Let $\{e_n\}$ be an orthonormal basis of a hilbert space $H$ and let $T_n:H\to H$ and $T:H\to H$ be given by $$T_n(x)=\sum_{j=1}^n\frac1{j}\langle x,e_j\rangle e_j\ \ T(x)=\sum_{n=1}^\infty\frac1{n}\langle x, e_n\rangle e_n$$ Then, is it true that $\|T_n-T\|$ does not converge to zero but $\forall x\in H$, $\|T_n(x)-T(x)\|\to 0$ as $n\to\infty$?
I think that both the operator and the image converge, because the definition of the operators seems same to me and partial sums of convergent series is convergent to the sum in a complete space. Any hints? Thanks beforehand.
For $x \in H$ we have $$ \|T_n(x)-T(x)\|^2 = \Big\|\sum_{j=n+1}^{\infty}\frac1{j}\langle x,e_j\rangle e_j \Big\|^2 = \sum_{j=n+1}^{\infty}\frac{1}{j^2}|\langle x,e_j\rangle |^2 \leq \Big(\sum_{j=n+1}^{\infty}\frac{1}{j^2} \Big) \|x\|^2,$$
Hence, $\|T_n-T\| \leq \big(\sum_{j=n+1}^{\infty}\frac{1}{j^2}\big)^{\frac{1}{2}}$. Since $ \big(\sum_{j=n+1}^{\infty}\frac{1}{j^2}\big)^{\frac{1}{2}} \to 0$ as $n \to \infty$, we derive $\|T_n-T\| \to 0$ as $n \to \infty$.