Let $E$ an **infinite dimensional** complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.

**Definition:** Let $T \in \mathcal{L}(E)$. The Moore-Penrose inverse of $T$, denoted by $T^{+}$, is defined as the unique linear extension of $(\bar{T})^{-1}$ in
$$D(T^{+}) = \mathcal{R}(T)+\mathcal{R}(T)^{\perp},$$
with $\mathcal{N}(T^{+}) = \mathcal{R}(T)^{\perp}$ and $\bar{T}$ is the isomorphism
$$\bar{T}:=T|_{{\mathcal{N}(T)}^{\perp}}: {\mathcal{N}(T)}^{\perp} \longrightarrow \mathcal{R}(T).$$
Moreover, $T^{+}$ is the unique solution of the four ''Moore-Penrose equations'':
$$TXT = T,\quad XTX = X,\quad XT = P_{N{(T)^{\bot}}}\,\,\mbox{and}\,\,\quad TX = P_{\overline{\mathcal{R}(T)}}{{|}_{D(T^{+})}}.$$

Here $\mathcal{R}(T)$ and $\mathcal{N}(T)$ denote respectively the range and the nullspace of $T$. Also $P_{F}$ denote the orthogonal projection onto $F$.

It is well know that $A^+$ is bounded iff $A$ has a closed range.

If $A$ is selfadjoint matrix, then $$ A^{+}= \lim_{t \to 0}(A^2+tI)^{-1} A.\;\;(1).$$

If $A$ is not selfadjoint, then $A^+=A^*(AA^*)^+$ (which is equal to $(A^*A)^+A^*$).

Is the formula $(1)$ true only for matrix or even for Hilbert space operators? Let for example $S$ an operator on $\ell^2$ given by $$S(x_1,x_2,\cdots)=(x_1,\tfrac{1}{2} x_2,\tfrac{1}{3} x_3,\cdots).$$ How can I find $S^+$?