For a arbitrary positive $\varepsilon$, is there a $n_\varepsilon\in\mathbb{N}$ such that $$\underset{(a_n)_{n=1}^{\infty}\in X}{\sup}\left(\displaystyle\sum_{n\geq n_\varepsilon}|a_n|\right)\leq \varepsilon\text{ },$$ where $X$ is bounded in $\ell_1$?
PS: $\ell_1$ is a sequence space such that:
$$\ell_n=\left\{(a_i)_{i=1}^{\infty};a_i\in\mathbb{K}, \forall i\in\mathbb{N} \text{ and }\displaystyle\sum_{i=1}^\infty|a_i|^n<\infty\right\},$$
where $\mathbb{K}$ is the scalar space. Also, the norm $||\cdot||_n$ is given by
$$||(a_i)_{i=1}^{\infty}||_n=\left(\displaystyle\sum_{i=1}^\infty|a_i|^n\right)^{\frac{1}{n}}.$$
My attempt is:
For all $a\in X$, there is a $M>0$ such that $||a||_1\leq M$. Than take a sequence $(a_j)_{j=1}^\infty$ in $X$ and we have $$\left(\displaystyle\sum_{j=1}^\infty|a_j|\right)=||(a_j)_{j=1}^\infty||_1\leq M.$$
How can I proceed?
A bit less formally, another way to think about the given condition is that it's asking whether the elements (that is, the sequences) of the bounded set $X$ form series with uniformly small tails. For any finite $X$, this is certainly true, and you might try proving it for sets like that first. The proof might suggest to you what could go wrong when $X$ is infinite.
A slightly easier problem where the same ideas are in play is to address whether, given a bounded $X\subset \ell^1$, there is an $n_0$ such that $$ \sup_{\{a_n\}\in X}\sum_{n\ge n_0}|a_n| = 0\ ? $$ This questions asks you if the elements of $X$ form series whose tails are all dead zero if you "wait" long enough. I think it is clearer what the answer is to this question, and it also suggests a possible approach to the problem you are considering.