A set $A \subset l_1$ is compact if and only if closed, bounded, and one other condition 
A set $A \subset \ell_1$ is compact if and only if $A$ is closed and bounded and given any $\epsilon >0$, there exists $n_0$ such that $\sum_{k=n}^{\infty} |x_k| < \epsilon$ for all $n> n_0$ and $x \in A$.

It's easy to prove the part assuming that $A \subset l_1$ is compact, but I am finding difficulty to prove the other part.
 A: Suppose that $A$ is closed, bounded, and has the following property:
$$\forall\varepsilon>0,\exists n_0\in\mathbb N:\sum_{k=n}^{\infty}|x_k|<\varepsilon\quad\forall x\in A\text{ and }\forall n>n_0.\tag{$\star$}$$
Since $A$ is closed and $\ell^1$ is a Banach space, it follows that $A$ is a complete subset. Hence, in order to show compactness, it is sufficient to show that $A$ is totally bounded; that is, it can be covered by finitely many balls of any prescribed positive radius, no matter how small that prescribed radius is. Note that total boundedness is more restrictive of a concept than plain boundedness in infinite-dimensional normed spaces (or in metric spaces in general), but the two concepts happen to coincide in finite-dimensional Euclidean spaces. This latter fact is exactly what I'm going to exploit.
Since $A$ is bounded, there exists some $C>0$ such that $x\in A$ implies that $\|x\|_{\ell^1}\leq C$. Fix $\varepsilon>0$ and let $n_0\in\mathbb N$ be a the natural number corresponding to ($\star$). Consider the space $\mathbb R^{n_0}$ and endow it with the $n_0$-dimensional Manhattan norm
$$(x_1,\ldots,x_{n_0})\mapsto\|(x_1,\ldots,x_{n_0})\|_{M}\equiv\sum_{k=1}^{n_0}|x_k|.$$
Define $$K\equiv\{x\in\mathbb R^{n_0}\,|\,\|x\|_M\leq C\},$$
which is the closed ball of radius $C$ in $\mathbb R^{n_0}$ with respect to the Manhattan norm. Then, $K$ is a Manhattan-bounded subset of $\mathbb R^{n_0}$, so it is also totally bounded in $\mathbb R^{n_0}$ given that $\mathbb R^{n_0}$ is finite-dimensional. This implies that there is a finite collection of Manhattan balls of radius $\varepsilon$ covering $K$. In order words, there exist $\{x^1,\ldots,x^{J}\}\subset\mathbb R^{n_0}$, where $J\in\mathbb N$, such that
$$K\subseteq \bigcup_{j=1}^J\{x\in\mathbb R^{n_0}\,|\,\|x-x^{j}\|_M<\varepsilon\}.$$
Now define, for each $j\in\{1,\ldots, J\}$,
$$y^{j}\equiv(x_1^{j},\ldots x_{n_0}^{j},0,0,0,\ldots)\in\ell^1.$$
I claim that
$$A\subseteq\bigcup_{j=1}^J\{y\in\ell^1\,|\,\|y-y^{j}\|_{\ell^1}<2\varepsilon\};$$
this will show that $A$ can be covered by finitely many $\ell^1$ balls of radius $2\varepsilon$, which, in turn, will imply that $A$ is totally bounded. Pick $y\in A$. Let $x\equiv(y_1,\ldots y_{n_0})\in\mathbb R^{n_0}$. Clearly, $\|x\|_M\leq\|y\|_{\ell_1}\leq C$, so that $x\in K$. Then, $\|x-x^{j}\|_M<\varepsilon$ for some $j\in\{1,\ldots,J\}$. This, together with ($\star$), implies that
\begin{align*}
\|y-y^{j}\|_{\ell^1}=\sum_{k=1}^{n_0}|y_k-y^{j}_k|+\sum_{k=n_0+1}^{\infty}|y_k-y^{j}_k|=\|x-x^{j}\|_M+\sum_{k=n_0+1}^{\infty}|y_k|<\varepsilon+\varepsilon=2\varepsilon.
\end{align*}
This completes the proof.
