Show that the sequence $\{x_n\}_{n=1}^\infty$ does not converge in $ \ell^2 (\mathbb{N}). $

Question

For each $n \in \mathbb{N}$, define a sequence $x_n=\{ x_n(k) \}_{k=1}^\infty$ by $$ x_n(k) = \begin{cases} \frac{1}{n} \;\;\; \text{ if $1\leq k \leq n^2$}\\ 0 \;\;\; \text{ otherwise.} \end{cases} $$ Show that the sequence $\{x_n\}_{n=1}^\infty$ does not converge in $ \ell^2 (\mathbb{N}). $

Note that $\{x_n (k) \}_{k=1}^\infty \rightarrow 0$ as $n \rightarrow \infty$.

Thus, $\|x_n - \{0\}\|_{2}^{2} = \|\{x_n (k) \}_{k=1}^\infty \|_{2}^{2} $

Im not sure what to do next. Please help.

Answer 1

Note that $$||x_n||_2^2 = \sum_{k=1}^\infty x_n(k)^2 = \sum_{k=1}^{n^2} \frac{1}{n^2} = 1,$$ hence $$||x_n-0||_2 = 1 \not\to 0\quad(n\to\infty).$$

Answer 2

HINT.-Taking the norm of each $x_n$ we have $$\|x_n||=\sqrt{\left(\frac1n\right)^2+\left(\frac1n\right)^2+\cdots+\left(\frac1n\right)^2}=\sqrt\frac{n^2}{n^2}=1$$ So what?

Answer 3

Suppose $x_n \to x$ in $l^2.$ Then for every $k,$ $x_n(k)\to x(k).$ But for any $k, x_n(k)\to 0.$ Thus $x$ is the zero element of $l^2.$ Therefore $\|x_n-x\|_2 = \|x_n\|_2 \to 0.$ But as others have pointed out, $\|x_n\|_2 = 1$ for every $n.$ This is a contradiction, hence $x_n$ does not converge in $l^2.$