Convergences in $l^{2}$

Question

Show that the sequence $\{x_{n}\}_{n\geq 1}$, where $x_{n}=\left(1,\dfrac{1}{2},...,\dfrac{1}{n},0,...\right)$ converges to $x=\left(1,\dfrac{1}{2},\dfrac{1}{3},...,\dfrac{1}{n},...\right)$ in $l^{2}$.

My approach. Note that, $d(x_{n},x)$, considerer $l^{2}$ with the norm $\vert\vert x-y\vert\vert=\sqrt{\sum_{i}{(x_{i}-y_{i})^{2}}}$, then $$d(x_{n},x)=\sqrt{\sum_{k=n+1}^{\infty}{\dfrac{1}{k^{2}}}}\to 0$$ when $n\to\infty$.It this sufficiently to show that $x_{n}$ converges to $x$ in $l^{2}$. Regards

The last line is right. The first line, where you say that $d(x_n, x)$ equals a sequence, doesn't make sense. And the fact that $l^2$ is complete is irrelevant here. — Nate Eldredge, Oct 20 '15 at 04:31
The distance between two elements is a number. You have written that a number equals a sequence. That does not make sense. — Nate Eldredge, Oct 20 '15 at 04:41

score 0 · Answer 1 · answered Oct 20 '15 at 13:31

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Yes, it is enough. To prove that $x_n \to x$ in ($l_p, ||\cdot ||_p$) the only thing that you have to prove is that $||x-x_n||_p \to 0$ and that $x \in (l_p, || \cdot ||_p$).

answered Oct 20 '15 at 13:31

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Convergences in $l^{2}$

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