Formalizing converging norm

Question

Let $x \in \ell^2$. Define $(u_N)$ as the sequence of vectors $u_N \in U$ such that $u_n = x_n$ if $n \le N$, $0$ otherwise.

Then $\lVert u_N \rVert$ is an increasing function which converges to finite $\lVert x \rVert$.

The statement seems obvious, but how can I make this claim more rigorous? Saying the norm of $u_N$ increases doesn't seem convincing.

I suppose it depends on how norm is defined. If defined as an infinite sum which is defined as the limit of a finite summation, then the result follows by definition ...

Answer 1

$\newcommand{\norm}[1]{\lVert #1 \rVert}$What I wanted to show is actually both $\norm{u_N} \to \norm{x}$ and $\norm{x - u_N} \to 0$.

Using the definition of norm as a finite value such that $$\norm{x}^2 = \sum_{n=1}^\infty x_n^2 = \lim_{N \to \infty} \sum_{n=1}^N x_n^2 = \lim_{N \to \infty} \norm{u_N}^2$$

we see clearly $\lVert u_N \rVert$ is an increasing function and $\norm{u_N} \to \norm{x}$. Therefore $$\norm{x - u_N}^2 = \sum_{n=N+1}^\infty x_n^2 = \sum_{n=1}^\infty x_n^2 - \sum_{n=1}^N x_n^2 = \norm{x}^2 - \norm{u_N}^2 \to 0$$

Answer 2

$$\|x\|^2\geq\sum_{n=1}^{N+1}x_n^2=\|u_{N+1}\|^2=x_1^2+\cdots + x_{N+1}^2\geq x_1^2+\cdots + x_{N}^2=\|u_N\|^2$$

Answer 3

$\sqrt {\sum\limits_{k=1}^{n} |x_k|^{2}}$ increases to $\sqrt {\sum\limits_{k=1}^{\infty} |x_k|^{2}}$. This is just deinition of the infinite sum $\sum\limits_{k=1}^{\infty} |x_k|^{2}$.