This question comes (but is different in purpose) from a previous question about subspaces of $\ell^2$. While the mentioned question was asking for a general answer now I want to know the specific details of an specific comment (Paul's first comment).
Let $\mathbf{x}^\alpha=(n^{-\alpha})_{n=1}^\infty$ so that $\{\mathbf{x}^\alpha\}_{0.5<\alpha<1}\subseteq\ell^2\setminus\ell^1$ and let $V=\langle\{\mathbf{x}^\alpha\}_{0.5<\alpha<1}\rangle$. The (accepted) answer (by David) explains why $V\cap\ell^1=\{\mathbf{0}\}$ and the argument is as follows: if $$x=\sum_{i=1}^n \lambda_i\mathbf{x}^{\alpha_i}$$ with $\alpha_1<\alpha_2<\dots<\alpha_n$ and $\lambda_i\ne0$ for every $i$ then the $m$-th term of $x$ is $$x_m=\sum_{i=1}^n \lambda_i\mathbf{x}^{\alpha_i}_m=\sum_{i=1}^n \lambda_im^{-\alpha_i}=m^{-\alpha_1}(\lambda_1+\lambda_2m^{\alpha_1-\alpha_2}+\dots+\lambda_nm^{\alpha_1-\alpha_n})$$ and, with $m$ big enough (bigger than some specified constant $N$ depending on the $\lambda$'s and $\alpha$'s) we can make sure that $$\vert \lambda_1+\lambda_2m^{\alpha_1-\alpha_2}+\dots+\lambda_nm^{\alpha_1-\alpha_n} \vert\ge c>0$$ so $$\Vert x\Vert_1=\sum_{m=1}^\infty\vert x_m\vert\ge\sum_{m=N}^\infty cm^{-\alpha_1}$$ which diverges.
What about $\overline{V}$? The same argument doesn't work: if we want to keep the $c$, we cannot control $N$ and vice versa. For a sequence $\{x^n\}_{n=1}^\infty\subseteq V$ such that $x^n\to x\in\ell^1$ i thought of something like taking an increasing sequence $\alpha_1<\alpha_2<\dots$ converging to $1$ and $$x^n=\sum_{i=1}^n (-1)^i(1-\alpha_i)\mathbf{x}^{\alpha_i}\quad\text{ or }\quad=\sum_{i=1}^n (-1)^i\frac{1-\alpha_i}{i^2}\mathbf{x}^{\alpha_i}$$ which I can show converge to some $x$ but cannot calculate its $1$-norm.
So any ideas/proofs/disproofs on $\overline{V}\cap\ell^1=\{\mathbf{0}\}$?
Thanks
