Let $p\in[1,\infty)$ Show that a subset $M\subseteq l^{p}(\mathbb{K})$ is totally bounded if and only if M:
$\underset{x\in M}{\text{sup}}$ $\sum\limits_{i=n}^{\infty}|x_{i}|^{p}\rightarrow 0$ for $n\rightarrow\infty$ and $M$ bounded.
Now I managed to show "$\Rightarrow$" (when M is totally bounded) but for the second one I have no idea how to get started. I'm sitting in a introductory class to functional analysis so it would be nice to consider that in your answers.
Let $\epsilon>0$. Since $\sum\limits_{i=n}^{\infty}|x_{i}|^{p}\rightarrow 0$, there exists $N$ such that:
$\sum\limits_{i=N+1}^{\infty}|x_{i}|^{p}< \dfrac{\epsilon}{3}$ for all $x\in M$.
Define the projection $\pi:l^p(\mathbb{K})\rightarrow \mathbb{K}^N$, by $\pi \Big( \big( x_i\big)_{i\in \mathbb{N}} \Big)=(x_1, ...,x_N)$.
Since bounded sets in $\mathbb{K}^N$ with the $l^p$ norm are totally bounded, and $\Vert \pi(x)\Vert_p \leq \Vert x\Vert_p \leq m$ for all $x\in M$, there exists a finite set of points $\{x^j \}_{j=1}^{N_0}$ such that $ \{ \pi(x^j) \}_{j=1}^{N_0}$ are an $\dfrac{\epsilon}{3}$-net for $\pi[M]$.
It remains to show that $\{x^j \}_{j=1}^{N_0}$ is an $\epsilon$-net in $M$. Let $x\in M$, we know that:
$\sum\limits_{i=1}^{N}|x^j_{i}-x_i|^{p} < \dfrac{\epsilon}{3}$ by choice of $\{x^j \}_{j=1}^{N_0}$. Then:
$\Vert x^j-x\Vert_p= \sum\limits_{i=1}^{\infty}|x^j_{i}-x_i|^{p}= \sum\limits_{i=1}^{N}|x^j_{i}-x_i|^{p}+ \sum\limits_{i=N+1}^{\infty}|x^j_{i}-x_i|^{p} \leq$
$\leq \sum\limits_{i=1}^{N}|x^j_{i}-x_i|^{p} + \sum\limits_{i=N+1}^{\infty}|x^j_{i}|^{p}+ \sum\limits_{i=N+1}^{\infty}|x_i|^{p} \overset{x,x^j\in M}{\leq} \dfrac{\epsilon}{3}+ \dfrac{\epsilon}{3}+ \dfrac{\epsilon}{3} $
And for a general $\epsilon$, you've found a finite cover of $M$ by the balls $B_{l^p}\big(x^j, \epsilon \big)$ (radius $\epsilon$).