If $\limsup\limits_{n\rightarrow \infty} a_n \leq a$ and $a>b>0,$ is it true that $\limsup\limits_{n\rightarrow \infty} a_n \leq b?$

Question

If $\limsup\limits_{n\rightarrow \infty} a_n \leq a$ where $a_n\geq 0$ for all $n\in \mathbb{N},$ and $a>b>0,$ is it true that $\limsup\limits_{n\rightarrow \infty} a_n \leq b?$

For some context, I am studying Murphy's book on C$^*$-Algebras and Operator Theory. He noted that if $0<\rho(a)\leq \frac{1}{r}$ and $\limsup\limits_{n\rightarrow \infty}\|a^n\|^{1/n}\leq \frac{1}{r},$ then this implies that $\limsup\limits_{n\rightarrow \infty}\|a^n\|^{1/n}\leq \rho(a).$ I have no idea how that happened. I just conjectured that the above statement might be true so that this deduction will make sense, however I cannot seem to prove it. I tried going back to the definition of limit superiors, but I'm still getting a blank. Anyone care to help? Even a simple counterexample will help me, as that will mean that I'm looking at this problem in a completely wrong way.

Answer 1

No. If each $a_n$ is equal to $2$, if $a=3$ and if $b=1$, then $\limsup_na_n=2<a$, but it is not true that $\limsup_na_n\leqslant b$.