Let $B$ be a unital Banach algebra with unit $e$. Let $T\in B$ and $(\lambda_n)$ a sequence of complex numbers not in the spectrum of $T$, but with limit $\lambda$ in the spectrum. Then, I want so show that $||\left(\lambda_n e - T \right)^{-1}|| \to \infty$.
I don't really know where to start. I've tried assuming for contradiction that the above sequence of norms is bounded, but I don't see how this is useful. Alternatively, I think it would suffice to show that $||\lambda_n e - T||\to 0$, but I don't think this is true in general.
This question has been asked If $(x_n)$ converges to a non-invertible element, then $\lim \| x_n^{-1} \| = \infty .$ here previously, but I don't know how to show/use the hint that is given.
The spectrum plays no real role here, in the sense that you want to show that if $T_n\to T$ with $T_n$ invertible for all $n$ and $T$ not invertible, then $\|T_n^{-1}\|\to\infty$.
Suppose that $\|T_n^{-1}\|\leq c$. You have $$ \|e-T_n^{-1}T\|=\|T_n^{-1}(T_n-T)\|\leq c\|T_n-T\|. $$ As long as $\|T_n-T\|<\frac1c$ (which happens for $n$ big enough) this implies that $T_n^{-1}T$ is invertible, and hence $T$ is invertible.
So we have have shown that $\sup_n\|T_n^{-1}\|=\infty$. As the above can be applied to any subsequence of $\{T_n^{-1}\}$, we also get that $\liminf\|T_n^{-1}\|=\infty$, and hence $\lim_n\|T_n^{-1}\|=\infty$.