# Determine residual spectrum of the operator $Tx=(0,\frac{x_1}1,\frac{x_2}2,...)$ on $\ell^2$

Let $$T: \ell^2 \rightarrow \ell^2$$ be an operator $$T(x)= (0, \frac{x_1}{1}, \frac{x_2}{2}, \frac{x_3}{3},...)$$, $$\forall x = (x_n)_{n \in \mathbb{N}} \in \ell^2$$. $$T$$ is a compact, bounded linear operator and $$\sigma_p (T)= \emptyset$$, so $$(T- \lambda I)^{-1}$$ exists for all $$\lambda$$. Now I have to prove that residual spectrum consists only of $$0$$. Residual spectrum are those $$\lambda \in \mathbb{C}$$ for which $$(T- \lambda I)^{-1}$$ exists but is not defined on the dense subset of $$\ell^2$$.

It can be proved that $$0 \in \sigma_r (T)$$ and now we consider $$\lambda \neq 0$$. My attempt is to find the vector $$x =(x_n) \in \ell^2$$ such that

$$(T- \lambda I )(x)= (\alpha_1, \alpha_2, ...,\alpha_n,0,0,0....)$$, since the set of such sequences is dense in $$\ell^2$$.

$$- \lambda x_1 = \alpha_1$$

$$x_1- \lambda x_2= \alpha_2$$

$$\frac{x_2}{2} - \lambda x_3 = \alpha _3$$

...

$$\frac{x_{n-1}}{n-1} - \lambda x_{n} = \alpha _n$$

$$\frac{x_{n}}{n} - \lambda x_{n+1} = 0$$

...

How can one prove from this system that $$x \in \ell ^2$$?

For $$k>n$$ we have $$x_k={\lambda^n\over \lambda^k}{(n-1)!\over (k-1)!}x_n$$ The sequence $$\lambda^{-k}((k-1)!)^{-1}$$ is absolutely summable (by for example the ratio test), hence square summable.
Another argument: for $$k>n$$ and $$k>2|\lambda|^{-1}$$ we have $$|x_{k+1}|={1\over |\lambda| k}|x_k|\le {1\over 2}|x_k|$$ Thus the sequence is square summable.