Showing compactness of operator We consider $X=l^p(\mathbb{Z},\mathbb{C})$ for $1\leq p\leq\infty$ and define $T((x_n)_{n\in\mathbb{Z}})= (a_n x_n)_{n\in\mathbb{Z}} :=(\frac{1}{n^2+1}x_n)_{n\in\mathbb{Z}}$. Let $a_n^N = a_n$ if $n\leq N$ and $a_n^N = 0$ else and $T_N(x_n) = (a_n^N x_n)$. Then $rk (T_N)\leq N$, thus $T_N$ is compact and $T_N \rightarrow T$ in $L(X)$. So: T is compact by a lemma which exactly states the limit situation as above.
Does everything goes through in my proof?

Your idea is good ! But you should explain in detail why $T_N \to T$ in $L(X)$.
Hence give a proof for $||T_N -T|| \to 0$
