\notin G_B(A)$.