I've been looking through proofs of $l^p$ spaces being complete. And in order to prove that the elementwise-limit sequence also belons to $l^p$, a partial sum is used. (See for example the answer here )
However, I do not understand why that is needed. Why can't one do the same directly with infinite sums? Basically, what I want to know is why is the following not okay:
Let $\{x^{(n)}\}_n$ be a Cauchy sequence from $l^p$, and $\epsilon >0$. And let $x^{\infty}$ be an elementwise limit (that is $x_k^{(n)} \to x_k^{\infty}$ as $n \to \infty$ for every k).
Since $\{x^{(n)}\}_n$ is Cauchy, we know that $\sum_k |x_k^n - x_k^m|^p < \epsilon^p$ for all $n,m\geq N(\epsilon)$. Therefore $\lim_{n\to\infty}\sum_k |x_k^n - x_k^m|^p = \sum_k |x_k^\infty - x_k^m|^p < \epsilon^p$.
I don't understand why this would not be okay - the only reason I could think of is the existence of the limit. But we know that that limit exists, since we know about the elementwise convergence....And it's kind of the same argumentation I guess as in the case of taking this same limit on the finite sum anyway... Right?
In general a limit and an infinite sum cannot be interchanged. So you have to say $\sum\limits_{k=1}^{N} |x_k^{n}-x_k^{m}| <\epsilon^{p}$ for each $N$ for $n,m \geq N(\epsilon)$ and take limit as $n \to \infty$ to get $\sum\limits_{k=1}^{N} |x_k-x_k^{m}| \geq \epsilon^{p}$ for each $N$ for $m \geq N(\epsilon)$. [Note that $<$ changes to $\leq $ when you take limits]. Now let $N \to \infty$ to finish the proof.