Show that the symmetric operator $Sx:=(nx_n)_{n\in\mathbb N}$ on $\mathcal D(S):=\{x\in H:(nx_n)_{n\in\mathbb N}\in\ell^2\}$ is even self-adjoint Let $H:=\ell^2(\mathbb N,\mathbb C)$ and $$Sx:=(nx_n)_{n\in\mathbb N}\;\;\;\text{for }x\in\mathcal D(S):=\left\{x\in H:(nx_n)_{n\in\mathbb N}\in H\right\}.$$ $S$ should be bijective and $T:=S^{-1}\in\mathfrak L(H)$ with $$Ty=\left(\frac1ny_n\right)_{n\in\mathbb N}\;\;\;\text{for all }y\in H.\tag1$$ It's easy to see that $S$ is symmetric, i.e. $$\langle y,Sx\rangle_H=\sum_{n=1}^\infty\overline{y_n}nx_n=\sum_{n=1}^\infty\overline{ny_n}x_n=\langle Sy,x\rangle_H\;\;\;\text{for all }x,y\in\mathcal D(S).\tag2$$
What's the easiest way to show that $S$ is even self-adjoint?
 A: Theorem: A densely-defined symmetric linear operator $S : \mathcal{D}(S)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ on a complex Hilbert space $\mathcal{H}$ is self-adjoint iff $S\pm iI$ are surjective.
This is easy to verify in your case:
$$
         (S\pm iI)\{ x_n \} = \{(n\pm I)x_n\} = \{ y_n \} \\
             \iff \{ x_n \} = \{ \frac{1}{n\pm i}y_n\}
$$
More precisely, if $\{ y_n \} \in \ell^2(\mathbb{N})$, then $\{x_n\}$ as given above are in $\ell^2(\mathbb{N})$ and the first equations hold. So $S\pm iI$ are surjective.
Everything comes back to this early theorem of von Neumann, which you have probably seen:
Theorem: Let $S : \mathcal{D}(S)\subseteq\mathcal{H}\rightarrow\mathcal{H}$ be a closed symmetric linear operator. Let $\mathcal{G}(S^*)$ and $\mathcal{G}(S)$ denote the graphs of $S^*$ and $S$ in $\mathcal{H}\times\mathcal{H}$. Then one has the orthogonal decomposition
$$
         \mathcal{G}(S^*)=\mathcal{G}(S)\oplus\mathcal{D}^+\oplus\mathcal{D}^-
$$
where $\mathcal{D}^{+}$  and $\mathcal{D}^{-}$ are the graphs of $\mathcal{G}(S^*)$ restricted to $\mathcal{N}(S^*-iI)$ and $\mathcal{N}(S^*+iI)$, respectively.
