= Solution
The weighted right shift $T$ is injective, so zero is not an eigenvalue. If $Ta=\lambda a$ with $\lambda\ne0$, the first coordinate forces $a_1=0$ and subsequent coordinates force every $a_j=0$. Thus its <point spectrum> is empty.
For $m\geq1$,
$$
T^me_j=\frac{e_{j+m}}{j(j+1)\cdots(j+m-1)},\qquad\|T^m\|=\frac1{m!},
$$
since the largest column sum is attained at $j=1$. Therefore the <spectral radius formula> gives $\nu(T)=\lim_m(m!)^{-1/m}=0$. The limit follows, for example, from $m!\geq(m/2)^{m/2}$ for large $m$. Zero is in the spectrum because the output first coordinate is always zero, so $T$ is not onto. Hence
$$
\boxed{p\sigma(T)=\varnothing,\qquad\sigma(T)=\{0\}.}
$$
This is a nonzero <quasinilpotent element>, specifically the <factorial weighted right shift>.
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