 Factorial weighted right shift

On the absolutely summable sequence space, this weighted shift has $T^m{e}_j=e_{j+m}/[j(j+1)\cdots (j+m-1)]$, so $\Vert T^m\Vert =1/m!$. The spectral radius formula gives radius zero. Its spectrum is exactly $\{0\}$ because it is not onto, while it has no eigenvalues: zero is excluded by injectivity, and any nonzero eigenvalue equation forces all coordinates to vanish starting at the first. Thus it is a nonzero quasinilpotent element of the operator algebra.