Solution
= Solution
Again the weights tend to zero, so finite-rank truncations prove compactness. Products of $m$ consecutive weights are bounded by $1/(m+1)!$, making the spectral radius zero. Therefore
$$
\boxed{\sigma(Q)=\{0\}.}
$$
= Solution
Again the weights tend to zero, so finite-rank truncations prove compactness. Products of $m$ consecutive weights are bounded by $1/(m+1)!$, making the spectral radius zero. Therefore
$$
\boxed{\sigma(Q)=\{0\}.}
$$