Closed subspace of $l_2$ Let $$X=\{ a \in l_2 : \sum_{n=1}^{\infty}{\frac{1}{\sqrt{n}}a_n = 0} \}$$

Show that X is not a closed subset of $l_2$

My general approach is to find a converging sequence $a_n \in X$, whose limit $a$ does not belong to $X$. Any hints to this problem as I fail to see what sequence I can construct here
Here is a proof, which assumes that the sum in the definition of $X$ has to converge absolutely.
The sequence $x\in l^2$ with $x_n = \frac1{\sqrt n\log n}$ has the property
that $\sum_n x_n \frac1{\sqrt n}=+\infty$.
Now let me 'double' this sequence, construct a sequence $y$ such that
$$
y_{2n+1}, \ y_{2n+2} \sim \frac1{\sqrt n\log n}
$$
and
$$
\frac{y_{2n+1}}{\sqrt{2n+1}} = -\frac{y_{2n+2}}{\sqrt{2n+2}},
$$
i.e.
$$
y_{2n+1}:= \frac1{\sqrt n\log n}, \ y_{2n+2}:=-\frac1{\sqrt n\log n} \frac{\sqrt{2n+2}}{\sqrt{2n+1}}.
$$
Then by construction
$$
\sum_{n=1}^{2m} y_n \frac1{\sqrt n}=0
$$
for all $m$. Hence
$$
y^m:=(y_1, \dots, y_{2m},0,\dots)
$$
is a sequence in $X$ such that $y^m \to y$ in $l^2$. But $y$ is not in $X$,
as $\sum_{n=1}^\infty y_n\frac1{\sqrt n}$ does not converge absolutely. It does converge conditionally.
There's a problem in the very definition of $X$. The sum $\sum n^{-1/2} a_n$ doesn't have to converge, for example it's infinite for $a_n = n^{-1/2} (\ln n)^{-1}$ (which still lies in $\ell_2$).
Once you fix the definition of $X$ (in one way or another), you should be able to construct a counterexample to closedness, based on the sequence $(a_n)$ given above.