Showing a subspace is not closed I have been asked the following problem which I have so far failed to solve rigourosly.
Let $M=\{ (a_n)_{n=1}^\infty\in l^2\vert \sum_{n=1}^\infty \frac{a_n}{\sqrt{n}}=0 \}$. Determine whether $M$ is a closed subspace of $l^2$.
My attempt:
I think that the answer is no because this seems like the kernel of a functional
$$ \phi(a):= \Big \langle a,\big(\frac{1}{\sqrt{n}}\big) \Big \rangle $$.
And since $(l^2)^*=l^2$, and $\Big( n^{-1/2} \Big)_{n=1}^\infty\notin l^2$, we should know that $\phi$ is not continuous and its kernel is dense, i.e., not closed.
However it is not clear that $\phi$ is defined and hence not even necessarily a functional.
My question is whether this argument is salvagable, and if not whether someone knows of a solution to this question?

For an explicit example, consider the sequence $a^{(n)}$ of $\ell^2$ defined as follows.


*

*For $k \in \{2, \dots, n-1 \}$, $a^{(n)}_k = \frac{1}{\sqrt{k}\log k}$.

*For $k \in \{n, \dots, r_n\}$, $a^{(n)}_k = - \frac{x_n}{\sqrt{k}\log k}$.

*Otherwise, $a^{(n)}_k = 0$.


We denote here $r_n = \lfloor e^{(\log n)^2} \rfloor$, and
$$
x_n = \frac{\sum_{k=2}^{n-1} \frac{1}{k\log k}}{\sum_{k=n}^{r_n} \frac{1}{k\log k}}.
$$
By comparison with integrals, we see that
$$
\sum_{k=2}^{n-1} \frac{1}{k \log k} = \log \log n + O(1),
$$
and thus a short computation shows that, by the very choice of $(r_n)$, we have $x_n \to 1$. Finally the sequence $(a^{(n)})$ has the following properties.


*

*By construction and definition of $x_n$, $a^{(n)} \in M$.

*If $(b_k)$ is the sequence defined by $b_k = \frac{1}{\sqrt{k}\log k}$ for $k \geq 1$, then
$$
\|a^{(n)}-b\|_{\ell^2} \leq x_n^2 \sum_{k \geq n} \frac{1}{k\log^2 k},
$$
and this tends to $0$ since $(x_n)$ is bounded and the series $\sum \frac{1}{k \log^2 k}$ converges.

*Yet, $(b_k)$ is obviously not in $M$.

