Show the existence of a weighted shift operator that converges to a unilateral weighted shift operator

Question

Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward weighted shift. i.e. $Te_n=t_n e_{n+1}$. And $t_n = (gcd(n, 2^n))^{-1}$

1. Show that there exist weight shifts $(W_n)_n$ such that $\lim_{n \rightarrow \infty } W_n= T $ with respect to the same orthonormal basis and if $W_n$ is of order $m$, then $W_n^{m-1 } \ne 0 =W_n^m$.

2. Conclude that the spectral radius $\rho:B(H) \rightarrow \mathbb{R}$ is not continuous.

I'm don't have any clues how to construct such weighted shift $W_n$.

I know that $T^me_k=T^{m-1}(t_ke_{k+1})=T^{m-2}t_k\cdot t_{k+1}e_{k+2}=\dots=(t_k\cdot t_{k+1}\cdots t_{k+m-1})e_{k+m}$.

Not sure if this helps.

Thanks in advance!

Answer 1

I am only answering with a hint.

Set $W_n:H\to H$ as $W_ne_k=t_ke_{k+1}$ when $k$ is NOT divided by $2^n$ and $W_ne_k=0$ when $k$ is divided by $2^n$. Show that $W_n^{2^n}=0$ and that $\|W_n-T\|=\frac{1}{2^n}\to0$. Justify why $\rho(W_n)=0$ for all $n$.