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Consider the following sequence of sequences :

$x_{0}:=\{0,0,0,0......\}$,$x_{1}:=\{0,\frac{1}{2},0,\frac{1}{2}.......\}$ (i.e $0,\frac{1}{2}$ repeated infinitely), $x_{2}:=\{0,\frac{1}{3},\frac{2}{3},0,\frac{1}{3},\frac{2}{3}.0.......\}$ (i.e $0,\frac{1}{3},\frac{2}{3}$ repeated infinitely),........,$x_{n}:=\{0,\frac{1}{n+1},\frac{2}{n+1},\frac{3}{n+1},...\frac{n}{n+1},0,.......\}$ (i.e $0,\frac{1}{n+1},\frac{2}{n+1},\frac{3}{n+1},...\frac{n}{n+1}$ repeated infinitely)

Define $\mathcal H:=$ Set of all complex valued sequences say $\{a_{n}\}$ such that $\sum_{n=1}^{\infty}\frac{|a_{n}|^{2}}{n(n+1)} < \infty$.

Is it true $\{1,1,1,...\}$ (i.e all 1's) belongs to closed linear span of $\{x_{n}\}\subset \mathcal H$

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    @Lord_Farin: Done2013-06-06

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This very belated answer is posted as per Lord_Farin's request - see the comments on the question.

  • If you're using the sup norm then $(1, 1, \dots, )$ doesn't belong to the closed linear span because all the $x_n$s have a $0$ in them and hence $\left\|x_n-(1,1,\dots)\right\| \ge 1$ (in fact, $=1$); and...

  • Even if you're using the Hilbert norm, since the first term of all the $x_n$s is $0$, the distance is still $\ge 1$.