Let $A$ be a Banach algebra and suppose $a \in A$, I'd like to prove
$ \lim_{n\to\infty}\lVert x^n\rVert^{\frac{1}{n}} = \inf_{n\geq 1}\{\lVert x^n\rVert^{\frac{1} {n}}\}$
by Fekete's Lemma, so at first, I'll mention this lemma as following
let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers such that $$a_{m+n}\leq a_m+a_n $$ for all $m, n\in \Bbb{N}$ if the sequence $(\frac{a_n}{n})_{n=1}^{\infty}$ is bounded below, then $$ \lim_{n\to\infty}\frac{a_n}{n} = \inf_{n\geq 1}(\frac{a_n}{n})$$
any help would gratly be appreciated. Thanks
Fisrt, notice that your statement is obviously true if $x =0$.
So suppose $x \neq 0$ and study $a_n = \ln \left\| x^n \right\|$. You have $$a_{n+m} = \ln \left\| x^{n+m} \right\| \leq \ln \left( \left\| x^n \right\| \left\| x^m \right\| \right) = \ln \left\| x^n \right\| + \ln \left\| x^m \right\| = a_n + a_m$$
So you know, by what you call Fekete's lemma, that $$\lim_{n \rightarrow +\infty} \frac{a_n}{n} = \inf_{n \in \mathbb{N}^*} \frac{a_n}{n}$$
(actually, this lemma is satisfied also in the case where $(a_n/n)$ is now bounded below and you authorize the limit/inf to be $-\infty$)
You deduce that $$e^{\lim_{n \rightarrow +\infty} \frac{a_n}{n}} = e^{\inf_{n \in \mathbb{N}^*} \frac{a_n}{n}}$$
so $$\lim_{n \rightarrow +\infty} e^{\frac{\ln \left\| x^n \right\|}{n}} = \inf_{n \in \mathbb{N}^*} e^{\frac{\ln \left\| x^n \right\|}{n}}$$
i.e. $$\lim_{n \rightarrow +\infty} \left\| x^n \right\| ^{\frac{1}{n}} = \inf_{n \in \mathbb{N}^*} \left\| x^n \right\| ^{\frac{1}{n}} $$