Let $H$ be Hilbert and $T\in B(H)$ the left shift operator w.r.t. the orthonormal basis $(e_{k})_{k\in\mathbb{N}}$. That is, $Te_{1}=0$ and $Te_{k}=e_{k-1}$ for $k>1$. Then how do I show that $T^{n}x\to0$ (as $n\to\infty$ for all $x\in H$, that is, $(T^{n})_{n\in\mathbb{N}}$ is strongly convergent to $0$ in $B(H)$ (according to the definition of Murphy in his book on C*-algebras).
Any $h\in H$ van be written as $h=\sum_{k\geq1}\langle h,e_{k}\rangle e_{k}$. So by continuity of $T$, for all $x\in H$ I am able to estimate
\begin{align*} \|T^{n}x\|&=\bigg\|T^{n}\sum_{k\geq1}\langle h,e_{k}\rangle e_{k}\bigg\|=\bigg\|\sum_{k\geq1}\langle h,e_{k}\rangle T^{n}e_{k}\bigg\|\\ &=\bigg\|\sum_{k>n}\langle x,e_{k}\rangle e_{k-n}\bigg\|\leq\sum_{k>n}|\langle x,e_{k}\rangle|. \end{align*}
But I don't see why the right hand side converges to $0$ as $n\to\infty$. I Know that $\|x\|^{2}=\sum_{k\geq1}|\langle x,e_{k}\rangle|^{2}$ by Parseval's identity, but it differs from what I have.
Any suggestions are greatly appreciated!
As you've correctly derived, we have $$ \Vert T^n x \Vert^2 = \left\Vert \sum_{k > n} \langle x, e_k \rangle e_{k - n} \right\Vert^2 \leq \sum_{k > n} |\langle x, e_k\rangle|^2. $$ As you've mentioned, $\Vert x \Vert^2 = \sum_{k} |\langle x, e_k\rangle|^2$. Therefore this sequence is convergent, which immediately yields that $\sum_{k > n} |\langle x, e_k\rangle|^2 \to 0$ as $n \to \infty$.