Define $T\colon c_0\rightarrow\ell^1$ by setting $$T(x)=\left(\frac{x_n}{n^2}\right),\quad \text{for each}\ x=(x_n)\in c_0.$$ Show that $T\in\mathcal{B}(c_0,\ell^1)$. If $B$ is the closed unit ball in $c_0$, show that $T(B)$ is not closed in $\ell^1$.
In the first part I argue in the following way: For each $x\in c_0$, we have that \begin{eqnarray}\|T(x)\|_1&=&\left\|\left(x_1,\frac{x_2}{2^2},...,\frac{x_n}{n^2},...\right)\right\|_1=\sum_{n=1}^\infty\left|\frac{x_n}{n^2}\right|\\ &\leqslant&\sum_{n=1}^\infty\frac{\sup_{n\in\Bbb{N}}|x_n|}{n^2}=\|x\|_\infty\sum_{n=1}^\infty\frac{1}{n^2}, \end{eqnarray} where the last term on expression is finite, since $x\in c_0$ and the other factor is a $p$-series with $p=2$. Therefore $T$ is bounded.
But I have no idea how to procede in the second part (show that $T(B)$ is not closed in $\ell^1$, where $B$ is the closed unit ball in $c_0$).
Take the point $y=(1,\frac{1}{4},\frac{1}{9},...,\frac{1}{n^2},...)$ and $x_n \in c_0$ such that $$x_1=(1,0,0........)$$ $$x_2=(1,1,0,0.....)$$ $$x_3=(1,1,1,0,..)$$ e.t.c. Then $x_n \in B, \forall n \in \Bbb{N}$ and $T(x_n) \to y$ thus $y \in \overline{T(B)}$
But if existed $z=(z_k) \in B$ such that $T(z)=y$ then we would have that $z_k=1,\forall k \in \Bbb{N} \Longrightarrow z \notin c_0$ which is a contradiction.