Let $1\leq p < \infty$ and $A \subseteq \ell^{p}$. Show that:
$A$ relatively compact $\iff$ $A$ is bounded and $\lim\limits_{n \to \infty} \sup\limits_{x \in A}(\sum\limits_{i=n}^{\infty}\vert x_{i}\vert^{p})^{\frac{1}{p}}=0$
"$\Rightarrow$" Assume $A$ is relatively compact, then $\overline{A}$ is compact and hence totally bounded, the fact that $A \subseteq \overline{A}$ immediately shows that $A$ is also totally bounded and therefore bounded.
And I think I need to solve $\lim\limits_{n \to \infty} \sup\limits_{x \in A}(\sum\limits_{i=n}^{\infty}\vert x_{i}\vert^{p})^{\frac{1}{p}}=0$ via contradiction. So, assume there exists $c>0$ so that $\lim\limits_{n \to \infty} \sup\limits_{x \in A}(\sum\limits_{i=n}^{\infty}\vert x_{i}\vert^{p})^{\frac{1}{p}}=c$ then we can find a sequence $(x^{m})_{m} \subseteq A$ so that $\lim\limits_{m \to \infty}\lim\limits_{n \to \infty}(\sum\limits_{i=n}^{\infty}\vert x_{i}^{m}\vert^{p})^{\frac{1}{p}}$ but his just makes it overly complicated. What am I missing?
I would prove the last part directly.
Let $\epsilon>0$.
Use the fact that $A$ is totally bounded to get a finite set of sequences $\{x^k\}$ such that $A\subset \bigcup_k B(x^k,\epsilon)$.
Define the tail norm as $\|\cdot \|_{p,n}$.
Then for each $k$, find $N_k$ such that $\|x^k\|_{p,N_k}<\epsilon$.
Let $N$ be the max of the $N_k$.
For every $x\in A$, let $k_x$ denote the index of the sequence $x^{k_x}$ that is closest to $x$ in the $p$-norm.
We then have $\|x-x^{k_x}\|_{p,N} \leq \|x-x^{k_x}\|_{p} < \epsilon$, so
\begin{align} \sup_{x\in A} \|x\|_{p,N} & \leq \sup_{x\in A} \|x-x^{k_x}\|_{p,N} + \|x^{k_x}\|_{p,N} \\ & \leq 2\epsilon \end{align}