Let $T: l_2(\mathbb{N}) \to l_2(\mathbb{N})$ be an operator given by
$$Tx = \left( \frac{n-1}{n+1} x_n \right)_{n \in \mathbb{N}} \text{ where } x = (x_n)_{n \in \mathbb{N}} \in l_2(\mathbb{N}).$$
My goal is it to determine the spectrum $\sigma(T)$, the point spectrum $\sigma_p(T)$, the continuous spectrum $\sigma_c(T)$ and the residual spectrum $\sigma_r(T)$.
Calculation:
Since $\langle Tx,y\rangle = \langle x,Ty\rangle$ for all $x,y \in l_2(\mathbb{N})$ we observe that $T=T^*$ is self-adjoint and hence closed. Therefore, we know that $\sigma(T)$ is the disjoint union of all the other spectra.
We define $M := \{ \frac{n-1}{n+1} | n \in \mathbb{N}\}$.
Consider the case $\lambda \in \mathbb{C} \setminus M$. Let $y \in l_2(\mathbb{N})$ be arbitrary. Then $$(T-\lambda)x = y \Longleftrightarrow \forall n \in \mathbb{N}: x_n = \left(\frac{n-1}{n+1} - \lambda \right)^{-1} y_n,$$ so we have a unique solution $x \in l_2(\mathbb{N})$. Therefore, $T-\lambda$ is bijective, so $M \subset \rho(T) := \mathbb{C} \setminus \sigma(T)$.
Now consider the case $\lambda \in M$, say $\lambda = \frac{k-1}{k+1}$ for a $k \in \mathbb{N}$. Then $$(T-\lambda)e^{(k)} = (T-\lambda)0 = 0$$ where $e^{(k)}$ denotes the $k$-th unit vector which is in $l_2(\mathbb{N})$. Since $e^{(k)} \neq 0$ we have $T-\lambda$ is not injective and we get $\sigma(T) = \sigma_p(T)=M$ and $\sigma_c(T) = \sigma_r(T) = \emptyset$.
Thank you!