Compact operator $T:l^1\to c_0$

Given the operator $$T:l^1\to c_0$$ with $$T(x=(x_k))=\left(\sum\limits_{k=1}^\infty x_k,\sum\limits_{k=2}^\infty x_k,...,\sum\limits_{k=n}^\infty x_k,...\right)$$.

I have to decide if this is a compact operator. I think it is, and to prove it I have the sequence of operators:

$$T_n(x=(x_k))=\left(\sum\limits_{k=1}^\infty x_k,\sum\limits_{k=2}^\infty x_k,...,\sum\limits_{k=n}^\infty x_k,0,0,...\right)$$

whose image is finite dimensional. So if I prove that $$T_n\to T$$, I have that $$T$$ is compact.

I get $$||T_n(x)-T(x)||\leq \sum\limits_{k=n+1}^\infty|x_k|$$, but I want to find $$||T_n(x)-T(x)||\leq \varepsilon_n||x||$$ where $$\varepsilon_n\to0$$. If I prove that, I have that $$||T_n-T||\leq\varepsilon_n\to 0$$ and I conclude my proof.

Can somebody help me to find that $$\varepsilon_n$$? Thank you

In fact it's pretty clear that $$||T_n-T||=1$$ (for example because the standard isomorphism of $$\ell_1^*$$ and $$\ell_\infty$$ is an isometry.) That doesn't prove that $$T$$ is not compact, but it's not; for example $$||Te_n-Te_m||=1$$ for $$n\ne m$$, so the sequence $$(Te_n)$$ has no convergent subsequence.

It doesn't exist, and this operator is not compact.

Consider elements $$x \in \ell^1$$ of the form $$(0,0,\dots, 0,-1,1,0,0,\dots)$$. They are all in the ball of radius 2. Look at what the corresponding elements $$T(x)$$ are, and show that this is an infinite set with no limit points.