I have trouble finding the spectra of the following operator $T$:
$T: \ell^2 \rightarrow \ell^2$, $x = (\xi_i)_i \in \ell^2$: \begin{equation} Tx = (0,\frac{\xi_1}{1}, \frac{\xi_2}{2}, \ldots) \end{equation}
Crearly it isn't invertible, so $0 \in \sigma(T)$. Also, i have proved that $\sigma_p(T)= \emptyset$. Why $\{0\} = \sigma(T) = \sigma_r(T)$?
For $n\in\mathbb{N}$ it is easy to see that $$T^nx = \left(\underbrace{0, \ldots, 0}_n, \frac{\xi_1}{n!}, \frac{\xi_2}{(n+1)!}, \ldots, \frac{\xi_k}{(n+k-1)!}, \ldots\right)$$
This implies $$\|T^nx\|_2^2 = \sum_{k=1}^\infty \left|\frac{\xi_k}{(n+k-1)!}\right|^2 \le \frac1{n!^2} \sum_{k=1}^\infty |\xi_k|^2 =\frac1{n!^2} \|x\|_2^2 $$ so $\|T^n\| \le \frac1{n!}$. Hence the spectral radius is $$r(T) = \lim_{n\to\infty} \|T^n\|^{\frac1n} \le \lim_{n\to\infty} \left(\frac1{n!}\right)^{\frac1n} = 0$$
so $\sigma(T) \subseteq \{0\}$. Since $0 \in \sigma(T)$ it follows $\sigma(T) = \{0\}$.
$T$ is injective so $0 \notin \sigma_p(T)$. Also, $\operatorname{Im} T \subseteq \{x = (\xi_i)_i \in \ell^2 : \xi_1 = 0\}$, which is not dense in $\ell^2$ so $0 \notin \sigma_c(T)$. Hence $0 \in \sigma_r(T)$.
Therefore $\sigma(T) = \sigma_r(T) = \{0\}$ and $\sigma_p(T) = \sigma_c(T) = \emptyset$.