Proving that certain subspace of $\ell_1$ is non closed

Question

I need to prove that

$$L= \left\{(x_i) \in\ell_1 : \sum_{i=1}^\infty ix_i= 0\right\}$$ is non-closed in $\ell_1$.

I can't really think of sequences of sequences that are in this subspace, much less one that converges specifically to a sequence not in this space, so help would be appreciated.

Answer 1

Just to clarify the hint julien gave in the comments above:

There is a sequence $\{x^{(k)}\}_{k\in\mathbb{N}}$ in $L$, converging in the $\ell_1$ norm to something not in $L$, whose first term is the element $x^{(1)}\in L$ defined by $$x^{(1)}_1=1,\quad x^{(1)}_2=-\frac{1}{2},\quad x^{(1)}_n=0\text{ for all }n\geq 3.$$

Answer 2

Zev's answer solves the problem. We can give a generalization:

Let $c:=\{c_n\}$ be a sequence of complex numbers. The subspace $$V_c:=\{x\in\ell^1,\sum_{n=1}^{+\infty}c_nx_n=0\}$$ is closed in $\ell^1$ (for its natural norm) if and only $c$ is bounded.