The problem is from Garth Dales, Introduction to Banach algebra, chapter 5 and Lemma 5.1.5
Lemma: Let $(A, \|.\|)$ be a unital Banach algebra, let $a\in A$ and let $\epsilon>0.$ then there is a norm $\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}\vertiii{.}$ on $A$ such that $\vertiii.$ is equivalent to $\|.\|,$ $\vertiii{e}=1$ and $\vertiii{a}\leq \nu(a)+\epsilon $ where $$\nu(a)=\lim_{n\to\infty}\|a^n\|^{\frac{1}{n}}$$
Proof: If we let $b=\frac {a}{\nu(a)+\epsilon}$ the $S=\{b^n: n\in \mathbb{Z^+}\}$ is bounded. For $c\in A$ let $$p(c)=\sup\{\|sc\|: s\in S\}, \vertiii{c}=\sup\{p(cd): d\in A, p(d)\leq 1\}$$
It is easy to prove that $S$ is bounded but I couldn't check $\vertiii{a}\leq \nu(a)+\epsilon $
Any piece of advice would be much appreciated