Spectrum of weighted right shift operator on $\ell^{\infty}(\mathbb N)$ – solution verification

Question

I'm interested in finding the spectrum of the following weighted right shift operator on $\ell^{\infty}(\mathbb N)$:
$$T(a_1,a_2,a_3,…)=\left(0,\frac{a_1}{1},\frac{a_2}{2},\frac{a_3}{3},…\right)$$
If first noticed that $T-\lambda I$ is always injective:
$$(T-\lambda I)(a_1,a_2,a_3,…)=\left(-\lambda a_1,\frac{a_1}{1}-\lambda a_2,\frac{a_2}{2}-\lambda a_3,\frac{a_3}{3}-\lambda a_4,…\right)$$
And this vector is equal to $0$ if and only if $0=a_1=a_2=a_3=\dotsb$

To find whether it is surjective, I ask myself if there is a solution to the problem:
$$ \left(-\lambda a_1,\frac{a_1}{1}-\lambda a_2,\frac{a_2}{2}-\lambda a_3,\frac{a_3}{3}-\lambda a_4,…\right)=(b_1,b_2,b_3,b_4,…)$$
Clearly, for $\lambda=0$ there isn't always a solution (so $0$ is definitely in the spectrum). For $\lambda \neq 0$, I can try to solve this equation by induction and get the solution via the following recursion relation:
$$a_1=-\frac{b_1}{\lambda},a_2=-\frac{b_2-a_1}{\lambda},…,a_n=-\frac{1}{\lambda}\left(b_n-\frac{a_{n-1}}{n-1}\right),…$$
Since we've found a solution for any $(b_n)$, the operator is presumably surjective. But the problem is, the solution can possibly be unbounded, and then $T$ cannot really be surjective (since the solution does not belong to our space).

Since our formula for $a_n$ above has a leading factor which behaves like $\frac{1}{\lambda^n}$ (at least it seems so), then for $|\lambda|<1$ the solution is unbounded and for $|\lambda|>1$ it is bounded. So the spectrum includes the open unit disk (since then $T$ is not surjective), and not the exterior of the closed unit disk. From compactness of the spectrum, it must be exactly the closed unit disk.

Does my solution seem correct? I'm not entirely sure about my work. If I'm not mistaken, the spectrum of the usual right shift operator is the closed unit disk as well, and intuitively I'd expect the coefficients to affect the spectrum somehow. But I guess it doesn't have to happen necessarily.

Thanks a lot in advance.

Answer 1

So thanks to the advice by Ruy, I saw a problem with my computation - the solution is actually bounded for $|\lambda|<1$ as well, so the spectrum does not contain the unit disk.

A simpler way to find the spectrum is noticing that the spectral radius is $0$. It's not too hard to see that $||T^n||=\frac{1}{n!}$, which means that the spectral radius is $lim_{n\rightarrow \infty}(||T^n||)^{1/n}=0$, which means that the spectrum must be $0$ (since it's non-empty and bounded by $0$).