I would like to know if my proof is correct about the fact that the space $\ell^{2}$ is a Banach space with the usual inner product !
Here is my attempt :
We consider a Cauchy sequence $\{x^{k}\}_{k\in\mathbb{N}}$ in $\ell^{2}$. This means that
$$ \forall\varepsilon>0, \exists N\in\mathbb{N} : n,m>N\implies \lVert x^{n} - x^{m}\rVert = \left(\sum_{i\geq 0} \lvert x_i^{n} - x_{i}^{m}\rvert^{2}\right)^{\frac{1}{2}} < \varepsilon $$
where the last inequality is equivalent to
$$ \lVert x^{n} - x^{m}\rVert^{2} = \sum_{i\geq 0} \lvert x_i^{n} - x_{i}^{m}\rvert^{2} <\varepsilon^2 $$
This implies that for all $i\in\mathbb{N}$ we have
$$ \forall n,m>N : \lvert x_{i}^{n} - x_{i}^{m}\rvert < \varepsilon $$
Which represents the fact that the sequence $(x_{i}^{m})_{m\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$. Since the latter is a Banach space for the absolute value, this implies that there exists $x_{i}\in\mathbb{R}$ which is the limit of the sequence $(x_{i}^{m})_{m\in\mathbb{N}}$, note that this holds for all $i\in\mathbb{N}$.
Now we notice that from the Cauchy property in $\ell^{2}$ we have for $\varepsilon>0$ and $n,m>N_{\varepsilon}$
$$ \forall p\in\mathbb{N} : \sum_{i=0}^{p}\lvert x_i^{n} - x_i^{m}\rvert^{2} < \varepsilon^2 $$
Moreover, since for all $i\in\mathbb{N}$ the sequence $(x_{i}^{m})_{m\in\mathbb{N}}$ has a limit $x_{i}$ in $\mathbb{R}$ we can find $N_{1}$ such that
$$ k>N_1\implies \lvert x_{i}^{m} - x_i\rvert < \eta_1 $$
In the same line, by what has been said earlier (the fact we have a Cauchy sequence), we can find $N_2$ such that
$$ n,m>N_2\implies \lvert x_{i}^{n} - x_{i}^{m}\rvert < \eta_2 $$
Now consider $\varepsilon>0$. We notice that for all $p\in\mathbb{N}$, taking $\eta = \eta_1 = \eta_2 =\left(\frac{\varepsilon^2}{2(p+2)}\right)^{1/2}$, and $n,m>N = max(N_1, N_2)$ we have
$$ \sum_{i=0}^{p}\lvert x_i^{n} - x_i\rvert^{2} \leq \sum_{i=0}^{p}\lvert x_i^{n} - x_i^{m}\rvert^{2} + 2\sum_{i=0}^{p}\lvert x_i^{n} - x_i^{m}\rvert\lvert x_i^{m} - x_i\rvert + \sum_{i=0}^{p} \lvert x_i^{m} - x_i\rvert^{2}\leq 2(p+2)\eta^2 = \varepsilon^2 $$
Since this is true for all $p\in\mathbb{N}$ we have
$$ \sum_{i\in\mathbb{N}}\lvert x_i^{n} - x_i\rvert^{2} < \varepsilon^2 $$
This shows that
$$ \forall\varepsilon>0, \exists N\in\mathbb{N} : n>N\implies \lVert x^{n} - x\rVert < \varepsilon $$
where $x$ is the sequence where each terms $i$ of the sequence corresponds to $x_i$. The last line shows that the sequence $y = x - x^{n}$ is in $\ell^2$. Using the vector space structure of $\ell^2$ and the fact that $x = y + x^{n}$ we see that $x\in \ell^{2}$ as the sum of two elements of $\ell^2$.
I would like to know if the proof is correct, if all my argument are clearly justified.. if not please don't hesitate to add a comment or an answer.
Thank you a lot !