Let $S$ be a complex Banach space and $A$ a bounded linear operator on $S$. Suppose $k_0\in\mathbb{C}$ is a boundary point of the spectrum of $A$.
How do I show that $$\lim\limits_{n\rightarrow\infty}||(A-k_n)^{-1}||=\infty$$
What I know:
Being a boundary point means that
a) $A-k_0$ is not invertible in $B(S)$
b) there is a sequence $\{k_n\}$ in $\mathbb{C}$ s.t. $\lim\limits_{n\rightarrow\infty} k_n=k_0$ and $A-k_n$ is invertible in $B(S)$ for all $n\in\mathbb{N}$.
Further we can use the following fact:
IF $T,T'\in B(S)$, $T$ invertible in $B(S)$ and $||T'-T||\leq\frac{1}{||T^{-1}||}$, THEN $T'$ is invertible in $B(S)$.
How do I use all these?
Suppose it is not true that the limit is $\infty$. Then there exists $M>0$ such that for all $N>0$ there exists $n>N$ such that $\|(A-k_n)^{-1}\|<M$. Let $N$ be such that $|k_0-k_n|<\dfrac1M$ when $n>N$, and then take a particular $n>N$ such that $\|(A-k_n)^{-1}\|<M$. Then from your "following fact" you will be able to conclude that $(A-k_0)$ is invertible.