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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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    What is your norm? If it's the sup norm, for instance, then $\{ 1, 1, 1, \dots \}$ certainly doesn't lie in the (closure of the) linear span of the $x_n$, since each $x_n$ has a zero entry and so lies at distance 1 from $\{ 1, 1, 1, \dots \}$.2012-05-16
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    I assume the norm is the weighted Hilbert space norm, i.e., $\|a\|^2=\sum |a_n|^2/(n(n+1))$. Clive's comment still applies: everything in the closed linear span will have the leading zero.2012-05-16
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    @Leonid Kovalev : Your assumption is correct norm is weighted Hilbert space norm.2012-05-17
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    @CliveNewstead Could you consider crafting that comment into an answer, so that this question may disappear from the Unanswered queue?2013-06-06
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    @Lord_Farin: Done2013-06-06

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