Prove that $\mathcal{B}((l^1,\|.\|_1),(s,\|.\|_1))$ is not a Banach space. Let $\mathcal{B}((l^1,\|.\|_1),(s,\|.\|_1))$ be the space of all bounded linear operators from $(l^1,\|.\|_1)$ to $(s,\|.\|_1)$, where $s$ is the space of all eventually zero sequences. It is required to prove that $\mathcal{B}((l^1,\|.\|_1),(s,\|.\|_1))$ is not a Banach space. The following is my attempt.
Fix $N\in\mathbb{Z^+}$. Define $T_N:l^1\to s$ by $T_N((x_n))=(x_1,x_2,\cdots,x_N,0,\cdots)$ for each $(x_n)\in l^1.$ $T_N$ is a linear operator and for each $(x_n)\in l^1,$ $$\|T_N((x_n))\|_1=\sum_{n=1}^N|x_n|\leq\sum_{n=1}^\infty|x_n|=\|(x_n)\|_1$$
and therefore $T_N$ is bounded. Hence $(T_N)_{N\in\mathbb{Z^+}}$ is a sequence in $\mathcal{B}((l^1,\|.\|_1),(s,\|.\|_1))$. 
For each $N>M\in\mathbb{Z^+}$ and $(x_n)\in l^1$ we have $$\|T_N((x_n))-T_M((x_n))\|_1=\|(0,\cdots,0,x_{M+1},\cdots,x_N,0,\cdots)\|_1=\sum_{n=N+1}^M|x_n|.$$
Therefore since for any $(x_n)\in l^1,\ (\sum_{k=1}^n|x_k|)_{n\in\mathbb{Z^+}}$ is a Cauchy sequence we have $(T_N)_{N\in\mathbb{Z^+}}$ is a Cauchy sequence.
Note that for each $x=(x_n)\in l^1$ and for each $N,\ \|T_N(x)-x\|_1=\sum_{n=N+1}^\infty|x_n|$. Therefore $\|T_N(x)-x\|_1\to 0$ as $N\to\infty$. Therefore $T_N\to I,$ where $I$ is the identity linear operator which maps each $x\in l^1$ to $x$. But $I\notin\ \mathcal{B}((l^1,\|.\|_1),(s,\|.\|_1))$. Hence $\mathcal{B}((l^1,\|.\|_1),(s,\|.\|_1))$ is not a Banach space.
Could someone please tell me if this proof is alright? Thanks.
 A: You prove that $\{T_N((x_n))\}_N$ is a Cauchy sequence, but you have not shown that $\{T_N\}$ is a Cauchy sequence. For $N>M$ we have $T_N(e_N)-T_M(e_N)=e_N$, where $e_n$ is $1$ in the $n^\text{th}$ position and $0$ otherwise. Hence $\|T_N-T_M\|\ge1$, and so in particular $\{T_N\}$ is not Cauchy.
A simple modification of $\{T_N\}$ works nicely. Let $S_N((x_n))=(x_1,\frac{x_2}2,\frac{x_3}3,\ldots,\frac{x_N}N,0,0,\ldots)$. Then for $N>M$ we have
$$\|S_N((x_n))-S_M((x_n))\|_1=\sum_{k=M+1}^N\left|\frac{x_k}k\right|\le\frac1M\sum_{k=M+1}^N|x_k|\le\frac1M\|(x_n)\|_1,$$
which shows $\|S_N-S_M\|\le\frac1M$. This is enough to deduce $\{S_N\}$ is Cauchy. Assume that $\{S_N\}$ converges. Then $\{S_N((x_n))\}$ converges in $(s,\|\cdot\|_1)$ for every $(x_n)\in\ell^1$. Take for instance $x_n=2^{-n}$ to reach a contradiction; we would have $S_N((x_n))=(2^{-1},\frac{2^{-2}}2,\ldots,\frac{2^{-N}}{N},0,0,\ldots)$ which converges in $\ell^1$ to the obvious limit which is not in $s$.
Overall, you had the right idea but need to be more careful with convergence of operators - $T_n\to T$ is not the same as $T_nx\to Tx$ for all $x$.
