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## Homework Statement

Let [itex]T: \ell^{2} \rightarrow \ell[/itex] be defined by

[itex]T(x)=x_{1},\frac{1}{2}x_{2},\frac{1}{3}x_{3},\frac{1}{4}x_{4},...}

Show that the range of T is not closed

## The Attempt at a Solution

I figure that I need to find some sequence of [itex]x_{n} \rightarrow x[/itex] such that [itex]T(x_{n})[/itex] is in [itex]\ell^{2}[/itex] but [itex]T(x)[/itex] is not.

I figured that if [itex]x=(\sqrt{1}, \sqrt{2}, \sqrt{3},...)[/itex] then [itex]T(x)[/itex] is the harmonic series, which is not square summable. The problem is, I can't think of any square summable [itex]x_{n}[/itex] which converge to that [itex]x[/itex].

I have a feeling that I'm taking the wrong approach to this. The thing is, I think the operator [itex]T(x)[/itex] must be bounded since [itex]k_{i}^{2} \geq \frac{1}{i^{2}}k_{i}^2[/itex]. Maybe I should be looking for some [itex]x_{n} \rightarrow x[/itex] such that [itex]T(x_{n})[/itex] converges to something other than [itex]T(x)[/itex].