How do I prove the completeness of $\ell^p$?
Say $\{x_n\}$ is Cauchy in $\ell^p$ and $x$ is its pointwise limit. To argue that $x \in \ell^p$ would the following be correct:
Let $\varepsilon > 0$ and let $N$ be s.t. $n,m > N$ $\Rightarrow$ $|x_n - x_m|_p < \varepsilon$. Then $\lim_{m \to \infty} |x_n - x_m|_p = |x-x_n|_p \le \varepsilon$.
I saw the following different argument: $|x_n - x_m|_p < \varepsilon$ implies $\left(\sum_{k=0}^M |(x_n - x_m)_k|^p\right)^{1/p} < \varepsilon$ for all $M$ therefore $\lim_{m \to \infty}\left(\sum_{k=0}^M |(x_n - x_m)_k|^p\right)^{1/p} = \left(\sum_{k=0}^M |(x_n - x)_k|^p\right)^{1/p} \le \varepsilon$ for all $M$ therefore $\lim_{M \to \infty} \left(\sum_{k=0}^M |(x_n - x)_k|^p\right)^{1/p} \le \varepsilon$.
The difference is to use a finite sum step in between. Is it correct to drop it? And if not: why not? Norm seems to be continuous so one should be able to exchange norm and limit. Thanks.
Edit: The second argument is correct, not the first one. But that's not the only thing you have to prove. There are three steps. See the usual elementary proof below.
Now if you really want to swap some limits, here is what you can do. Fix $\epsilon>0$ and take $N$ such that $\|x_n-x_m\|_p\leq \epsilon$ for all $n,m\geq N$. Now fix $n\geq N$. And consider the sequence $y_m=x_n-x_m$ which converges pointwise to $x_n-x$, so that $\liminf y_m(k)=x_n(k)-x(k)$ for all $k$. We have $$ \|y_m\|_p\leq \epsilon\qquad\forall m\geq N \qquad\Rightarrow\qquad \liminf_m\|y_m\|_p\leq \epsilon. $$ Now by Fatou's lemma (which is easily proved in $\ell^p$), you have $$ \|x_n-x\|_p=\|\liminf_m y_m\|_p\leq \liminf_m \|y_m\|_p\leq \epsilon $$ for all $n\geq N$. So this proves everything at the same time.
Usual elementary proof:
Find a pointwise limit $x$ by pointwise completeness. You've done that already.
Prove $x$ belongs to $\ell^p$. You've done that also.
Check that $x_n$ tends to $x$ for the $\ell^p$ norm.
ALthough the technique is essentially the same, one needs to treat 2 and 3 separately.
About 2. There is actually an $\epsilon$ free argument at this point, see the note at the end. Now to prove this the way you did, one usually picks $\epsilon=1$. Then there is $N$ such that for all $n,m\geq N$, $\|x_n-x_m\|_p\leq 1$. In particular, for all $K$ and all $m\geq N$: $$ \left(\sum_{k=1}^K|x_m(k)|^p\right)^{1/p}\leq \left(\sum_{k=1}^K|x_N(k)|^p\right)^{1/p}+\left(\sum_{k=1}^K|x_m(k)-x_N(k)|^p\right)^{1/p} $$ $$ \leq \|x_N\|_p+\|x_m-x_N\|_p\leq \|x_N\|_p+1. $$ Now let $m$ tend to $+\infty$. And finally $K$ to $+\infty$. You get $$ \|x\|_p\leq \|x_N\|_p+1 $$ so $x\in\ell^p$.
For 3., now. Fix $\epsilon>0$. Take $N$ such that $\|x_n-x_m\|_p\leq \epsilon$ for all $n,m\geq N$. Then for $n,m\geq N$ and all $K$: $$ \left(\sum_{k=1}^K|x_m(k)-x_n(k)|^p\right)^{1/p}\leq \|x_m-x_n\|_p\leq \epsilon. $$ Let $m$ tends to $+\infty$. Then $K$. You get $\|x-x_n\|_p\leq \epsilon$ for all $n\geq N$.
Note: Proof of 2 without $\epsilon$. Since $x_n$ is Cauchy, it is bounded, say by $M$. Then for all $K$ and all $n$: $$ \left(\sum_{k=1}^K|x_n(k)|^p\right)^{1/p}\leq \|x_n\|_p\leq M. $$ Now let $n$ tend to $=\infty$. And then $K$. You get $\|x\|_p\leq M$. So $x\in\ell^p$.
It's not correct to drop the finite sum step. Your equation $$\lim_{m\to\infty} |x_n-x_m|_p = |x-x_n|_p$$ assumes already that $x_m\to x$ in $\ell^p$. The finite sum step is a trick to allow you to take a limit. (It's always OK to take a termwise limit of a finite sum, but it's not always OK for an infinite sum.)