Stuck while trying to show $(\mathbb{R}^d, || \cdot ||_p)$ is complete

Question

I want to show that $(\mathbb{R}^d, || \cdot ||_p)$ is complete where

$$|| \cdot ||_p = \bigg(\sum_{j=1}^\infty |x_j|^p \bigg)^{\frac{1}{p}}.$$

Let $(x^n)$ be a Cauchy sequence in $\mathbb{R}^d$. Then for all $\epsilon > 0, \exists N \in \mathbb{N}$ such that for $n \ge N$

$$||x^n - x^m||_p^p = \sum_{j=1}^\infty |x_j^n - x_j^m|^p < \epsilon.$$

In particular we have $|x_j^n - x_j^m|^p < \epsilon$ for fixed $j$. So $x_j^n$ is a Cauchy sequence over the real numbers and by the completeness of $\mathbb{R}$ it converges to some $x_j \in \mathbb{R}$.

So we have that for all $\epsilon > 0, \exists N \in \mathbb{N}$ such that for $n \ge N$, $|x_j^n - x_j| < \epsilon$. Consider $x = (x_1, x_2, \dots, x_n) \in \mathbb{R}^d.$ I claim that $x^n$ converges to this $x$.

But I don't know where to go from here, how do I use this to show that $x^n$ converges to $x$ using the precise definition of convergence? That is, how do I show that for all $\epsilon > 0, \exists N \in \mathbb{N}$ such that for $n \ge N$, $||x^n - x|| < \epsilon$?

Answer 1

$$\| x^n -x\|_p^p = \sum_j |x^n_j - x_j|^p \le \sum_j \varepsilon^p = d \varepsilon^p$$ so that $\| x^n -x\|_p \le d^{1/p} \varepsilon$. Since $d^{1/p}$ is a constant independent by $n$, the norm $\| x^n -x\|_p$ gets arbitrarily smaller.