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