Let $H$ be a Hilbert space , $T\in L(H)$ (set of all continuous linear functions of $H$ in $H$), $(z_{n})$ a othonormal sequence on $H$, and $(\lambda_{n})$ a sequence in $\mathbb{K}$ (it is $\mathbb{R}$ or $\mathbb{C}$) such that $T^{*}(z_{n})=\lambda_{n}z_{n}$ (where $T^{*}$ is the adjoint map of $T$) and $\sum_{n=1}^{\infty}|\lambda_{n}|^{2}<\infty$.
Show that $$\sum_{n=1}^{\infty}\left|\langle y,z_{n}\rangle\right|<\infty$$ for all $y\in T(H)$.
Remark: I am distrusting the hypothesis $\sum_{n=1}^{\infty}|\lambda_{n}|^{2}<\infty$. Note that if we replace $\sum_{n=1}^{\infty}|\lambda_{n}|^{2}<\infty$ for $\sum_{n=1}^{\infty}|\lambda_{n}|<\infty$ then the problem is easier to show, in fact, for $y=T(x)\in T(H)$ we would have $$|\langle y,z_{n}\rangle|=|\langle T(x),z_{n}\rangle|=|\langle x,T^{*}(z_{n})\rangle|=|\langle x,\lambda_{n}z_{n}\rangle|\leq|\lambda_{n}|\|x\|.$$ Therefore $$\sum_{n=1}^{\infty}|\langle y,z_{n}\rangle|\leq \|x\|\sum_{n=1}^{\infty}|\lambda_{n}|<\infty.$$ I do not see how to prove it if the hypothesis is $\sum_{n=1}^{\infty}|\lambda_{n}|^{2}<\infty$.
We have
$$\lvert\langle y, z_n\rangle\rvert = \lvert \lambda_n\rvert\cdot \lvert \langle x,z_n\rangle\rvert,$$
and by Bessel's inequality
$$\sum_{n = 1}^{\infty} \lvert \langle x,z_n\rangle\rvert^2 \leqslant \lVert x\rVert^2.$$
And by Cauchy-Schwarz
$$\sum_{n = 1}^{\infty} \lvert \lambda_n\rvert\cdot \lvert \langle x,z_n\rangle\rvert \leqslant \sqrt{\sum_{n = 1}^{\infty} \lvert \lambda_n\rvert^2} \cdot \sqrt{\sum_{n = 1}^{\infty} \lvert \langle x, z_n\rangle\rvert^2}\,.$$