I was reading a paper where it was said that for a sequence $(x_n)_n$ of elements in $\ell^1$, the closed, convex hull of the sequence is
$$\left\{ \sum_{n \geq 1} \lambda_nx_n : \lambda_n \geq 0 \text{ and } \sum_{n \geq 1} \lambda_n \leq 1 \right\}$$
Is this generally true, or is it particular to $\ell^1$?
This doesn't make much sense though if the sequence $(x_n)$ is not bounded. For example, set $e_n$ to be the sequence that is $0$ everywhere except the $n$-th slot where it is $1$. Then set $x_n=2^ne_n$ and $\lambda_n=\frac{1}{2^n}$ (note that $\sum_{n}\lambda_n=1$. We have that $\sum_{n=1}^N\lambda_nx_n=(1,1,\dots,1,0,0,0,\dots)$, so $\|\sum_{n=1}^N\lambda_nx_n\|_1=N\to\infty$, so the series $\sum_{n=1}^N\lambda_nx_n$ does not converge in $\ell^1$.
If the sequence $(x_n)$ is bounded, then it makes sense but it fails for trivial reasons. Take for example the following sequence in $\ell^1$: $x_1=(\frac{1}{n^2})_{n=1}^\infty$ and $x_2=x_3=\dots=(\frac{1}{2^n})_{n=1}^\infty$. Then obviously the closed convex hull of $\{x_n\}$ is the line segment $[x_1,x_2]$, i.e. $$\overline{\text{co}}(\{x_n\}_{n=1}^\infty)=\{tx_1+(1-t)x_2: t\in[0,1]\}$$ Set $A:=\{\sum_{n=1}^\infty\lambda_nx_n:\lambda_n\geq0,\sum_{n\geq1}\lambda_n\leq1\}$. Note that for $\lambda_n=0$ for all $n$ we get that $0\in A$. On the other hand, $0\not\in\overline{\text{co}}(\{x_n\}_{n=1}^\infty)$, because if this wasn't the case then for some $t_o\in(0,1)$ it would be $x_1=-\frac{1-t_o}{t_o}x_2$ and $x_1,x_2$ would be scales of each other, which isn't the case of course.
I can give more examples: take $\lambda_1=t_o\in(0,1)$ and $\lambda_2=\lambda_3=\dots=0$. Then this gives $t_ox_1\in A$, but it is $t_ox_1\not\in\overline{\text{co}}(\{x_n\}_{n=1}^\infty)$: If this wasn't the case then $t_ox_1=t_1x_1+(1-t_1)x_2$ for some $t_1\in(0,1)$ so $(t_o-t_1)x_1=(1-t_1)x_2$, a contradiction.
Maybe this is true for the particular sequence your paper mentions, but in general this is not true.