Let $T_n: l^2 \rightarrow l^2$ be defined by $$T_nx = (\zeta_1,\zeta_2,\zeta_3, ...,\zeta_n,0,0,0,...)$$ where $x = (\zeta_1, \zeta_2, \zeta_3,...)$. Prove that $(T_n)$ is a sequence of bounded linear operators such that $(T_n(x))$ converges for every $x \in l^2$. However, show that $|| T_n-T||=1$ for all $n \in N$.
Now I have showed that the sequence is a sequence of bounded linear operators and converges for every $x \in l^2$. I am having trouble with the last part. Since we are using the $l^2$ norm, how can I approach this proof? Thank you for your help!
For all $x\in l^2$ we have $||(T-T_n)(x)||^2=\sum_{k=n+1}^\infty |\zeta_k|^2\leq \sum_{k=1}^\infty |\zeta_k|^2=||x||^2$, which shows that $||T_n-T||\leq 1$. On the other hand, take any vector $0\ne x\in l^2$ with its first $n-1$ coordinates being $0$. Then $||T_n-T||\geq \frac{||(T-T_n)(x)||}{||x||}=1$.