= Solution
Suppose the inverse norms did not tend to infinity. Some subsequence would satisfy $\|a_n^{-1}\|\leq C$. Since $a_n\to a$, for its large indices $\|a_n^{-1}(a-a_n)\|<1$. Then
$$
a=a_n\bigl(1+a_n^{-1}(a-a_n)\bigr)
$$
is invertible in $A$ by the <Neumann series>, contradicting the hypothesis. Hence \b[$\boxed{\|a_n^{-1}\|\to\infty}$], 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 $\|a_n^{-1}-a^{-1}\|\leq\|a_n^{-1}\|\|a-a_n\|\|a^{-1}\|$, 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 \b[$a\notin G_B(A)$].
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