Consider the following linear map from $l^{2}$ to $l^{2}$:
$T(x)_{n} = \frac{x_{n}}{n}$
I know that this operator is the pointwise limit of the operators $T_{n}$ that output $(x_{1}, \dots, \frac{x_{n}}{n}, 0, 0, \dots)$. How do I show that $||T-T_{n}|| \to 0$, where $||\cdot||$ is the operator norm on $l^{2}$, defined by $$ ||T|| = \sup_{||x||=1} ||T(x)|| $$ I'm struggling with showing that the convergence is uniform on the unit sphere.
For $x=(x_{1},x_{2},...)$, we have $|x_{k}|\leq\|x\|_{2}$, $k=1,2,...$, so \begin{align*} \|(T-T_{n})x\|_{2}&=\left(\sum_{k\geq n+1}\dfrac{x_{k}^{2}}{k^{2}}\right)^{1/2}\\ &\leq\left(\sum_{k\geq n+1}\dfrac{\|x\|_{2}^{2}}{k^{2}}\right)^{1/2}\\ &\leq\|x\|_{2}\left(\sum_{k\geq n+1}\dfrac{1}{k^{2}}\right)^{1/2}, \end{align*} so \begin{align*} \|T-T_{n}\|\leq\left(\sum_{k\geq n+1}\dfrac{1}{k^{2}}\right)^{1/2}\rightarrow 0. \end{align*}