Let $S$ be the right shift operator on $\ell^2$, that is, $$Sx=y,\ x=(\xi_1,\xi_2,\cdots),\ y=(0,\xi_1,\xi_2,\cdots).$$ Then it is OK that $||S||=1$. For $a\in\Bbb C$ with $|a|>1$, then $a\in \rho(S)$, and how can we prove that $$||(a I-S)^{-1}||=(|a|-1)^{-1}.$$
Clearly, if $(aI-S)^{-1}x=y$, then $$||x||=||(aI-S)y|| >=|a| ||y||-||Sy||=(|a|-1)||y||.$$ From this, we know $$||(a I-S)^{-1}||\leq (|a|-1)^{-1}.$$ Now hwo can I have prove that $$||(a I-S)^{-1}||>=(|a|-1)^{-1}.$$
Take $y=(1,\lambda,\cdots,\lambda^n,\cdots)$ where $|\lambda|<1$. One may immediately check that $y=\lambda Sy+(1,0,0,\cdots)$. Hence $(aI-S)y=ay-Sy=(a\lambda-1)Sy+(a,0,0,\cdots)$. As a consequence, $$|a\lambda-1|\cdot\|y\|-|a|\leq\|(aI-S)y\|\leq |a\lambda-1|\cdot\|y\|+|a|$$ Let $x=(aI-S)y$. Since $\|y\|$ can be arbitrarily large(just take $|\lambda|\to 1^-$), the above inequality implies that $\|x\|$ can be arbitrarily large, while $\|(aI-S)^{-1}x\|\geq|a\lambda-1|^{-1}\|x\|-|a|/|a\lambda-1|$. Hence $$\|(aI-S)^{-1}\|\geq|a\lambda-1|^{-1}-\frac{|a|}{|a\lambda-1|\cdot\|x\|}$$ Now choose appropriate $\theta\in\mathbb{R}$ such that $ae^{i\theta}=|a|$. Take $\lambda=re^{i\theta}$ and let $r\to 1^-$. The condition $|a|>1$ is applied so that $|a\lambda-1|$ won't be zero as $r\to 1^-$.