I'm trying to prove this theorem, which I encounter at this Wikipedia's page.
Theorem: For $1 \le p \in \mathbb R$, we define the normed vector space $\langle \ell^p, \|\cdot\|_p \rangle$ by $$\ell^p = \left \{x \in {\mathbb K}^{\mathbb N} \,\middle\vert\, \sum |x_k|^p < \infty \right\}, \quad \|x\|_p = \left(\sum |x_k|^{p}\right)^{1 / p}, \quad x \in \ell^p$$
Then $\langle \ell^p, \|\cdot\|_p \rangle$ is a Banach space.
Could you please verify if my attempt is fine or it contains logical gaps/mistakes? I'm grateful for your help!
My attempt:
$\ell^p$ is clearly closed under scalar multiplication. Let $(x_k)_{k \in \mathbb N}, (y_k)_{k \in \mathbb N} \in \ell^p$. Then $(x_k)_{k \in \mathbb N}+ (y_k)_{k \in \mathbb N} = (x_k+y_k)_{k \in \mathbb N}$.
Lemma: For $1 \le p \in \mathbb R$, there exists $M \in \mathbb R$ such that$$\forall a,b \ge 0:(a+b)^p \le M(a^p +b^p)$$
By Lemma, we have $|x_k+y_k|^p \le (|x_k| + |y_k|)^p \le M(|x_k|^p + |y_k|^p)$. On the other hand, $ \sum |x_k|^p < \infty$ and $\sum |y_k|^p < \infty$. As such, $ \sum |x_k+y_k|^p < \infty$ and thus $(x_k+y_k) \in \ell^p$. Hence $\ell^p$ is closed under addition.
Let $(u_n)_{n \in \mathbb N}$ be a Cauchy sequence in $\ell^p$. Each $u_n \in \ell^p$ has the form $(u^k_n)_{k \in \mathbb N}$ where $u^k_n \in \mathbb K$ for all $k,n \in \mathbb N$.
There exists $N \in \mathbb N$ such that $$\| u_n - u_m\|_p < \varepsilon, \quad m,n \ge N$$ or equivalently $$\left(\sum_{k=0}^\infty \left |u^k_n -u^k_m \right|^{p}\right)^{1 / p} < \varepsilon, \quad m,n \ge N \tag{1}$$
Consequently, $$\left |u^k_n -u^k_m \right| < \varepsilon, \quad k\in\mathbb N, \quad m,n \ge N$$
As such, $(u^k_n)_{n \in \mathbb N}$ is a Cauchy sequence in $\mathbb K$ for each $k \in \mathbb N$. Since $\mathbb K$ is a Banach space, $(u^k_n)_{n \in \mathbb N}$ is convergent for each $k \in \mathbb N$. We define $v \in {\mathbb K}^{\mathbb N}$ by $$v_k = \lim_{n \to \infty} u^k_n, \quad k \in \mathbb N$$
Next we prove that $v \in \ell^p$. Take the limit $m \to \infty$ in $(1)$, we get $$\left(\sum_{k=0}^\infty \left |u^k_n - v_k \right|^{p}\right)^{1 / p} \le \varepsilon, \quad n \ge N \tag{2}$$ and consequently $$\sum_{k=0}^\infty \left |u^k_N - v_k \right|^{p} \le \varepsilon^p$$
We have $$\begin{aligned} \sum |v_k|^p &= \sum_{k=0}^\infty \left |v_k \right|^{p} &&= \sum_{k=0}^\infty \left |(v_k - u^k_N) +u^k_N \right|^{p}\\ &\le \sum_{k=0}^\infty \left( \left |v_k - u^k_N\right| +\left|u^k_N \right| \right)^{p} &&\le \sum_{k=0}^\infty M \left( \left |v_k - u^k_N \right|^p + \left|u^k_N \right|^{p}\right)\\& = M \sum_{k=0}^\infty \left( \left |v_k - u^k_N \right|^p + \left|u^k_N \right|^{p}\right) \end{aligned}$$
It follows from $u_N \in \ell^p$ that $\sum_{k=0}^\infty \left|u^k_N \right|^{p} < \infty$. We already show that $\sum_{k=0}^\infty \left |u^k_N - v_k \right|^{p} \le \varepsilon^p$. As such, $$M \sum_{k=0}^\infty \left( \left |v_k - u^k_N \right|^p + \left|u^k_N \right|^{p}\right) < \infty$$ and consequently, $\sum |v_k|^p < \infty$. Hence $v \in \ell^p$. It follows directly from $(2)$ that $\| u_n -v \|_p \le \varepsilon$ for all $n \ge N$. Consequently $$\lim_{n \to \infty} u_n = v$$
This completes the proof.