Prove that $\ell^2(\mathbb R)$ is complete Let $$X:=\ell^2(\mathbb R)=\left\{x=(x_n) _ {n\in\mathbb N}\in \mathbb R^{\mathbb N}\mid\sum_{n=1}^\infty x_n^2<\infty \right\}$$
and $$\|x\| _ X=\left(\sum_{k=1}^\infty x_k^2\right)^\frac{1}{2}.$$
$(X,\|\cdot \|_X)$ is a normed space. Prove that it's complete.
Let $\varepsilon >0$ and $(x^n)_{n\in\mathbb N}\in X^{\mathbb N}$ a Cauchy sequence where $x^n:=(x^{n} _k) _ {k\in\mathbb N}$.
Since $(x^n)$ is a Cauchy sequence, then so is $(x_k^n)_{n\in\mathbb N}$ for all $k\in\mathbb N$. In particular, there is $x_k\in \mathbb R$ s.t. $x_k^n\to x_k$ whenever $n\to \infty $.
Since for all $n\in\mathbb N$, $$\sum_{k=N}^\infty (x^n_k-x_k)^2\underset{N\to \infty }{\longrightarrow }0,$$
if $n\in\mathbb N$ is fixed, there is $N_n\in\mathbb N$, s.t. $$\sum_{k=N_n+1}^\infty (x_k^n-x_k)^2<\frac{\varepsilon }{2}.$$
We have that \begin{align*}
\|x^n-x\| _ X^2&=\sum_{k=1}^{N_n}(x_k^n-x)^2+\sum_{k=N_n+1}^\infty (x_k^n-x_k)^2\\
&\leq \sum_{k=1}^{N_n}(x_k^n-x)^2+\frac{\varepsilon }{2}.
\end{align*}
Question Since $N_n$ depend on $n$, I can't find $M\in\mathbb N$ s.t. $|x^n_k-x_k|<\frac{\varepsilon }{2N_n}$ for all $k=1,...,N_n$. Can someone give a hint on how to solve this problem ?
 A: Fix $\varepsilon > 0$. Let $N = N(\varepsilon)$ be such that
$$n, m \ge N \implies \sum_{k=1}^\infty (x^n_k - x^m_k)^2 < \frac{\varepsilon^2}{4}.$$
This implies that
$$n, m \ge N \implies (x^n_k - x^m_k)^2 < \frac{\varepsilon^2}{4},$$
for any given $k$. This proves $(x^n_k)_n$ is Cauchy, and hence convergent, as you have already detailed.
I claim that $\sum_{k=1}^\infty (x_k^n - x_k)^2$ exists, and is less than or equal to $\frac{\varepsilon^2}{4}$. Note that it simply suffices to show that the partial sums are bounded above by $\frac{\varepsilon^2}{4}$; the fact that each term is non-negative means that the series must converge, by the monotone convergence theorem.
For any $M \in \Bbb{N}$, and $n \ge N$, we have,
$$\sum_{k=1}^M (x_k^n - x_k)^2 = \sum_{k=1}^M \lim_{m \to \infty} (x^n_k - x^m_k)^2 = \lim_{m \to \infty} \sum_{k=1}^M (x^n_k - x^m_k)^2,$$
where the latter equality is the algebra of limits on a (finite) sum. But then,
$$m \ge N \implies \sum_{k=1}^M (x^n_k - x^m_k)^2 \le \sum_{k=1}^\infty (x^n_k - x^m_k)^2 < \frac{\varepsilon^2}{4},$$
hence when taking the limit as $m \to \infty$,
$$\sum_{k=1}^M (x_k^n - x_k)^2 \le \frac{\varepsilon^2}{4},$$
for all $M \in \Bbb{N}$. Then, taking the limit as $M \to \infty$,
$$\sum_{k=1}^\infty (x_k^n - x_k)^2 \le \frac{\varepsilon^2}{4}.$$
Therefore,
$$n \ge N \implies \|x^n - x\| \le \frac{\varepsilon}{2} < \varepsilon.$$
