Finding a "backward orthogonalization" in $\ell_2$ 
Let $\mathbf{H}$ be a Hilbert space, and $\left (x_n\right )_{n=1}^\infty$ a set in $\mathbf{H}$. We say that an orthogonal set $\left (u_n\right )_{n=1}^\infty$ is a backward orthogonalization of $\left (x_n\right )_{n=1}^\infty$ if $\overline{\mathrm{Sp}}\left \{ x_n, x_{n+1}, \ldots \right \} = \overline{\mathrm{Sp}}\left \{ u_n, u_{n+1}, \ldots \right \}$ for all $n$.
Let $x_1 = e_1,\space x_2=e_1+2e_2,$ and for all $k\geq3, \space x_k=e_k+e_{k+1}$, where $\left (e_k\right )_{k=1}^\infty$ is the standard basis of $\ell_2$. Find a backward ortogonalization of $\left (x_k\right )_{k=1}^\infty$

I'm stuck on this question for an entire day. I couldn't find any resources online about this "backward orthogonalization".
My first attempt was using $\left (u_n\right )_{n=1}^\infty=\{x_1, u_2, e_3, e_4, e_5, \ldots \}$ where $u_2$ was some orthogonal vector to $x_1$ which is still in $\mathrm{Sp}\{x_1, x_2\}$.  I then said: for all $k\geq3$:
$$ e_k = \lim_{n\to \infty} (x_k-x_{k+1}+x_{k+2}-\cdots \pm x_{k+n})$$
but I later realized this is incorrect, as the expression inside the limit is always equal to $e_k\pm e_{k+n+1}$ which does not converge to $e_k$.
How can I find such a backward orthogonalization?
 A: Your attempt works. The vector $e_k$ is in the closure of the linear span of the set $S = \{x_{k+j} : j\ge 0\}$. This means, by definition, that $e_k$ can be approximated by linear combinations of vectors in $S$. Notably, this does not mean that one can write $e_k$ as some infinite sum of scalar multiples of vectors in $S$. The latter is a much stronger property. (See: Schauder basis versus a set with a dense span.) 
To prove the claim, pick $\alpha\in (-1, 1)$ and $n\in \mathbb N$ and consider the vector
$$
\sum_{j=0}^n \alpha^j x_{k+j} = e_k + (1+\alpha)e_{k+1} + \alpha (1+\alpha)e_{k+2} + \dots + \alpha^{n-1} (1+\alpha)e_{k+n} + \alpha^n x_{k+n+1}
$$
Subtract $e_k$ from the above, and compute the squared norm of this difference:
$$
(1+\alpha)^2 +  \alpha^2 (1+\alpha)^2 + \dots + \alpha^{2n-2} (1+\alpha)^2 + \alpha^{2n}
$$ 
As $n\to \infty$, this sum converges to 
$$
(1+\alpha)^2 \sum_{j=0}^\infty \alpha^{2j} = \frac{(1+\alpha)^2}{1-\alpha^2} = \frac{1+\alpha}{1-\alpha}
$$
and the latter can be made arbitrarily small by picking $\alpha$ close to $-1$. 
