No closed set in $\ell^2$

Question

Let $$V=\left\{x=\{x_k\}\in \ell^2\; |\; x_k=0\quad\forall k> k_0, k_0\in\mathbb{N}\right\}.$$ I have to prove that $V$ is not closed.

Let $\left\{ x^{(n)} \right\}$ be a sequence in $V$ defined as follows:

$$x^{(1)}=\left \{ x_k^{(1)}\right\}_k=\left\{1,0,\dots\right\}$$ $$x^{(2)}=\left \{ x_k^{(2)}\right\}_k=\left\{1,\frac{1}{2},0,\dots\right\}$$ $$\vdots$$ $$x^{(n)}=\left \{ x_k^{(n)}\right\}_k=\left\{1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},0,\dots\right\}$$

$$\vdots$$

We set $x:=\left\{\frac{1}{n}\right\}_n$, I must prove that $$x^{(n)}\to x\quad\text{in}\;\ell^2\;\text{for}\; n\to \infty.$$

We observe that fixed $k\in \mathbb{N}$ the sewuence $\left\{ x_k^{(n)}\right\}_n$ converges to $\frac{1}{k}$ when $n\to \infty$ and then $$\left\lvert x_k^{(n)}-\frac{1}{k}\right\rvert<\varepsilon\quad \forall n>n_0.$$

Question How can I continue from here?

Answer 1

The norm in $\ell^2$ is

$$\lVert y \rVert_2 = \left(\sum_{k=1}^\infty y_n^2\right)^{\frac{1}{2}}$$ for $y = \{y_n\}\in \ell^2$.

Hence $$\lVert x - x^{(n)} \rVert_2^2 = \sum_{k=n+1}^\infty \frac{1}{k^2}$$ converges to zero with $n \to \infty$ as the series $\sum_{k=0}^\infty \frac{1}{k^2}$ is convergent. This gives the desired result.