= Solution
First give the general definitions and radius facts. The <spectrum of a bounded operator> is $\sigma(A)=\{\lambda:A-\lambda I\text{ has no bounded inverse}\}$, and the <point spectrum> is $p\sigma(A)=\{\lambda:\ker(A-\lambda I)\ne0\}$. If $|\lambda|>\|A\|$, the <Neumann series> inverts $I-A/\lambda$, hence also $A-\lambda I=-\lambda(I-A/\lambda)$. Thus $\sigma(A)$ is contained in the closed disc of radius $\|A\|$ and is bounded.
The <spectral radius> is $\nu(A)=\sup_{\lambda\in\sigma(A)}|\lambda|$. The <spectral radius formula> is
$$
\boxed{\nu(A)=\lim_{m\to\infty}\|A^m\|^{1/m}=\inf_{m\geq1}\|A^m\|^{1/m}.}
$$
For the left shift $R$, an eigenvector satisfies $a_{j+1}=\lambda a_j$, so $a_j=\lambda^{j-1}a_1$. A nonzero such sequence is in $\ell_1$ exactly when $|\lambda|<1$. Since $\|R\|=1$, its spectrum is contained in the closed unit disc. It contains the open unit disc as <point spectrum> and therefore its closure by closedness of the spectrum. Hence
$$
\boxed{p\sigma(R)=\{\lambda:|\lambda|<1\},\qquad\sigma(R)=\{\lambda:|\lambda|\leq1\}.}
$$
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