Non-standard basis for $\ell^2$

Question

We have $\{e_1, e_2,...,e_n,...\}$ the standard orthonormal basis on $\ell^2$ , the space of square summable sequences. We construct the following set $$u_1 = \frac {e_0 -e_1}{\sqrt{2}} ; \ \ \ \ u_2= \frac {e_0 +e_1 - 2e_2}{\sqrt{6}} \ \ ; \ \ \dots \ \ u_n = \frac {e_0 +e_1 + e_2 + ... +e_{n-1} -ne_n}{\sqrt{n(n+1)}} $$ Showing that these vectors form an orthonormal basis is quite straight forward, however, I'm having trouble figuring out exactly what their linear span is, my notes say that the following set $$\mathcal S = \Bigl\{ x\in \ell^2 , \sum_{k=0}^\infty x_k = 0 \Bigr \}$$ Is dense in $\ell^2$. I cannot see why $\mathcal S$ is the linear span of this basis, because if we only allow finite linear combinations of $u_i$, then we would have a sequence with only finite non-zero entries. Are infinite linear combinations allowed in this case? Is density shown differently? Thanks.

Answer 1

Your definition of $S$ requires a bit of elucidation, because in general for $x \in \ell^2$, $\sum_{k=0}^\infty x_k$ might not converge. In fact, the set of $x$ with all but finitely many elements $0$ and $\sum_{k=0}^\infty x_k = 0$ is dense. I believe you'll find that any such $x$ is a (finite) linear combination of your basis elements.