Compactness of Subset of $\ell^1$. Let $X$ be a finite-dimensional normed space and let $\ell^1(X)$ denote the Banach space of all sequences in $X$ for which $\sum_{n=1}^{\infty} \|x_n\|_X<\infty$.  For $s\geq 0$, is the set
$
C = \left\{
(x_n)\in \ell^1(X):\, \sum_{n=1}^{\infty} n\|x_n\|\leq s
\right\}
$
compact?
What I've tried:
I tried using the Grothendieck compactness principle to show that $C$ is contained in the closed convex-hull of the sequence $(\frac{s}{n})$ but I'm not sure this is the case..
 A: Let $(x_n^{(i)})$ be  as sequence in $C$. For each $n$ we have $n \|x_n^{(i)}\|^{2} \leq s$. So $(x_n^{(i)})_i$ is bounded. Since $X $ is finite dimensional any bounded sequence in it has a convergent subsequence. By cantor's diagonal procedure we can find $i_1,i_2<...$ such that $\lim_{k \to \infty} x_n^{(i_k)} =x_n$ (say) exists for each $n$.Now $\sum _n n\|x_n^{(i_j)})\|^{2} \leq s$ for each $j$ implies that $\sum n\|x_n\|^{2} \leq s$ by Fatous Lemma.  Hence, every sequence in $C$ has a convergent subsequence.
I leave it to you to check that $\|x^{(i)}-x\|_{\ell^{1}(X)} \to 0$.
[Use the fact that $ \sum\limits_{k=N}^{\infty} \|x_n\| \leq\frac s N$ for all $(x_n) \in C$].
A: I think it's important to understand the motivation behind why this result holds, so I'll be a little verbose with my answer, perhaps excessively so, but bear with me.
Since $\ell^1 (X)$ is a Banach space, by the Heine-Borel theorem we want to show $C$ is totally bounded and closed. Closed is obvious, since it is the pre-image under a continuous map of a closed set,
$$ C = \bigcap_{N \geq 1} C_N, \qquad C_N = \left\{ x \in \ell^1 : \sum_{n = 1}^N n |x_n| \leq s \right\}. $$
The idea behind showing $C$ is totally bounded: the factor of $n$ gives a quantitative decay condition on the tail of the $\ell^1$ norm of points in $C$, so we'd like to control the tail uniformly, and reduce to proving total boundedness on $X^M$ for some positive integer $M \in \mathbb N$, which is a nice finite dimensional Banach space.
To get a good intuition, think about the case $X = \mathbb R$. Compactness is easy to prove in $\mathbb R^M$ (closed and bounded), and $\mathbb R^M$ is embedded in $\ell^1 (\mathbb R)$ by just adding zeros or whatever. We are moving in the opposite direction, noting the tail contribution is small and projecting $C \subseteq \ell^1 (\mathbb R)$ down to $\mathbb R^M$. To put it as a moral, infinite dimensions bad (in particular, totally bounded $\neq$ bounded), but finite dimensions good (total bounded $=$ bounded)!
Controlling the tail uniformly: Let $\epsilon > 0$ and, without loss of generality, suppose $s = 1$. Then we can find $N_\epsilon \in \mathbb N$ such that $N_\epsilon \geq 2/\epsilon$. It follows that for any $x \in C$,
$$ \sum_{n > N_\epsilon} |x_n| \leq  \frac{1}{N_\epsilon}\sum_{n > N_\epsilon} n|x_n|  \leq \frac{1}{N_\epsilon} < \frac{\epsilon}{2}. $$
Reduce to finite dimensional case: We now project $C$ down to these first $N_\epsilon$ terms, namely
$$C^{N_\epsilon} = \{ (x_1, \dots, x_{N_\epsilon}) \in X^{N_\epsilon} : x \in C\}. $$
$C^{N_\epsilon} \subseteq X^{N_\epsilon}$ is bounded by definition of $C$ and, since $X^{N_\epsilon}$ is a finite dimensional vector space, therefore totally bounded, so there exist $k_\epsilon \in \mathbb N$ and $\widehat{x^{(1)}}, \dots, \widehat{x^{(k_\epsilon)}} \in X^{N_\epsilon}$ such that
$$ C^{N_\epsilon} \subseteq \bigcup_{k = 1}^{k_\epsilon} B_{\epsilon/2}  \left(\widehat{x^{(k)}}\right), \qquad B_{\epsilon/2}  \left(\widehat{x^{(k)}}\right) = \left\{ y \in X^{N_\epsilon} : |y - \widehat{x^{(k)}}| \leq \epsilon/2\right\} $$
Use the embedding of finite dimension into $\ell^1 (X)$ to conclude:
We extend each $\widehat{x^{(k)}}$ to an element $x^{(k)} = \{x_n^{(k)}\}_{n \geq 1} \in C$ be setting $x_n^{(k)} = \widehat{x_n^{(k)}}$ for $n \leq N_\epsilon$ and $x_n^{(k)} = 0$ for $n > N_\epsilon$. Let $x \in C$, then by total boundedness in $X^{N_\epsilon}$ we can find $1 \leq k \leq k_\epsilon$ such that
$$||x - x^{(k)}||_1 = \sum_{n = 1}^{N_\epsilon} |x_n - x_n^{(k)}| + \sum_{n > N_\epsilon} |x_n| < \epsilon.$$
We can conclude that $C$ is totally bounded. With appropriate modifications, you can extend this proof to $C$ defined similarly for $\ell^p (X)$ for any $1 \leq p < \infty$.
On the other hand, the closed ball $\{ x \in \ell^1(X) : ||x||_1 \leq 1 \}$ is not compact because we don't have a quantitative control over the tail, so you can always "move" mass to infinity, which is why total boundedness fails for balls in infinite dimensional Banach spaces. No matter how many balls you cover it with, these finitely many balls can't "capture" the mass at infinity.
A: Another different proof may be to use this nice and not so well-known criterion, and apply for $p=1$:
\begin{aligned}
&\text { Theorem. For any } p, 1 \leq p<\infty, \text { a bounded subset } K \text { of } \ell^{p} \text { is relatively compact if and }\\
&\text { only if } \lim _{n_{\epsilon}} \sum_{i=n_{\epsilon}}^{\infty}\left||k_{i}|\right|^{p}=0, \text { uniformly for } k \in K
\end{aligned}
If you need a proof of the theorem, you can click here.
