I need to show that the operator $\displaystyle T_n:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C})$ defined by $\displaystyle (T_nx)(j)=\frac{1}{2^j}x(j)$ for $\displaystyle j\leq n$ and $\displaystyle (T_nx)(j)=0$ for $\displaystyle j>n$ converges to $\displaystyle T$ in the operator norm $\displaystyle \|T_n-T\|_{op}\rightarrow 0$ where $\displaystyle (Tx)(j)=\frac{1}{2^j}x(j)$ for all $\displaystyle j$
I need to show that the operator $\displaystyle T_n:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C})$ defined by $\displaystyle (T_nx)(j)=\frac{1}{2^j}x(j)$ for $\displaystyle j\leq n$ and $\displaystyle (T_nx)(j)=0$ for $\displaystyle j>n$ converges to $\displaystyle T$ in the operator norm $\displaystyle \|T_n-T\|_{op}\rightarrow 0$ where $\displaystyle (Tx)(j)=\frac{1}{2^j}x(j)$ for all $\displaystyle j$
Notice that if $\displaystyle y = (T-T_n)x$ then $\displaystyle y(j) = 0$ if $\displaystyle j\leqslant n$ and $\displaystyle y(j) = 2^{-j}x(j)$ for j>n. So $\displaystyle \|y\|^2 = \sum_{j=n+1}^\infty 2^{-2j}|x(j)|^2 < 2^{-2n}\sum_{j=n+1}^\infty |x(j)|^2 < 2^{-2n}\|x\|^2$. Hence $\displaystyle \|T-T_n\| \leqslant 2^{-n}$.