What will be $\sigma (T)\ $? 
Let $T$ be an operator on $\ell^2$ given by $Te_n = d_n e_n,$ where $$
d_n = \left\{
  \begin{array}{lr} 
      \frac {1} {2}, & \text {if}\ n \equiv 0\ (\text {mod}\ 3) \\
      \frac {n} {n+1}, & \text {if}\ n \equiv 1\ (\text {mod}\ 3) \\
      \frac {1} {n}, & \text {if}\ n \equiv 2\ (\text {mod}\ 3)  
      \end{array}
\right.
$$ Then what will be the $\sigma(T),$ the spectrum of $T\ $ and $\sigma_c(T),$ the continuous spectrum of $T\ $?

Clearly $\overline {\{d_n\ |\ n \geq 1 \}} \subseteq \sigma (T).$  Hence $\{d_n\ |\ n \geq 1 \} \cup \{0,1\} \subseteq \sigma (T).$ Will there be any other $\lambda \in \sigma (T)\ $?  Also how do I find $\sigma_c(T)\ $? It is clear that $\sigma_c (T) \cap \{d_n\ |\ n \geq 1\} = \varnothing.$ But I don't have any other idea about that.
Any help will be warmly appreciated. Thanks in advance.
 A: Hint Let $\lambda \notin \{d_n\ |\ n \geq 1 \} \cup \{0,1\}$.
Define
$$
Se_j=\frac{1}{\lambda - d_j}e_j
$$
Show that $S$ is bounded and that $S^{-1}=\lambda I -T$.
Next, you obviously have $\{d_n\ |\ n \geq 1 \} \subseteq \sigma_p(T)$. Show that this is equality:
If $a$ is an eigenvalue you have
$$
T(x)=ax
$$
for some $x=\sum a_i e_i$. Calculate $T(x)$ and compare.
A: As you have correctly concluded, the elements $d_n$ for $n \geq 1$ are elements of the point spectrum rather than the continuous spectrum (i.e. they are eigenvalues of $T$). The only question that remains is what to do with the elements $0$ and $1$.
As it turns out, we have $\sigma_c(T) = \{0,1\}$. You have already established that $0,1$ are elements of $\sigma(T)$, and it is easy to see that these are not eigenvalues. To see that the image of $T$ and $T - I$ is dense in $\ell^2$, it suffices to note that the range contains each of the elements $e_n$ for $n \geq 1$.
One way to verify that there are no other elements in the spectrum is to explicitly compute the inverse of $T - \lambda I$, noting that the sequence of multipliers that results is bounded. Alternatively: note that for all other $\lambda$, there exists a $c > 0$ such that $\|Tx\| \geq c \|x\|$.
