# Let x not be quasinilpotent, so l i m n &#x2192;<!-- → -->

Let x not be quasinilpotent, so $li{m}_{n\to \mathrm{\infty }}||{x}^{n}|{|}^{1/n}=\lambda \ne 0$. Let $\pi \left(x\right)=\overline{x}$, where $\pi \left(x\right):A\to A/rad\left(A\right)$ is the canonical quotient map. Suppose that $||{\overline{x}}^{n}|{|}^{1/n}=0$. Then it's spectrum $\sigma \left(\overline{x}\right)=0$, so $\overline{x}$ is not invertible in $A/rad\left(A\right)$, and thus generates a proper ideal in $A/rad\left(A\right)$. So then ${\pi }^{-1}\left(\overline{x}\right)$ generates a proper ideal in $A$ containing $rad\left(A\right)$.
From here, if $rad\left(A\right)$ is maximal I think I'd have a contradiction, but I don't know if that's true. If not, does anyone have another strategy I could try?
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Bentyrchjurvk
Assume $x\in A$ such that $x+rad\left(A\right)$ is nilpotent in $A/rad\left(A\right)$. This implies that for all $\epsilon >0$ there is an $n\in \mathbb{N}$ and $r\in rad\left(A\right)$ for which $|{x}^{n}-r|\le {\epsilon }^{n}$.
$r$ is quasi-nilpotent so there exists $M\in \mathbb{N}$ s.t. for all $m>M$, $|{r}^{m}|\le {\epsilon }^{nm}$
We then show by induction that there exists $A>0$ such that $|{x}^{nm}|\le A{2}^{m}{\epsilon }^{nm}$ for all $m$. We choose $A\ge 1$ such that this is true for all $m\in \left[0,M\right]$.
Then if $m>M$, we have
${x}^{nm}-{r}^{m}=\left({x}^{n}-r\right)\left({x}^{n\left(m-1\right)}+\dots +{r}^{m-1}\right),$
so
$|{x}^{nm}|\le {\epsilon }^{n}\left(A{2}^{m-1}{\epsilon }^{n\left(m-1\right)}+\dots +A{\epsilon }^{n\left(m-1\right)}\right)+{\epsilon }^{nm}\le A{\epsilon }^{nm}\left(1+1+\dots {2}^{m-1}\right)=A{2}^{m}{\epsilon }^{nm}.$
So we can conclude that $\mathrm{\forall }\epsilon >0\phantom{\rule{mediummathspace}{0ex}}lim inf|{x}^{n}{|}^{1/n}<2\epsilon$, which means that $x\in rad\left(A\right)$.