Suppose $S:H \to H$ is positive self adjoint operator on Hilbert space $H$, with $$Sx=\sum_n a_n \langle x,e_n\rangle e_n,\quad a_n >0$$ for all $n$. Show that the range of $\sqrt S$, $\sqrt S(H)$ equals $$ \sqrt S(H)=\left\{ x\in H: x= \sum_nb_n e_n: \sum_n \frac{b_n^2}{a_n}<\infty \right\}$$
Suppse $y\in \sqrt S(H)$, then $y=\sum_n \sqrt a_n \langle x,e_n\rangle e_n$ for some $x\in H$. This is getting me nowhere
For $y\in\sqrt{S}(H)$, then $y=\displaystyle\sum_{n}\sqrt a_{n}(x,e_{n})e_{n}$, and $\displaystyle\sum_{n}\dfrac{|\sqrt{a_{n}}(x,e_{n})|^{2}}{a_{n}}=\sum_{n}|(x,e_{n})|^{2}\leq\|x\|^{2}<\infty$.
For $y=\displaystyle\sum_{n}b_{n}e_{n}$ be such that $\displaystyle\sum_{n}\dfrac{b_{n}^{2}}{a_{n}}$, we let $x=\displaystyle\sum_{n}\dfrac{b_{n}}{\sqrt{a_{n}}}e_{n}$, this $x$ exists. And we check that $\sqrt{S}(x)=y$.